Answer:
given,
mass of the block = m₁
mass of the another block = 3 m₁
initial Amplitude, A = 6 cm
final amplitude = 6 cm
total mechanical energy = 12 J
total energy of the block spring
[tex]E = \dfrac{1}{2}kA^2[/tex]
A is the amplitude and k is spring constant
initial energy is equal to 12 J
from the above expression we can say that
Energy of the given system depends up on the magnitude of spring constant and the amplitude.
so, energy of both the system will be same.
we know,
[tex]E = \dfrac{1}{2}mv^2[/tex]
[tex]12= \dfrac{1}{2}\times 3 m_1 v^2[/tex]
[tex]v^2 = \dfrac{8}{m_1}[/tex]
[tex]v = \sqrt{ \dfrac{8}{m_1}}[/tex]
The concept of critical density is just like the idea of escape velocity for a projectile launched from the earth. An object launched with a velocity less than the escape velocity will fall back to the earth, and an object launched with one greater than the escape velocity can escape the earth's gravitational field. Find an expression for vesc, the escape velocity of a projectile launched from the earth's surface. Express your answer in terms of the universal gravitational constant G, the mass of the earth Me, and the radius of the earth Re.
Answer:
v = √ 2 G M/ [tex]R_{e}[/tex]
Explanation:
To find the escape velocity we can use the concept of mechanical energy, where the initial point is the surface of the earth and the end point is at the maximum distance from the projectile to the Earth.
Initial
Em₀ = K + U₀
Final
[tex]Em_{f}[/tex] = [tex]U_{f}[/tex]
The kinetic energy is k = ½ m v²
The gravitational potential energy is U = - G m M / r
r is the distance measured from the center of the Earth
How energy is conserved
Em₀ = [tex]Em_{f}[/tex]
½ mv² - GmM / [tex]R_{e}[/tex] = -GmM / r
v² = 2 G M (1 / [tex]R_{e}[/tex] – 1 / r)
v = √ 2GM (1 / [tex]R_{e}[/tex] – 1 / r)
The escape velocity is that necessary to take the rocket to an infinite distance (r = ∞), whereby 1 /∞ = 0
v = √ 2GM / [tex]R_{e}[/tex]
The escape velocity of an object from Earth is calculated by equating the object's kinetic energy to its gravitational potential energy, which yields the formula v_esc = sqrt(2GM_e/R_e), where G is the universal gravitational constant, M_e is Earth's mass, and R_e is Earth's radius.
The escape velocity (vesc) is the speed at which an object must travel to break free from the gravitational pull of a celestial body, in this case, Earth. To find the escape velocity, we use the principle that the kinetic energy at launch should be equal to the gravitational potential energy. The formula for escape velocity is derived using the equation KE = GPE, where KE stands for kinetic energy and GPE for gravitational potential energy. The kinetic energy of the object is given by (1/2)mvesc2 and the gravitational potential energy by -G(Mem)/Re, where m is the mass of the object, G is the universal gravitational constant, Me is the mass of the earth, and Re is the radius of the earth. Setting these two energies equal to each other and solving for vesc while cancelling out the mass of the object (since it appears on both sides of the equation), we obtain vesc = sqrt(2GMe/Re).
In Millikan’s oil-drop experiment, one looks at a small oil drop held motionless between two plates. Take the voltage between the plates to be 2210 V, and the plate separation to be 1.87 cm. The oil drop (of density 0.816 g/cm^3) has a diameter of 3.3x 10^−6 m . Calculate the charge in terms of the number of elementary charges (1.6 x 10^-19).
Answer:
q = 1,297 10⁻¹⁹ C , n=1
Explanation:
For this problem we will use Newton's second law in the case of equilibrium
[tex]F_{e}[/tex] + B - W = 0
Where [tex]F_{e}[/tex] is the electrical force up, B the thrust and W the weight of the drop.
Let's look for weight and thrust
oil
ρ = m / V
m = ρ V
Air
B = [tex]\rho _{air}[/tex] g V
Electric force
[tex]F_{e}[/tex] = qE
E = V / d
[tex]F_{e}[/tex] = q V/d
Let's replace
q V / d + [tex]\rho _{air}[/tex] g V - ρ V g = 0
qV / d = (4/3 π r³) g (ρ –[tex]\rho _{air}[/tex])
q = 4/3 π r³ (ρ –[tex]\rho _{air}[/tex]) d / V
Reduce to SI units
d = 1.87 cm (1m / 100cm) = 1.87 10⁻² m
ρ= 0.816 r / cm3 (1kg / 1000g) (102cm / 1m)³ = 816 kg / m³
[tex]\rho_{air}[/tex] = 1.28 kg / m³
Let's calculate the charge
r = d / 2 = 3.3 10⁻⁶ m
r = 1.65 10⁻⁶ m
q = 4/3 π (1.65 10⁻⁶)³ (816 - 1.28) 0.0187 / 2210
q = 12.9717 10⁻²⁰ C
q = 1,297 10⁻¹⁹ C
If we assume that the load is
q = n e
In this case n = 1
Problem 4: A mass m = 1.5 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring constant k = 95 N/m and negligible mass. The mass undergoes simple harmonic motion when placed in vertical motion, with its position given as a function of time by y(t) = A cos(ωt – φ), with the positive y-axis pointing upward. At time t = 0 the mass is observed to be at a distance d = 0.35 m below its equilibrium height with an upward speed of v0 = 6 m/s.
Answer:
The angular frequency of spring-mass system is 7.95 rad/s.
The phase angle is 65.11 degree.
Explanation:
Given data:
The mass is, [tex]m=1.5 \;\rm kg[/tex].
The value of spring constant is, [tex]k=95 \;\rm N/m[/tex].
Distance covered at t=0 is, [tex]y (t=0) = 0.35 \;\rm m[/tex]
Upward speed is, [tex]v_{0}=6 \;\rm m/s[/tex].
The position of mass is,
[tex]y(t)=Acos(\omega t-\phi)[/tex]
Here, A is amplitude of vibration, [tex]\omega[/tex] is the angular frequency, t is time and [tex]\phi[/tex] is the phase angle.
(a)
The expression for the angular frequency of spring-mass system is:
[tex]\omega =\sqrt\frac{k}{m}[/tex]
Substitute the values as,
[tex]\omega =\sqrt\dfrac{95}{1.5}\\[/tex]
[tex]\omega = 7.95 \;\rm rad/s[/tex]
Thus, the value of angular frequency is 7.95 rad/s.
(b)
The position of mass is, [tex]y(t)=Acos(\omega t-\phi)[/tex].
Differentiating the above equation to obtain the speed as,
[tex]y(t)=Acos(\omega t-\phi)\\y'(t)= v_{0}=-A\omega sin(-\phi)[/tex]
At time t=0,
[tex]y(t=0)=A cos(\omega (0)-\phi)\\y(t=0)=A cos(-\phi)\\45= A cos(-\phi) .....................................................(1)[/tex]
Speed of system is,
[tex]6=-A\omega sin(-\phi) ............................................(2)[/tex]
Taking ratio of equation (2) and (1) as,
[tex]\dfrac{-A \omega sin(- \phi)}{Acos(\phi)} = \dfrac{6}{0.35} \\\\7.95 \times tan(\phi) = \dfrac{6}{0.35} \\\phi = tan^{-1}(\dfrac{6}{0.35 \times 7.95}) \\\phi =65.11 \;\rm degree[/tex]
Thus, the phase angle is 65.11 degree.
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The question involves solving for properties of simple harmonic motion, such as force exerted by a spring, equilibrium position, oscillation amplitude, and maximum velocity for a mass-spring system. These are common problems in high school physics involving mechanics and wave motion.
Explanation:The student's question pertains to simple harmonic motion (SHM) exhibited by a mass attached to a vertical spring. Given the mass m, the spring constant k, the displacement d below the equilibrium position, and the upward speed v0, we are tasked with discussing characteristics such as force exerted by the spring, the new equilibrium position, the amplitude of oscillations, and the maximum velocity of the mass.
Step-By-Step Explanation:To find the force exerted by the spring at the lowest point, apply Hooke's law F = -kx, where x is the displacement from the equilibrium position.To determine the new equilibrium position of the spring when a mass is attached, use the force balance where the spring force equals the gravitational force mg.Calculate the amplitude of oscillations by considering the total energy in the system or the maximum displacement of the mass from its equilibrium position.The maximum velocity can be found by using conservation of energy principles or by referring to the maximum speed in SHM as v_max = Aω, where ω is the angular frequency.The solutions to these problems involve a blend of mechanics and oscillatory motion principles and are emblematic of problems found in high school or introductory college-level physics classes.
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a)A certain medical machine emits x-rays with a minimum wavelength of 0.024 nm. One day, the machine has an electrical problem and the voltage applied to the x-ray tube decreases to 76% of its normal value. Now what is the minimum x-ray wavelength produced by the machine?
lmin = ___________nm
(b) What is the maximum x-ray energy this machine (with electrical problems) can produce?
Emax =___________ eV
(c) The atomic number of an element is 82. According to the Bohr model, what is the energy of a Ka x-ray photon?
E =_______________ eV
Answer:
(a) 0.032 nm
(b) 39,235 eV
(c) 70,267.8 eV
Explanation:
(a) The energy of a photon can be calculated using:
E = hc/λ equation (1)
where:
h = 4.13*10^-15 eV.s
c = 3*10^8 m/s
λ = 0.024*10^-9 m
Thus:
E = (4.13*10^-15)*(3*10^8)/0.024*10^-9 = 51,625 eV
Then we calculate 76% of this estimated energy and determine the new wavelength:
[tex]E_{new} = 0.76*51625 = 39,235 eV[/tex]
Using equation (1) to determine the new wavelength:
λ[tex]_{new} = \frac{h*c}{E_{new} }[/tex]
λ[tex]_{new}[/tex] = (4.13*10^-15)*(3*10^8)/39235 = 3.15*10^-11 m = 0.032 nm
(b) As calculated in part (a), the maximum x-ray energy this machine can produce is [tex]E_{new} = 0.76*51625 = 39,235 eV[/tex]
(c) The energy of a Ka x-ray photon can be estimated using:
[tex]E_{ka} = (10.2 eV)*(Z-1)^{2}[/tex]
where Z is the atomic number = 84.
[tex]E_{ka} = (10.2 eV)*(84-1)^{2}[/tex] = 70,267.8 eV
A single-turn circular loop of wire of radius 50 mm lies in a plane perpendicular to a spatially uniform magnetic field. During a 0.10-s time interval, the magnitude of the field increases uniformly from 200 to 300 mT. (a) Determine the emf induced in the loop. (b) If the magnetic field is directed out of the page, what is the direction of the current induced in the loop
The magnitude of the induced emf in the loop can be calculated using Faraday's law of electromagnetic induction. The direction of the induced current that would oppose the change in flux can be determined using Lenz's law and the right-hand rule, which in this case would be clockwise.
Explanation:To answer your questions: (a) emf induced in the loop and (b) the direction of the current induced in the loop, we must understand Faraday's law of electromagnetic induction and Lenz's law.
Using Faraday's law, the magnitude of the induced emf is given as ε = |døm/dt|, where øm is the magnetic flux.
The magnetic flux through a loop is given by the product of the magnetic field B, the area A (Which can be calculated using the formula for the area of a circle, A = πr², where r is the radius of the loop), and the cosine of the angle Ө between the magnetic field lines and the perpendicular to the plane of the loop.
However, since the loop is perpendicular to the magnetic field, cosӨ = 1. Also, because the magnetic field B is spatially uniform and changes uniformly with time, we can find the average magnetic field over the 0.1 s interval as B = (200mT + 300mT) / 2 = 250mT = 0.25 T. Therefore, the change in magnetic flux over the time interval is Δøm = BA = (0.25 T)(π(0.05 m)²). Substituting these into Faraday's law gives us the induced emf.
According to Lenz's law, the direction of the induced emf (and hence the current) is such that it produces a magnetic field that opposes the change in the magnetic flux. Since the original field is directed out of the page and is increasing, the induced current must flow in a way that it produces a field into the page. Hence, according to the right-hand rule, the current would flow clockwise when viewed from above.
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By applying Faraday's Law of electromagnetic induction, the induced electromotive force (emf) in the loop is calculated to be approximately 7.85 mV. Following Lenz's Law, the induced current flows in a clockwise direction to oppose the increase in the magnetic field.
To address the student's question, we can use Faraday's Law of electromagnetic induction which states that the induced electromotive force (emf) in any closed circuit is equal to the negative of the time rate of change of the magnetic flux through the circuit.
(a) To find the induced emf, use the formula:
E = -dΦ/dt
where Φ is the magnetic flux given by:
Φ = BA cos(θ)
B is the magnetic field's magnitude, A is the area of the loop, and θ is the angle between the field and normal to the loop. In this case, θ is 0 degrees because the field is perpendicular to the plane of the loop.
Φ = B(πr²)
Given that r = 0.05 m, Binitial = 200 mT, Bfinal = 300 mT, and Δt = 0.10 s, we have:
ΔΦ = π(0.05)²(300 mT - 200 mT)
ΔΦ = π(0.05)²(100 mT) = π(0.0025 m²)(100 x 10⁻³ T) = π(0.00025 T·m²)
E = -ΔΦ/Δt = -π(0.00025)/0.10 = -0.00785 V
The magnitude of the induced emf is approximately |0.00785| V or 7.85 mV.
(b) According to Lenz's Law, the direction of the induced current will be such that it opposes the change in flux. Since the magnetic field is increasing and directed out of the page, the induced current will create a magnetic field into the page to oppose the increase. Using the right-hand rule, we can determine that the current flows in a clockwise direction when viewed from above.
2. At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (nautical miles per hour; a nautical mile is 2000 yards) and continued to do so all day. Ship B was sailing east at 9 knots and continued to do so all day. The visibility was 5 nautical miles. Did the ships ever sight each other?
Answer:
No
Explanation:
Let the reference origin be location of ship B in the beginning. We can then create the equation of motion for ship A and ship B in term of time t (hour):
A = 12 - 12t
B = 9t
Since the 2 ship motions are perpendicular with each other, we can calculate the distance between 2 ships in term of t
[tex]d = \sqrt{A^2 + B^2} = \sqrt{(12 - 12t)^2 + (9t)^2}[/tex]
For the ships to sight each other, distance must be 5 or smaller
[tex] d \leq 5[/tex]
[tex]\sqrt{(12 - 12t)^2 + (9t)^2} \leq 5[/tex]
[tex](12 - 12t)^2 + (9t)^2 \leq 25[/tex]
[tex]144t^2 - 288t + 144 + 81t^2 - 25 \leq 0[/tex]
[tex]225t^2 - 288t + 119 \leq 0[/tex]
[tex](15t)^2 - (2*15*9.6)t + 9.6^2 + 26.84 \leq 0[/tex]
[tex](15t^2 - 9.6)^2 + 26.84 \leq 0[/tex]
Since [tex](15t^2 - 9.6)^2 \geq 0[/tex] then
[tex](15t^2 - 9.6)^2 + 26.84 > 0[/tex]
So our equation has no solution, the answer is no, the 2 ships never sight each other.
Suppose you have two meter sticks, one made of steel and one made of invar (an alloy of iron and nickel), which are the same length (1.00 m) at 0°C. The coefficients of volume expansion for steel and invar are 3.6 × 10-5 /°C and 2.7 × 10-6 /°C respectively.What is their difference in length, in meters, at 20.5°C ?Repeat the calculation for two 30.0-m-long surveyor’s tapes
Answer:
Difference in Length of steel is 0.000246m
Difference in Length of invar is 0.00001845m
Difference in Length of steel surveyor's tape is 0.00738m
Difference in Length of invar surveyor's tape is 0.0005535m
Explanation:
Linear expansivity of steel is volume expansivity ÷ 3
Linear expansivity of steel = 0.000036/°C ÷ 3 = 0.000012/°C
Difference in Length of steel= Linear expansivity × initial length × temperature change
= 0.000012 × 1 × (20.5-0)
= 0.000012×1×20.5 = 0.000246m
Linear expansivity of invar = volume expansivity of invar ÷ 3
Linear expansivity of invar= 0.0000027/°C ÷ 3= 0.0000009/°C
Difference in Length of invar = 0.0000009×1×(20.5-0) = 0.00001845m
Difference in Length of steel surveyor's tape= 0.000012×30×(20.5-0) = 0.00738m
Difference in Length of invar surveyor's tape = 0.0000009×30×(20.5-0) = 0.0005535m
The difference in length between a steel and invar meter stick at 20.5°C is 6.83 mm, and for two 30-m-long surveyor's tapes, the difference at the same temperature is 20.5 cm.
Explanation:When a steel meter stick and an invar meter stick both expand due to an increase in temperature, we can determine their difference in length using the coefficients of linear expansion for each material.
The difference in length at 20.5°C can be determined using the formula:
ΔL = αLΔT
For steel, the coefficient of linear expansion (α) is 3.6 × 10^-5 /°C, and for invar it is 2.7 × 10^-6 /°C. Plugging in the values:
ΔL_steel = 3.6 × 10^-5 /°C × 1.00 m × 20.5 °C = 7.38 × 10^-3 mΔL_invar = 2.7 × 10^-6 /°C × 1.00 m × 20.5 °C = 5.54 × 10^-4 mThe difference in length is therefore 7.38 × 10^-3 m - 5.54 × 10^-4 m = 6.83 × 10^-3 m or 6.83 mm.
For two 30.00-m-long surveyor’s tapes:
ΔL_steel = 3.6 × 10^-5 /°C × 30.00 m × 20.5 °C = 2.21 × 10^-1 mΔL_invar = 2.7 × 10^-6 /°C × 30.00 m × 20.5 °C = 1.66 × 10^-2 mThe difference in length for the surveyor's tapes is 2.21 × 10^-1 m - 1.66 × 10^-2 m = 2.05 × 10^-1 m or 20.5 cm.
In an RL series circuit, an inductor of 3.54 H and a resistor of 7.76 Ω are connected to a 26.6 V battery. The switch of the circuit is initially open. Next close the switch and wait for a long time. Eventually the current reaches its equilibrium value. At this time, what is the corresponding energy stored in the inductor?
Answer:
Energy stored in inductor will be 20.797 J
Explanation:
We have given inductance L = 3.54 H
And resistance R = 7.76 ohm
Battery voltage V = 26.6 VOLT
After very long time means at steady state inductor behaves as short circuit
So current [tex]i=\frac{V}{R}=\frac{26.6}{7.76}=3.427Amp[/tex]
Now energy stored in inductor [tex]E=\frac{1}{2}Li^2=\frac{1}{2}\times 3.54\times 3.427^2=20.797J[/tex]
So energy stored in inductor will be 20.797 J
To find the energy stored in an inductor in an RL circuit at equilibrium, use Ohm's law to calculate the steady current, and then apply the energy formula E = (1/2)LI^2.
Explanation:When the switch in an RL series circuit is closed for a long time, the current in the circuit reaches an equilibrium value due to the voltage provided by the battery. The current, I, can be calculated using Ohm's law, I = V/R, where V is the voltage of the battery and R is the resistance in the circuit. In this case, with a 26.6 V battery and a 7.76 Ω resistor, the current would be I = 26.6 V / 7.76 Ω. After calculating the current, the energy stored in the inductor at equilibrium can be found using the formula for energy stored in an inductor, which is E = (1/2)LI^2, where L is the inductance and I is the current.
The corresponding energy stored in the inductor after a long time can be determined using these calculations.
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A stone is thrown upward from the top of a building at an angle of 30° to the horizontal and with an initial speed of 20 m/s. The point of release is 45 m above the ground. How long does it take for the stone to hit the ground? What is the stone's speed?
The time taken for the stone to hit the ground can be calculated using vertical motion equations, and the stone's speed at impact can be determined using the vertical velocity equation.
Time taken for the stone to hit the ground can be calculated using the vertical motion equation: t = sqrt(2h/g). Substituting the values, t = √(2×45/9.81) ≈ 3.0 seconds.
Stone's speed at impact can be calculated using the vertical velocity equation: vf = u + gt, where vf is the final velocity (0 m/s at impact), u is the initial vertical speed, and g is the acceleration due to gravity.
With the given information, the stone's speed at impact is approximately 29.4 m/s.
A firm wants to determine the amount of frictional torque in their current line of grindstones, so they can redesign them to be more energy efficient. To do this, they ask you to test the best-selling model, which is basically a diskshaped grindstone of mass 1.5 kg and radius 7.40 cm which operates at 730 rev/min. When the power is shut off, you time the grindstone and find it takes 38.8 s for it to stop rotating.
(a) What is the angular acceleration of the grindstone in rad/s^2? (Assume constant angular acceleration.)
(b) What is the frictional torque exerted on the grindstone in N·m?
Answer:
a) -1.97 rad/sec² b) -8.09*10⁻³ N.m
Explanation:
a) Assuming a constant angular acceleration, we can apply the definition of angular acceleration, as follows:
γ = (ωf -ω₀) / t
We know that the final state of the grindstone is at rest, so ωf =0
In order to be consistent in terms of units, we can convert ω₀ from rev/min to rad/sec:
ω₀ = 730 rev/min* (1 min/60 sec)* (2*π rad / 1 rev) = 73/3*π
⇒ γ = (0-73/3*π) / 38.8 sec = - 1.97 rad/sec²
b) In order to get the value of the frictional torque exerted on the grindstone, that caused it to stop, we can apply the rotational equivalent of the Newton's 2nd law, as follows:
τ = I * γ (1)
As the grindstone can be approximated by a solid disk, the rotational inertia I can be expressed as follows:
I = m*r² / 2, where m=1.5 kg and r = 0.074 m.
Replacing in (1) , m. r and γ (the one we calculated in a)), we get:
τ = (1.5 kg* (0.074)² m² / 2) * -1.97 rad/sec = -8.09*10⁻³ N.m
(The negative sign implies that the frictional torque opposes to the rotation of the grindstone).
Air resistance is a nonconservative force. It always opposes the motion of an object. An airplane flies from New York to Atlanta and then returns to its point of departure. The net work done by air resistance during this round trip ___________.a. is negative for slower speeds and positive for higher speedsb. is positivec. is negatived. is zeroe. is negative for higher speeds and positive for slower speeds
Answer:
c. is negative
Explanation:
When the magnitude of force is multiplied with the force vector's projection along the direction of the vector of displacement which is negative as it is a resistive force we get the work done.
As the wind is acting in opposite direction of the force which is being applied by the plane the work done will be negative. Also, the net work will be the sum of many smaller negative quantities.
Hence, the answer here is negative.
Seat belts and air bags save lives by reducing the forces exerted on the driver and passengers in an automobile collision. Cars are designed with a "crumple zone" in the front of the car. In the event of an impact, the passenger compartment decelerates with the car. By contrast, an unrestrained occupant keeps moving forward with no loss of speed (Newton's first law!) until hitting the dashboard or windshield, as we saw in Figure 4.2. These are unyielding surfaces, and the unfortunate occupant then decelerates over a distance of only about 5 mm.
a. A 60 kg person is in a head-on collision. The car's speed at impact is 15 m/s.
Estimate the net force on the person if he or she is wearing a seat belt and if the air
bag deploys.
b. Estimate the net force that ultimately stops the person if he or she is not restrained
by a seat belt or air bag.
c. How does these two forces compare to the person's weight
Answer:
a) [tex]F_{belt}[/tex] = 1.35 10⁴ N , b) F = 1.35 10⁶ N, c) F / [tex]F_{belt}[/tex] = 100
Explanation:
a) Let's start by calculating the aceleration it takes for the person with a seat belt and air bag to stop, the approximate distance between the floor and the steering wheel is about 50 cm, let's use kinematics to calculate the acceleration
v² = v₀² - 2 a x
0 = v₀² - 2 a x
a = v₀² / 2 x
a = 15² / (2 0.50)
a = 225 m / s²
We calculate the force with Newton's second law
F = m a
F = 60 225
[tex]F_{belt}[/tex] = 1.35 10⁴ N
b) we perform the same calculation for a person without a belt
a = v₀² / 2x
a = 15² / (2 0.005)
a = 22500 m / s²
F = m a
F = 60 22500
F = 1.35 10⁶ N
c) let's calculate the relationship between these two forces
F / [tex]F_{belt}[/tex] = 1.35 10⁶ / 1.35 10⁴
F / [tex]F_{belt}[/tex] = 10²
Two long, parallel wires are attracted to each other by a force per unit length of 350 µN/m. One wire carries a current of 22.5 A to the right and is located along the line y = 0.420 m. The second wire lies along the x axis. Determine the value of y for the line in the plane of the two wires along which the total magnetic field is zero.
Answer
given,
force per unit length = 350 µN/m
current, I = 22.5 A
y = y = 0.420 m
[tex]\dfrac{F}{L}= \dfrac{KI_1I_2}{d}[/tex]
[tex]I_2 = \dfrac{F}{L}\dfrac{d}{KI_1}[/tex]
[tex]I_2 = 350\times 10^{-6}\times \dfrac{0.42}{2 \times 10^{-7}\times 22.5}[/tex]
I₂ = 32.67 A
distance where the magnetic field is zero
[tex]\dfrac{4\pi \times 10^{-7}\times 32.67}{2\pi y_1}=\dfrac{4\pi \times 10^{-7}\times 22.5}{2\pi (0.42-y_1)}[/tex]
[tex]y_1 = 0.248\ m[/tex]
there the distance at which the magnetic field is zero in the two wire is at 0.248 m.
A neutral copper block is polarized as shown in the figure below, due to an electric field made by external charges (notshown). Which arrow (a–j) in the figure below best indicates the direction of the net electric field at location B, which isinside the copper block?
Answer:
As point B is located inside the copper block so net electric field at point B is j.
Explanation:
Consider the figure attached below. The net electric field at location B,that is inside the copper block is zero because when a conductor is charged or placed in an electric field of external charges, net charge lies on the surface of conductor and there is no electric field inside the conductor. As point B is located inside the copper block so net electric field at point B is zero as well direction of net electric field at point B is zero.
Two coils, held in fixed positions, have a mutual inductance of M = 0.0034 H. The current in the first coil is I(t) = I0sin(ωt), where I0 = 5.4 A, ω = 143 rad/s.
M = 0.0034 H I0 = 5.4 A ω = 143 rad/s
(a) Express the magnitude of the induced emf in the second coil, ε2, in terms of M and I.
(b) Express the magnitude of ε2 in terms of M, Io, and ω.
(c) Express the maximum value of |ε2|, εmax, in terms of M, I0, and ω.
(d) Calculate the numerical value of εmax in V.
The magnitude of the induced emf in the second coil can be expressed using the mutual inductance, maximum current in the first coil, and angular frequency.
Explanation:(a) The magnitude of the induced emf in the second coil, ε2, can be expressed as ε2 = M * (dI2/dt), where M is the mutual inductance and dI2/dt is the rate of change of current in the second coil.
(b) The magnitude of ε2 can also be expressed as ε2 = M * I0 * ω * cos(ωt), where I0 is the maximum current in the first coil and ω is the angular frequency.
(c) The maximum value of |ε2|, εmax, can be calculated by taking the maximum value of the function ε2 = M * I0 * ω * cos(ωt) over one period.
(d) To calculate the numerical value of εmax, substitute the values of M, I0, and ω into the equation ε2 = M * I0 * ω * cos(ωt) and evaluate it at the maximum value of cos(ωt), which is 1.
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Some curious students hold a rolling race by rolling four items down a steep hill. The four items are a solid homogeneous sphere, a thin spherical shell, a solid homogeneous cylinder and a hoop with all its mass concentrated on the hoop's perimeter. All of the objects have the same mass and start from rest. Assume that the objects roll without slipping and that air resistance and rolling resistance are negligible. For each statement below, select True or False.
a) Upon reaching the bottom of the hill, the hoop will have a larger rotational kinetic energy than any of the other objects will when they reach the bottom of the hill.
A: True B: False
b) The hoop reaches the bottom of the hill before the homogeneous cylinder.
A: True B: False
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
An electric eel (Electrophorus electricus) can produce a shock of up to 600 V and a current of 1 A for a duration of 2 ms, which is used for hunting and self-defense. To perform this feat, approximately 80% of its body is filled with organs made up by electrocytes. These electrocytes act as self-charging capacitors and are lined up so that a current of ions can easily flow through them.
a) How much charge flows through the electrocytes in that amount of time?
b) If each electrocyte can maintain a potential of 100 mV, how many electrocytes must be in series to produce the maximum shock?
c) How much energy is released when the electric eel delivers a shock?
d) With the given information, estimate the equivalent capacitance of all the electrocyte cells in the electric eel.
Answer:
[tex]2\times 10^{-3}\ C[/tex]
6000
1.2 J
[tex]3.33\times 10^{-6}\ F[/tex]
Explanation:
I = Current = 1 A
t = Time = 2 ms
n = Number of electrocyte
V = Voltage = 100 mV
Charge is given by
[tex]Q=It\\\Rightarrow Q=1\times 2\times 10^{-3}\\\Rightarrow Q=2\times 10^{-3}\ C[/tex]
The charge flowing through the electrocytes in that amount of time is [tex]2\times 10^{-3}\ C[/tex]
The maximum potential is given by
[tex]V_m=nV\\\Rightarrow n=\dfrac{V_m}{V}\\\Rightarrow n=\dfrac{600}{100\times 10^{-3}}\\\Rightarrow n=6000[/tex]
The number of electrolytes is 6000
Energy is given by
[tex]E=Pt\\\Rightarrow E=V_mIt\\\Rightarrow E=600\times 1\times 2\times 10^{-3}\\\Rightarrow E=1.2\ J[/tex]
The energy released when the electric eel delivers a shock is 1.2 J
Equivalent capacitance is given by
[tex]C_e=\dfrac{Q}{V_m}\\\Rightarrow C_e=\dfrac{2\times 10^{-3}}{600}\\\Rightarrow C_e=3.33\times 10^{-6}\ F[/tex]
The equivalent capacitance of all the electrocyte cells in the electric eel is [tex]3.33\times 10^{-6}\ F[/tex]
a) The charge that flows through the electrocytes is [tex]\( 2 \text{ mC} \).[/tex]
b) 6000 electrolytes must be in series to produce the maximum shock. c) the energy released when the electric eel delivers a shock is [tex]\( 0.6 \text{ J} \).[/tex]
d) the equivalent capacitance of all the electrocyte cells in the electric eel is approximately [tex]\( 3.33 \mu\text{F} \)[/tex]
To answer the questions about the electric eel's shock, we'll use the provided information and relevant physics formulas.
Part (a): We know:
- Current [tex]\( I = 1 \text{ A} \)[/tex]
- Time [tex]\( t = 2 \text{ ms} = 2 \times 10^{-3} \text{ s} \)[/tex]
The charge Q that flows can be found using the relation:
[tex]\[ Q = I \times t \][/tex]
Plugging in the values:
[tex]\[ Q = 1 \text{ A} \times 2 \times 10^{-3} \text{ s} \][/tex]
[tex]\[ Q = 2 \times 10^{-3} \text{ C} \][/tex]
[tex]\[ Q = 2 \text{ mC} \][/tex]
Part (b): We know:
- Maximum potential [tex]\( V = 600 \text{ V} \)[/tex]
- Potential of each electrocyte [tex]\( V_{\text{single}} = 100 \text{ mV} = 0.1 \text{ V} \)[/tex]
The number of electrocytes n in series required to produce 600 V can be found by:
[tex]\[ n = \frac{V}{V_{\text{single}}} \][/tex]
Plugging in the values:
[tex]\[ n = \frac{600 \text{ V}}{0.1 \text{ V}} \][/tex]
[tex]\[ n = 6000 \][/tex]
Part (c): We know:
- Voltage [tex]\( V = 600 \text{ V} \)[/tex]
- Charge [tex]\( Q = 2 \text{ mC} = 2 \times 10^{-3} \text{ C} \)[/tex]
The energy E released can be found using the relation:
[tex]\[ E = \frac{1}{2} Q V \][/tex]
Plugging in the values:
[tex]\[ E = \frac{1}{2} \times 2 \times 10^{-3} \text{ C} \times 600 \text{ V} \][/tex]
[tex]\[ E = \frac{1}{2} \times 1.2 \text{ J} \][/tex]
[tex]\[ E = 0.6 \text{ J} \][/tex]
Part (d): Given:
- Total voltage [tex]\( V = 600 \text{ V} \)[/tex]
- Charge [tex]\( Q = 2 \times 10^{-3} \text{ C} \)[/tex]
The capacitance C can be found using the relation:
[tex]\[ C = \frac{Q}{V} \][/tex]
Plugging in the values:
[tex]\[ C = \frac{2 \times 10^{-3} \text{ C}}{600 \text{ V}} \][/tex]
[tex]\[ C = \frac{2 \times 10^{-3}}{600} \text{ F} \][/tex]
[tex]\[ C = \frac{1}{300} \times 10^{-3} \text{ F} \][/tex]
[tex]\[ C \approx 3.33 \times 10^{-6} \text{ F} \][/tex]
[tex]\[ C = 3.33 \mu\text{F} \][/tex]
A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A constant tangential force of 210 N applied to its edge causes the wheel to have an angular acceleration of 0.932 rad/s2. What is the moment of inertia of the wheel? (Pick the answer closest to the true value.)A. 27.3 kg m2B. 42.4 kg m2C. 54.9 kg m2D. 74.4 kg m2E. 98.5 kg m2
Answer:
Moment of inertia will be [tex]I=74.356kgm^2[/tex]
So option (d) will be the correct answer
Explanation:
We have given radius of solid cylinder r = 0.330 m
Constant tangential force F = 210 N
Angular acceleration [tex]\alpha =0.932rad/sec^2[/tex]
We know that torque [tex]\tau =Fr=210\times 0.330=69.3Nm[/tex]
We also know that torque is given by [tex]\tau =I\alpha[/tex]
So [tex]69.3=I\times 0.932[/tex]
[tex]I=74.356kgm^2[/tex]
So option (d) will be the correct answer
Angular and Linear Quantities: A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 rad/s2. The child is standing 4.65 m from the center of the merry-go-round. What is the magnitude of the linear acceleration of the child? Angular and Linear Quantities: A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 rad/s2. The child is standing 4.65 m from the center of the merry-go-round. What is the magnitude of the linear acceleration of the child?
a. 3.46 m/s2
b.4.10 m/s2
c. 8.05 m/s2
d. 7.27 m/s2
e. 2.58 m/s2
To solve this problem we will use the kinematic equations of angular motion in relation to those of linear / tangential motion.
We will proceed to find the centripetal acceleration (From the ratio of the radius and angular velocity to the linear velocity) and the tangential acceleration to finally find the total acceleration of the body.
Our data is given as:
[tex]\omega = 1.25 rad/s \rightarrow[/tex] The angular speed
[tex]\alpha = 0.745 rad/s2 \rightarrow[/tex] The angular acceleration
[tex]r = 4.65 m \rightarrow[/tex] The distance
The relation between the linear velocity and angular velocity is
[tex]v = r\omega[/tex]
Where,
r = Radius
[tex]\omega =[/tex] Angular velocity
At the same time we have that the centripetal acceleration is
[tex]a_c = \frac{v^2}{r}[/tex]
[tex]a_c = \frac{(r\omega)^2}{r}[/tex]
[tex]a_c = \frac{r^2\omega^2}{r}[/tex]
[tex]a_c = r \omega^2[/tex]
[tex]a_c = (4.65 )(1.25 rad/s)^2[/tex]
[tex]a_c = 7.265625 m/s^2[/tex]
Now the tangential acceleration is given as,
[tex]a_t = \alpha r[/tex]
Here,
[tex]\alpha =[/tex] Angular acceleration
r = Radius
[tex]\alpha = (0.745)(4.65)[/tex]
[tex]\alpha = 3.46425 m/s^2[/tex]
Finally using the properties of the vectors, we will have that the resulting component of the acceleration would be
[tex]|a| = \sqrt{a_c^2+a_t^2}[/tex]
[tex]|a| = \sqrt{(7.265625)^2+(3.46425)^2}[/tex]
[tex]|a| = 8.049 m/s^2 \approx 8.05 m/s2[/tex]
Therefore the correct answer is C.
A large fake cookie sliding on a horizontal surface is attached to one end of a horizontal spring with spring constant k = 375 N/m; the other end of the spring is fixed in place. The cookie has a kinetic energy of 19.0 J as it passes through the position where the spring is unstretched. As the cookie slides, a frictional force of magnitude 10.0 N acts on it. (a) How far will the cookie slide from the position where the spring is unstretched before coming momentarily to rest? m (b) What will be the kinetic energy of the cookie as it slides back through the position where the spring is unstretched?
Answer:
The conservation of energy should be used to answer this question.
a)
At the position where the spring is unstretched, the elastic potential energy of the spring is zero.
[tex]K_1 + U_1 - W_f = K_2 +U_2\\K_1 - W_f = U_2[/tex]
since [tex]U_1[/tex] and [tex]K_2[/tex] is equal to zero.
[tex]W_f = F_fx\\\\U_2 = \frac{1}{2}kx^2\\\\19 - (10)x = \frac{1}{2}(375)x^2\\\\375x^2 + 20x - 38 = 0[/tex]
The roots of this quadratic equation can be solved by using discriminant.
[tex]\Delta = b^2 - 4ac\\x_{1,2} = \frac{-b \pm \sqrt{\Delta}}{2a}[/tex]
[tex]x_1 = -0.346\\x_2 = 0.292[/tex]
We should use the positive root, so
x = 0.292 m.
b)
We should use energy conservation between the point where the spring is momentarily at rest, and the point where the spring is unstretched.
[tex]K_2 + U_2 - W_f = K_3 + U_3\\U_2 - W_f = K_3[/tex]
since the kinetic energy at point 2 and the potential energy at point 3 is equal to zero.
[tex]\frac{1}{2}kx^2 - F_fx = K_3\\K_3 = 15.987 - 2.92 = 13.067 J[/tex]
Explanation:
In questions with springs, the important thing is to figure out the points where kinetic or potential energy terms would be zero. When the spring is unstretched, the elastic potential energy is zero. And when the spring is at rest, naturally the kinetic energy is equal to zero.
In part b) the cookie slides back to its original position, so the distance traveled, x, is equal to the distance in part a). The frictional force is constant in the system, so it is quite simple to solve part b) after solving part a).
Astronaut Rob leaves Earth in a spaceship at a speed of 0.960crelative to an observer onEarth. Rob's destination is a star system 14.4 light-years away (one light-year is the distance lighttravels in one year). Relative to a frame of reference that is fixed with respect to Earth, how longdoes it take Rob to complete the trip?
To solve this problem we will use the mathematical definition of the light years in metric terms, from there, through the kinematic equations of motion we will find the distance traveled as a function of the speed in proportion to the elapsed time. Therefore we have to
[tex]1Ly =9.4605284*10^{15}m \rightarrow 'Ly'[/tex]means Light Year
Then
[tex]14.4Ly = 1.36231609*10^{17} m[/tex]
If we have that
[tex]v= \frac{x}{t} \rightarrow t = \frac{x}{t}[/tex]
Where,
v = Velocity
x = Displacement
t = Time
We have that
[tex]t = \frac{1.36231609*10^{17}}{0.96c} \rightarrow c[/tex]= Speed of light
[tex]t = \frac{1.36231609*10^{17}}{0.96(3*10^8)}[/tex]
[tex]t= 454105363 s (\frac{1hour}{3600s})[/tex]
[tex]t= 126140 hours(\frac{1day}{24hours})[/tex]
[tex]t= 5255.85 days(\frac{1 year}{365days})[/tex]
[tex]t = 14.399 years[/tex]
Therefore will take 14.399 years
It takes Rob approximately 15.0 years to complete the trip relative to a frame of reference fixed with respect to Earth.
To solve this problem, we can use the concept of time dilation from the theory of special relativity. According to this concept, time measured in a frame of reference moving at a high velocity relative to a stationary observer (in this case, Earth) will be dilated, or ""slowed down,"" compared to the time measured by the stationary observer.
The time dilation formula is given by:
[tex]\[ t' = \frac{t}{\sqrt{1 - \frac{v^2}{c^2}}} \][/tex]
where:
- [tex]\( t' \)[/tex]is the time interval measured by the moving observer (Rob),
- [tex]\( t \)[/tex]is the time interval measured by the stationary observer (Earth),
-[tex]\( v \)[/tex] is the relative velocity between the two frames of reference, and
- [tex]\( c \)[/tex] is the speed of light in a vacuum.
Given that Rob's speed is [tex]\( 0.960c \)[/tex]and the distance to the star system is [tex]\( 14.4 \)[/tex] light-years, we can calculate the time it takes for Rob to complete the trip according to Earth's frame of reference.
First, we calculate the Lorentz factor [tex]\( \gamma \),[/tex] which is the factor by which time is dilated:
[tex]\[ \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \][/tex]
Substituting \( v = 0.960c \):
[tex]\[ \gamma = \frac{1}{\sqrt{1 - (0.960)^2}} \][/tex]
[tex]\[ \gamma = \frac{1}{\sqrt{1 - 0.9216}} \][/tex]
[tex]\[ \gamma = \frac{1}{\sqrt{0.0784}} \][/tex]
[tex]\[ \gamma = \frac{1}{0.280} \][/tex]
[tex]\[ \gamma \approx 3.5714 \][/tex]
Now, we calculate the time [tex]\( t \)[/tex] it takes for Rob to travel 14.4 light-years at a speed of [tex]\( 0.960c \):[/tex]
[tex]\[ t = \frac{d}{v} \][/tex]
[tex]\[ t = \frac{14.4 \text{ light-years}}{0.960c} \][/tex]
[tex]\[ t = \frac{14.4}{0.960} \text{ years} \][/tex]
[tex]\[ t = 15 \text{ years} \][/tex]
Since the Lorentz factor [tex]\( \gamma \)[/tex] is approximately 3.5714, the time measured on Earth for Rob's journey, taking into account time dilation, would be:
[tex]\[ t' = \gamma \cdot t \][/tex]
[tex]\[ t' \approx 3.5714 \cdot 15 \text{ years} \][/tex]
[tex]\[ t' \approx 53.571 \text{ years} \][/tex]
However, since the question asks for the time according to the frame of reference fixed with respect to Earth, we do not need to multiply by the Lorentz factor. The time [tex]\( t \)[/tex] is already the time measured by the stationary observer on Earth.
Therefore, the time it takes Rob to complete the trip relative to Earth is approximately 15.0 years.
Bob and Lily are riding on a merry-go-round. Bob rides a horse on the outer edge of the circular platform and Lily rides a horse near the center. When the merry-go-round is rotating at a constant angular speed w, Bob's speed isa) exactly half as much as Lily'sb) exactly twice as much as Lily'sc) smaller than Lily'sd) larger than Lily'se) same as Lily's
Answer:
d) larger than Lily's
Explanation:
If the merry-go-round is rotating at a constant angular speed, this means that any two points, located at different positions along a radius, rotate at the same angular speed, which means that they sweep the same angle at a given time interval.
In order to both points keep aligned along the same radius, we have a single choice (assuming that we are talking about a rigid body) to meet this premise:
The point farther of the center (Bob) must have a linear speed greater than a point closer to the center (Lily).
Mathematically, we can explain this result as follows:
ω = Δθ / Δt (by definition of angular velocity) (1)
but, by definition of angle, we can say the following:
θ = s/r , where s is the arc along the circumference, and r, the radius.
⇒Δθ = Δs /r
Replacing in (1) we have:
ω = (Δs /Δt) / r
By definition, Δs/Δt = v, so, arranging terms, we get:
v = ω*r
If ω=constant, if r increases, v increases.
So, as Bob is at a distance r from the center larger than Lily's, Bob's speed must be larger than Lily's.
Regardless of their position, Bob and Lily will rotate at the same angular speed. However, due to being further from the center of rotation, Bob's linear speed will be greater than Lily's.
Explanation:The subject matter of the question revolves around rotational motion, specifically concerning angular speed, and how location on a rotating reference frame, such as a merry-go-round, affects linear speed. Both Bob and Lily, being on the same merry-go-round, share the same angular speed (w), regardless of their location on the platform.
The linear speed of an object in rotational motion is given by: Speed = radius x angular speed. Given that Bob is on the outer edge (larger radius) while Lily is near the center (smaller radius), it's evident that Bob's linear or tangential speed (how fast he's actually moving along a path) will be larger than Lily's for the same angular speed. Therefore, the answer is d) Bob's speed is larger than Lily's.
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A flexible shaft consists of a 3 mm diameter steel wire in a flexible hollow tube which imposes a frictional torque of 0.04 N m per meter when the shaft rotates. The shaft is to be used for applying a torque of 0.1 N m to actuate a switch. What is the maximum length of shaft that may be used if the shear stress in the shaft is not to exceed 50 MPa?
Answer:
2.5m
Explanation:
Torque is defined as the rotational effect of a force on a body.
The torque T for the maximum shear stress is given as 0.1 Nm
Frictional torque is the torque caused by a frictional force
The frictional torque F is given as 0.04 Nm/m
The maximum length of the shaft is thus given as
L = T / F
= 0.1/0.04
L= 2.5 m
If a source radiates sound uniformly in all directions and you triple your distance from the sound source, what happens to the sound intensity at your new position?
a. The sound intensity drops to 1 / 27 of its original value.b. The sound intensity increases to three times its original value.c. The sound intensity drops to 1 / 3 of its original value.d. The sound intensity drops to 1 / 9 of its original value.e. The sound intensity does not change.
Sound intensity (energy) falls inversely proportional to the square of the distance from the sound:
[tex]I \propto \frac{1}{r^2}[/tex]
Therefore if we have two values of intensities we have
[tex]\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}[/tex]
As we have that
[tex]r_1 = 3 r_2[/tex]
Then we have that
[tex]\frac{I_1}{I_2} = \frac{r_2^2}{(3r_2)^2}[/tex]
[tex]\frac{I_1}{I_2} = \frac{1}{9}[/tex]
Therefore the correct answer is D. The sound intensity drops to 1 / 9 of its original value.
Suppose the velocity of an electron in an atom is known to an accuracy of 2.0×103m/s (reasonably accurate compared with orbital velocities). What is the electron’s minimum uncertainty in position, and how does this compare with the approximate 0.1-nm size of the atom?
Answer:
289.714 times bigger
Explanation:
[tex]\Delta x[/tex] = Uncertainty in position
[tex]\Delta p[/tex] = Uncertainty in momentum = [tex]\Delta v m[/tex]
[tex]\Delta v[/tex] = Uncertainty in velocity = [tex]2\times 10^3\ m/s[/tex]
h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]
m = Mass of electron = [tex]9.1\times 10^{-31}\ kg[/tex]
From the Heisenberg uncertainty principle we have
[tex]\Delta x\Delta p=\dfrac{h}{4\pi}\\\Rightarrow \Delta x\Delta v m=\dfrac{h}{4\pi}\\\Rightarrow \Delta x=\dfrac{h}{4\pi\Delta v m}\\\Rightarrow \Delta x=\dfrac{6.626\times 10^{-34}}{4\pi \times 2\times 10^3\times 9.1\times 10^{-31}}\\\Rightarrow \Delta x=2.89714\times 10^{-8}\ m[/tex]
Comparing with 0.1 nm size atom
[tex]\dfrac{\Delta x}{x}=\dfrac{2.89714\times 10^{-8}}{0.1\times 10^{-9}}\\\Rightarrow \dfrac{\Delta x}{x}=289.714[/tex]
So, the electron’s minimum uncertainty in position is 289.714 times bigger than an atom of size 0.1 nm
The Heisenberg Uncertainty Principle in quantum physics states a limit to how precisely the position and momentum of a particle, such as an electron, can be known simultaneously. In the given context, though there's a large uncertainty in the electron's velocity, the uncertainty in its position within the atom remains significantly small, thus reflecting the inverse relationship between the precision of these two measurements.
Explanation:The concept involved in this question is called the Heisenberg Uncertainty Principle which is a fundamental theory in quantum mechanics. This theory describes a limit to how precisely we can know both the simultaneous position of an object (such as an electron) and its momentum.
Using the principles of uncertainty (ΔxΔp ≥ h/4π), where h is the Planck constant, the uncertainty in position (Δx) is given. If the given velocity (v) of the electron is 2.0×103m/s with accuracy or uncertainty (Δv), the minimum uncertainty in position can be calculated, where the product of the uncertainties in position and velocity equal to or greater than Planck's constant divided by 4π, i.e. Δx ≥ h / (4πmΔv).
After calculation, it shows that the uncertainty in the electron's position within the atom is very small. Even though the uncertainty in velocity is large, the uncertainty in position remains smaller compared to the size of the atom. Hence, this represents the principle that increasing precision in measuring one quantity leads to greater uncertainty in the measurement of the other quantity.
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Two small spheres, each carrying a net positive charge, are separated by 0.400 m. You have been asked to perform measurements that will allow you to determine the charge on each sphere. You set up a coordinate system with one sphere (charge q1) at the origin and the other sphere (charge q2) at x = +0.400 m. Available to you are a third sphere with net charge q3 = 3.00×10−6 C and an apparatus that can accurately measure the location of this sphere and the net force on it. First you place the third sphere on the x-axis at x = 0.200 m; you measure the net force on it to be 6.50 N in the +x-direction. Then you move the third sphere to x = +0.600 m and measure the net force on it now to be 3.50 N in the +x-direction.
Part A) Calculate q1.
Part B) Calculate q2.
Answer:
a) q₁ = 15. 28 10⁻⁶C, b) q₂ = 5.64 10⁻⁶ C
Explanation:
For this exercise we use Newton's second law where force is Coulomb's electric force
Case 1. Distance (x₁ = 0.200 m) from the third sphere
F₁ = F₁₃ - F₂₃
F₁ = k q₁q₃ / x₁² - k q₂ q₃ / (0.4 - x₁)²
F₁ = k q₃ (q₁ / x₁² - q₂ / (0.4- x₁)²
Case2 Distance (x₂ = 0.6 m) from the third sphere
F₂ = F₁₃ + F₂₃
F₂ = k q₁q₃ / x₂² + k q₂q₃ / (0.4- x₂)²
F₂ = k q₃ (q₁ / x₂² + q₂ / (0.4-x₂)²
The distance is between the spheres, in the annex you can see the configuration of the charge and forces
Let's replace the values
F₁ = 8.99 10⁹ 3.00 10⁻⁶⁶ (q₁ / 0.2² - q₂ / (0.4-0.2)²
F₂ = 8.99 10⁹ 3.00 10⁻⁶ (q₁ / 0.6² + q₂ / (0.4-0.6)²
6.50 = 674. 25 10³ (q₁ –q₂)
3.50 = 26.97 10³ (q₁ / 0.36 + q₂ / 0.04)
We have a system of two equations with two unknowns, let's solve it. Let's clear q1 in the first and substitute in the second
q₁ = q₂ + 6.50 / 674 10³
3.50 / 26.97 10³ = (q₂ + 9.64 10⁻⁶) /0.36 + q₂ / 0.04
1.2978 10⁻⁴ = q₂ / 0.36 + q₂ / 0.04 + 26.77 10⁻⁶
q₂ (1 / 0.36 + 1 / 0.04) = 129.78 10⁻⁶ + 26.77 10⁻⁶
q₂ 27,777 = 156,557 10⁻⁶
q₂ = 156.557 10-6 /27.777
q₂ = 5.636 10⁻⁶ C
We look for q1 in the other equation
q₁ = q₂ + 6.50 / 674 10³
q₁ = 5.636 10⁻⁶ + 9.6439 10⁻⁶
q₁ = 15. 28 10⁻⁶C
Final answer:
To find the charges on the two spheres, we can use Coulomb's law. Calculations show that q1 is approximately 4.333 × 10^-8 C and q2 is approximately 1.111 × 10^-7 C.
Explanation:
To determine the charges on the two spheres, we can use Coulomb's law, which states that the force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Let's calculate the charges:
Part A) Calculating q1:
Using the information given, we can set up the following equation:
F1 = k * (q1 * q3) / (0.200)², where F1 is the net force at x = 0.200 m.
Substituting the given values, we get:
6.50 = (9 × 10^9) * (q1 * (3 × 10^-6)) / (0.200)².
Simplifying the equation, we find that q1 = 4.333 × 10^-8 C.
Part B) Calculating q2:
Using the same equation, but with F2, the net force at x = 0.600 m, we have:
3.50 = (9 × 10^9) * (q2 * (3 × 10^-6)) / (0.600)².
Simplifying, we find that q2 = 1.111 × 10^-7 C.#SPJ11
A car travels 40 miles in 30 minutes.
a) What is the average velocity in kilometers/hour?
b) If the car weighs 1.5 tons, what is its kinetic energy in joules (Note: you will need to convert your velocity to m/s)?
c) When the driver applies the brake, it takes 15 seconds to stop. How far does the car travel (in meters) while stopping?
d) What is the average acceleration of the car (in m/s2) during braking?
Answer:
a) v = 160.9 Km / h, b) K = 9,988 10⁵ J , c) d = 335.2 m , d) a = - 2.9796 m / s²
Explanation:
a) The average speed is defined as the distance traveled in the time interval
v = d / t
Let's reduce the magnitudes to the SI system
t = 30 min (60 s / 1min) = 1800 s
d = 50 mile (1609 m / 1mile) = 80450 m
v = 80450/1800
v = 44.6944 m / s
v = 44.6944 m / s (1 km / 1000m) (3600 s / 1h)
v = 160.9 Km / h
b) W = 2 ton (1000 kg / 1 ton) = 1000 kg
K = ½ m v²
K = ½ 1000 44.694²
K = 9,988 10⁵ J
c) take t = 15 s to stop
v = v₀ - at
v = 0
a = v₀ / t
a = 44,694 / 15
a = 2.9796 m / s²
v² = v₀² - 2 a d
v = 0
d = v₀² / 2 a
d = 44.694² / (2 2.9796)
d = 335.2 m
d) the average acceleration is the change of speed in the time interval
a = (v - v₀) / (t -t₀)
a = (0 - 44.694) / 15
a = - 2.9796 m / s²
A spider hangs by a strand of silk at an eye level 20 cm in front of a plane mirror. You are behind the spider, 65 cm from the mirror. Find the distance between your eye and the image of the spider in the mirror.
The distance between your eye and the image of the spider in the mirror is 85 cm. This is found by adding the distance from the spider to the mirror and from your eyes to the mirror.
Explanation:The situation regards the properties of reflection in a mirror. In a plane mirror, the image of an object appears to be the same distance behind the mirror as the object is in front of the mirror. So the image of the spider in the mirror is 20cm behind the mirror. The eye's distance from the mirror is 65cm. Therefore, the distance from your eye to the image of the spider in the mirror is the sum of these two distances: 20cm (distance from the spider to the mirror) + 65cm (distance from your eye to the mirror) = 85cm.
Learn more about Reflections in Mirrors here:https://brainly.com/question/10507280
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The parking brake on a 1000 kg Cadillac has failed, and it is rolling slowly, at 1 mph , toward a group of small children. Seeing the situation, you realize you have just enough time to drive your 2000 kg Volkswagen head-on into the Cadillac and save the children. You may want to review (Page 269) . Part A With what speed should you impact the Cadillac to bring it to a halt?
Answer:
0.5 mph in the opposite direction
Explanation:
[tex]m_1[/tex] = Mass of Cadillac = 1000 kg
[tex]v_1[/tex] = Velocity of Cadillac = 1 mph
[tex]m_2[/tex] = Mass of Volkswagen = 2000 kg
[tex]v_2[/tex] = Velocity of Volkswagen
In order to know the speed the system must have the momentum exchange
As the linear momentum of the system is conserved
[tex]m_1v_1+m_2v_2=0\\\Rightarrow v_2=-\dfrac{m_1v_1}{m_2}\\\Rightarrow v_2=\dfrac{1000\times 1}{2000}\\\Rightarrow v_2=-0.5\ mph[/tex]
The speed of the impact is given by 0.5 mph in the opposite direction
Final answer:
To bring the Cadillac to a halt using a 2000 kg Volkswagen, we can utilize the conservation of momentum. The necessary impact speed for the Volkswagen is calculated to be approximately 0.22352 m/s, opposing the direction of the Cadillac's motion.
Explanation:
To stop the Cadillac using a head-on collision with the Volkswagen, we must apply the principle of conservation of momentum which states that the total momentum of a system remains constant if no external forces act upon it. Given that both vehicles will come to a halt after the collision, we can set their combined momentum to zero.
Calculating Impact Speed
The Cadillac's momentum is its mass times its velocity (mass of Cadillac ×speed of Cadillac). Converting 1 mph to meters per second (approximately 0.44704 m/s), we get the Cadillac's momentum as 1000 kg × 0.44704 m/s.
The Volkswagen's mass is 2000 kg and we want to find out with what speed it should hit the Cadillac to bring both to a halt. Let's denote this unknown speed as v. The momentum of the Volkswagen right before the impact is 2000 kg × v.
Using conservation of momentum:
Total momentum before collision = Total momentum after collision
(1000 kg × 0.44704 m/s) + (2000 kg × v) = 0
447.04 kg× m/s + 2000 kg × v = 0
To find the speed v, we'll solve the equation:
2000 kg × v = -447.04 kg× m/s
v = -447.04 kg× m/s / 2000 kg
v = -0.22352 m/s
The negative sign indicates that the Volkswagen must be traveling in the opposite direction of the Cadillac's motion, which we already know. Thus, the required speed for the Volkswagen for a head-on impact to bring the Cadillac to a halt is approximately 0.22352 m/s.
Two blocks of ice , one four times as heavy as the other, aren at rest on a frozen lake. a person pushes each block the same distance d.ignore friction and assume that an equal force F is exerted on each block. which of the following statements is true about the kinetic energy of the heavier block after the push?A) it is smaller than kinetic energy of the lighter blockB) it is equal to the kinetic energy of the lighter blockC) it is larger than the kinetic energy of the lighter blockD) it cannot be determined without knowing the force and the mass of each block
Answer:
The correct answer is B: the kinetic energy of the heavier block is equal to the kinetic energy of the lighter block.
Explanation:
Hi there!
The work done on each block is calculated as follows:
W = F · d
Since the two blocks were pushed the same distance with the same force, the work done on each object is the same.
Using the work-energy theorem, we know that the work done on an object is equal to its change in kinetic energy (KE):
W = ΔKE
W = final KE - initial KE
Since the objects are at rest, initial KE = 0, then:
W = final KE
Since the work done on each block is the same, so will be its final kinetic energy.
The correct answer is B: the kinetic energy of the heavier block is equal to the kinetic energy of the lighter block.
Final answer:
When an equal force is applied over the same distance to two blocks of ice, with one being four times heavier than the other, the kinetic energy gained by both blocks is equal, because the work done on them is the same.
Explanation:
The question involves two blocks of ice on a frozen lake, with one block being four times as heavy as the other. When an equal force F is exerted on each block, pushing them the same distance d, we are asked which statement is true about the kinetic energy of the heavier block after the push compared to the kinetic energy of the lighter block. Since the work done (work = force × distance) on both blocks is the same and work done on an object is equal to the change in its kinetic energy (Work-Energy Principle), both blocks will have the same increase in kinetic energy. Therefore, without considering initial kinetic energies (since both start from rest), the correct answer is that it is equal to the kinetic energy of the lighter block (Option B). This is because the amount of work done on both blocks is the same, leading to an equal increase in kinetic energy.