Answer:
0.0549 m
Explanation:
Given that
equation y(x,t)=Acos(kx−ωt)
speed v = 8.5 m/s
amplitude A = 5.5*10^−2 m
wavelength λ = 0.5 m
transverse displacement = ?
v = angular frequency / wave number
and
wave number = 2π/ λ
wave number = 2 * 3.142 / 0.5
wave number = 12.568
angular frequency = v k
angular frequency = 8.5 * 12.568
angular frequency = 106.828 rad/sec ~= 107 rad/sec
so
equation y(x,t)=Acos(kx−ωt)
y(x,t)= 5.5*10^−2 cos(12.568 x−107t)
when x =0 and and t = 0
maximum y(x,t)= 5.5*10^−2 cos(12.568 (0) − 107 (0))
maximum y(x,t)= 5.5*10^−2 m
and when x = x = 1.52 m and t = 0.150 s
y(x,t)= 5.5*10^−2 cos(12.568 (1.52) −107(0.150) )
y(x,t)= 5.5*10^−2 × (0.9986)
y(x,t) = 0.0549 m
so the transverse displacement is 0.0549 m
This force will cause the path of the particle to curve. Therefore, at a later time, the direction of the force wil At a given moment the particle is moving in the +x direction (and the magnetic field is always in the +z direction). If q is positive, what is the direction of the force on the particle due to the magnetic field?
Answer:
force is -y direction
Explanation:
The magnetic force is given by the equation
F = q v x B
where bold letters indicate vectors.
The direction of this force can be found by the right-hand rule, where the thumb points in the direction of speed, the other fingers extend in the direction of the magnetic field and the palm is in the direction of force for a positive charge
let's apply this to our case
the speed is in the direction + x the thumb
the magnetic field in the + z direction fingers extended
the palm is in the direction - y
force is -y direction
Final answer:
The force due to the magnetic field on a positively charged particle moving in the +x direction and perpendicular to the magnetic field will be in the +-y direction.
Explanation:
At a given moment the particle is moving in the +x direction (and the magnetic field is always in the +z direction). If q is positive, the direction of the force on the particle due to the magnetic field will be in the +-y direction.
The force P is applied to the 59-kg block when it is at rest. Determine the magnitude and direction of the friction force exerted by the surface on the block if (a) P = 0, (b) P = 198 N, and (c) P = 288 N. (d) What value of P is required to initiate motion up the incline? The static and kinetic coefficients of friction between the block and the incline are μs = 0.20 and μk = 0.16, respectively. The friction force is positive if up the incline, negative if down the incline.
Answer:
Explanation:
Coefficient of static friction(μs) = 0.20 Coefficient of kinetic friction (μk) = 0.16
Mass of block (m) = 59kg
acceleration due to gravity (g) = 9.8m/s^2
Normal force = weight of block = mg
Weight of block(W) = 59kg × 9.8m/s^2 = 578.2N
N = 578.2 N
Maximum force of static friction Fs(max) is given by ;
Coefficient of static friction (μs) × N
0.20 × 578.2N = 115.64
As long as P is < 115.64, then the block will remain stationary and thus static frictional force = P (Fs = P)
magnitude and direction of the friction force exerted by the surface on the block if;
(A.) P = 0, Fs = 0
B.) P = 198N;
Fk = μk × N = 0.16 × 578.2 = 92.512N
Fs =
C.) P = 288;
Fk = μk × N = 0.16 × 578.2 =
D.) P value required to initiate motion;
0.2 × 59kg × 9.8m/s^2 = 115.64N
The correct answer is option (a) 0 N (indeterminate direction), (b) 198 N (negative direction, down the incline), (c) 92.6064 N (negative direction, down the incline) and (d) The value of [tex]\( P \)[/tex] required to initiate motion up the incline is 115.758 N.
Let's solve the problem step by step for each case:
(a) When [tex]\( P = 0 \)[/tex]:
The magnitude of the static friction force is equal to the applied force required to start motion, which is the maximum static friction force. This force is given by [tex]\( f_{s,\text{max}} = \mu_s N \)[/tex], where [tex]\( N = mg \)[/tex].
Given m = 59 kg and g = 9.81 m/s², we have:
[tex]\( N = 59 \text{ kg} \times 9.81 \text{ m/s}^2 \)[/tex]
[tex]\( N = 578.79 \text{ N} \)[/tex]
The maximum static friction force is:
[tex]\( f_{s,\text{max}} = 0.20 \times 578.79 \text{ N} \)[/tex]
[tex]\( f_{s,\text{max}} = 115.758 \text{ N} \)[/tex]
Since the block is at rest and no force is applied, the frictional force is exactly what is needed to balance any other forces acting on the block, which in this case is zero.
(b) When [tex]\( P = 198 \)[/tex] N:
The frictional force will exactly oppose the applied force [tex]\( P \)[/tex], so, the magnitude of the frictional force is 198 N, and it acts in the direction opposite to [tex]\( P \)[/tex], which is negative (down the incline).
(c) When [tex]\( P = 288 \)[/tex] N:
If the applied force [tex]\( P \)[/tex] exceeds the maximum static friction force, the block will start to move, and the frictional force will be kinetic. The magnitude of the kinetic friction force is given by [tex]\( f_k = \mu_k N \)[/tex].
[tex]\( f_k = 0.16 \times 578.79 \text{ N} \)[/tex]
[tex]\( f_k = 92.6064 \text{ N} \)[/tex]
(d) To initiate motion up the incline:
To move the block up the incline, the applied force [tex]\( P \)[/tex] must be equal to the maximum static friction force plus any additional force to overcome other forces such as gravity along the incline (which is not provided in the problem).
[tex]\( P = f_{s,\text{max}} \)[/tex]
[tex]\( P = 115.758 \text{ N} \)[/tex]
A 24.0-V battery is connected in series with a resistor and an inductor, with R = 9.40 Ω and L = 6.40 H, respectively. (a) Find the energy stored in the inductor when the current reaches its maximum value. J (b) Find the energy stored in the inductor at an instant that is a time interval of one time constant after the switch
Answer:
a) 20.81 J
b) 8.29 J
Explanation:
V = iR + L di/dt
where
i = a(1-e^-kt)
for large t
i = V/R
i = 24 / 9.4
i = 2.55 A
so
i = 2.55(1-e^-kt)
di/dt = 2.55 k e^-kt
24 = 24-24e^-kt + 6.4(2.55)k e^-kt
24 = 6.4(2.55) k
k = 24 / (6.4 * 2.55)
k = 24 / 16.32
k = 1.47 = R/L
so
i = 2.55(1-e^-(Rt/L))
current is maximum at great t
i max = 2.55 - 0
energy = (1/2) L i^2
E = (1/2)(6.4)2.55^2
E = 20.81 Joules
one time constant T = L/R and e^-(Rt/L) = 1/e = .368
i = 2.55 (1 - 0.368)
i = 2.55 * 0.632
i = 1.61 amps
energy = (1/2)(6.4)1.61^2
E = 8.29 Joules
The energy stored in the inductor when the current reaches its maximum value is 20.80 J. After one time constant, the energy stored is approximately 8.25 J.
we need to determine the energy stored in the inductor at two different points in time.
a) The energy stored in an inductor is given by the formula:
U = (1/2) × L × I²
Maximum Current
In an RL circuit, the maximum current is achieved when the circuit reaches a steady state. The maximum current is given by:
I_max = V/R, where V is the voltage and R is the resistance.
Given V = 24.0 V and R = 9.40 Ω, we have:
I_max = 24.0 V / 9.40 Ω = 2.55 A
Now, the energy stored in the inductor at the maximum current is:
U_max = (1/2) × L × I_max²
Given L = 6.40 H, we get:
U_max = (1/2) × 6.40 H × (2.55 A)² = 20.80 J
b) Energy After One Time Constant
The time constant for an RL circuit is defined as:
τ = L/R
Given L = 6.40 H and R = 9.40 Ω, we calculate:
τ = 6.40 H / 9.40 Ω ≈ 0.68 s
At one time constant, the current has reached approximately 63.2% of its maximum value:
I_τ ≈ 0.632 × I_max = 0.632 × 2.55 A ≈ 1.61 A
The energy stored in the inductor at this time is:
U_τ = (1/2) × L × I_τ² = (1/2) × 6.40 H × (1.61 A)² ≈ 8.25 J
complete question:
A 24.0-V battery is connected in series with a resistor and an inductor, with R = 9.40 Ω and L = 6.40 H, respectively.
(a) Find the energy stored in the inductor when the current reaches its maximum value. J
(b) Find the energy stored in the inductor at an instant that is a time interval of one time constant after the switch is closed. j
how much work is done to move 3c of charge through a potential difference of 1.5v
Answer:4.5 joules
Explanation:
charge=3c
Potential difference=1.5v
Work=potential difference x charge
Work=1.5 x 3
Work=4.5
Work=4.5 joules
Two concavemirrors are kept in a straight line as shown in fig. Mirror A forms 2 times enlarged real image of an object kept in front of it on the same straight line. The focal length of mirror A is given as 10cm
i.Using sign conventions calculate the object position and image position in the case of mirror A
Answer:
Explanation:
Mirror formation in mirror A
focal length f = - 10 cm ( concave mirror )
Let u be object distance , v is image distance .
In case image is virtual
v / u = 2
v = 2u
Mirror formula
[tex]\frac{1}{v} +\frac{1}{u} =\frac{1}{f}[/tex]
Putting the values
[tex]-\frac{1}{2u} +\frac{1}{u} =-\frac{1}{10}[/tex]
[tex]\frac{1}{2u}[/tex] = [tex]-\frac{1}{10}[/tex]
u = -5 cm
In case image is real
[tex]\frac{1}{v} +\frac{1}{u} =\frac{1}{f}[/tex]
[tex]\frac{1}{2u} +\frac{1}{u} =-\frac{1}{10}[/tex]
[tex]\frac{3}{2u} =-\frac{1}{10}[/tex]
u = [tex]-\frac{30}{2}[/tex]
u = - 15 cm
Occasionally, the electric circuit that produces a coordinated pumping of blood from the four chambers of the heart becomes disturbed. Ventricular fibrillation can occur-random muscle contractions that produce little or no blood pumping. To stop the fibrillation, two defibrillator pads are placed on the chest and a large current (about 20 amps) is sent through the heart, restarting its normal rhythmic pattern. The current lasts 5 ms and transfers 200 J of electric energy to the body. Find the potential difference between the defibrillator pads. Express your answer with the appropriate units.
Answer:
potential difference between the defibrillator pads = 2000 V
Explanation:
We are given;
Current;I = 20A
Time;t = 5 m.s = 5 x 10^(-3) s
Electrical energy;W = 200 J
Now let's find the value of the electric charge from the formula;.
q = It
Where q is charge, I is current and t is time.
Thus; q = 20 x 5 x 10^(-3)
q = 0.1 C
Now, the potential difference between the defibrillator pads would be calculated from;
V = W/q
Where V is potential difference and W and q are as stated earlier.
Plugging in the relevant values gotten earlier, we have;
V = 200/0.1
V = 2000 V
Finally, the boy decides that he has had enough fun. He decides to crawl to the outer edge of the merry-go-round and jump off. Somehow, he manages to jump in such a way that he hits the ground with zero velocity with respect to the ground. What is the angular speed in radians/second of the merry-go-round after the boy jumps off?
The angular speed of the merry-go-round would increase when the boy jumps off due to the conservation of angular momentum. However, without specific values like the weights and distances involved, the precise speed cannot be calculated.
Explanation:The speed of the merry-go-round after the boy jumps off cannot be determined from the given information. However, we can use the principle of conservation of angular momentum to analyze the situation; this principle states that the initial angular momentum (while the boy was still on the merry-go-round) should be equal to the final angular momentum (after the boy jumps off). The angular momentum is the product of moment of inertia and angular speed. When the boy was sitting on the edge of the merry-go-round, he had a certain moment of inertia due to his mass and the distance from the center. Once he jumps off, the moment of inertia decreases, hence according to the conservation law, the angular speed must increase to keep the momentum constant.
However, the exact increase in angular speed could only be calculated if we knew the weight of the boy and of the merry-go-round, as well as the radius of the merry-go-round (to calculate moments of inertia), and the initial angular speed. Another simplification made in this situation is that the friction between the merry-go-round and its axle is negligible, which might not be true in real life scenarios.
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When the boy jumps off the merry-go-round, the total angular momentum of the system should conserve. As the boy's leaving reduces the moment of inertia, the angular speed of the merry-go-round must increase to preserve angular momentum. To calculate the exact increase, we would need specific values for the moment of inertia before and after the jump.
Explanation:The question asked is associated with angular velocity and moment of inertia in the context of a merry-go-round. As the boy jumps off the merry-go-round, the system's total angular momentum must be conserved, because there are no external torques acting on it.
When the boy was still on the merry-go-round, he was part of the system, contributing to the total angular momentum. But when he departs, he no longer contributes to the system's moment of inertia, effectively reducing it. As a result, the angular speed of the merry-go-round must increase to conserve angular momentum (since angular momentum = moment of inertia x angular speed).
Unfortunately, without specific values for the moment of inertia both before and after the boy jumps off, we can't calculate the exact increase in angular speed. However, the key concept is the conservation of angular momentum: when the moment of inertia decreases, the angular speed must increase to maintain constant total angular momentum.
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The hydronium ion concentration in an aqueous solution is [H+] = 5.3 x10-4 M. What is the hydroxide ion concentration? Is this solution acidic or basic?
A.) [OH-] = 1.9 x10-11 M, acidic
B.) [OH-] = 4.7 x10-6 M, basic
C.) [OH-] = 1.1 x10-9 M, acidic
D.) [OH-] = 3.2 x10-8 M, acidic
Answer:a
Explanation:
Given
[tex][H^+]=5.3\times 10^{-4}\ M[/tex]
and [tex]pH+pOH=14[/tex]
Also [tex]pH=-\log [H^+][/tex]
therefore [tex]pH=-\log (5.3\times 10^{-3})[/tex]
[tex]pH=-(-4-\log (5.3))[/tex]
[tex]pH=3.275[/tex]
Thus [tex]pOH=14-3.275[/tex]
[tex]pOH=10.275[/tex]
and [tex]pOH=-\log [OH^{-}][/tex]
[tex][OH]^{-1}=10^{-10.25}[/tex]
[tex][OH]^{-1}=1.88\times 10^{-11}\ M[/tex]
As the pH is less than 7 therefore solution is acidic
A very long, straight solenoid with a cross-sectional area of 2.16 cm^2 is wound with 85.7 turns of wire per centimeter. Starting at t = 0, the current in the solenoid is increasing according to i(t)= (( 0.175 A/s^2 )t^2). A secondary winding of 5 turns encircles the solenoid at its center, such that the secondary winding has the same cross-sectional area as the solenoid. What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.2A ?
Answer:
1.74x10⁻⁵ V
Explanation:
n = 85.7 turns/cm => 8570 turns/metre
The field inside the long solenoid is given by B = μ₀ni
B = 4πx10⁻⁷ x 8570 x 0.175t² = 1.884x10⁻³ t²
dB/dt = 3.78x10⁻³ t
Cross-sectional Area'A'= 2.16 cm²=> 2.16 x [tex]10^{-4}[/tex] m²
Now, rate of change of flux linkage '|Emf|' is given by:
|Emf| = d(NAB)/dt = NA dB/dt
|Emf| = 5 x 2.16 x [tex]10^{-4}[/tex] x 3.78x10⁻³ t
|Emf| = 4.0824x10⁻⁶ t
Considering time 't' at which the current = 3.2A , we have
3.2 = 0.175T²
T² = 3.2/0.175
T = 4.28 s
|emf| = 4.0824x10⁻⁶ t => 4.0824x10⁻⁶ x4.28
|emf|= 1.74x10⁻⁵ V
Therefore,the magnitude of the emf induced in the secondary winding is 1.74x10⁻⁵ V
In a shipping company distribution center, an open cart of mass 50 kg is rolling to the left at a speed of 5 m/s. You can ignore friction between the cart and the floor. A 15 kg package slides down a chute that makes an angle of 27 degrees below the horizontal. The package leaves the chute with a speed of 3 m/s, and lands in the cart after falling for 0.75 seconds. The package comes to a stop in the cart after 4 seconds. What is:
Answer:
(1). 9.35 m/s.
(2).V2 = 3.22 m/s to the left.
Explanation:
So, we are given the following data or parameters or infomation in the question above as;
(1). "open cart of mass 50 kg is rolling to the left at a speed of 5 m/s. "
(2). "A 15 kg package slides down a chute that makes an angle of 27 degrees below the horizontal."
(3). "The package leaves the chute with a speed of 3 m/s, and lands in the cart after falling for 0.75 seconds."
(4). "The package comes to a stop in the cart after 4 seconds."
(a). So, in order to solve this question, we will be making use of the equations below;
(1). Gravitational potential energy = mass × acceleration due to gravity × height.
(2). Kinetic energy = 1/2 × mass × (velocity)^2.
(3). Kinetic energy (initial) + potential energy (initial) = kinetic energy (final) + potential energy (final).
So, starting from equation (1) above;
Gravitational potential energy = mass × acceleration due to gravity × height.
Gravitational potential energy = 15 × 9.8 × 4 = 588 J.
Kinetic energy = 1/2 × mass × (velocity)^2. = 1/2 × 15 × (3)^2.= 67.5 J.
kinetic energy (final) = 1/2 × 15 × v^2. = 7.5v^2.
Hence, Kinetic energy (initial) + potential energy (initial) = kinetic energy (final) + potential energy (final).
[potential energy (final) = 0].
67.5 + 588 = 7.5v^2.
v = 9.35 m/s
(b). -3 cos (27°) = -2.67 m/s.
15 × -2.67 + 50 × 5 =( 15 + 50) × v2.
=> 209.5 = 65v2.
=> V2 = 3.22 m/s to the left.
(1) The speed of the package before it lands on the cart is 9.35 m/s.
(2) The final speed of the cart is 3.22 m/s
Conservation of energy and momentum:The gravitational potential energy of a system is given by
PE = mgh
where m is the mass
g is the acceleration due to gravity
and h is the height.
The kinetic energy is given by:
KE = (1/2) mv²
According to the law of conservation of energy:
Kinetic energy (initial) + potential energy (initial) = kinetic energy (final) + potential energy (final).
The final potential energy will be zero since taking the cart as the reference as ground.
PE(initial) = 15 × 9.8 × 4 = 588 J
KE(initial) = (1/2) mv² = 1/2 × 15 × 3² = 67.5 J
KE(final) = 1/2 × 15 × v² = 7.5v²
So,
67.5 + 588 = 7.5v²
v = 9.35 m/s
(b) The package leave at an angle of 27° down the horizontal with velocity 3 m/s, so the horizontal speed is:
v ' = -3 cos (27°) = -2.67 m/s.
Negative sign, as the package is moving opposite to the cart
After the package lands on the cart, both move together, so from conservation of momentum:
15 × (-2.67) + 50 × 5 =( 15 + 50) × V
where V is the speed of cart and package together
209.5 = 65 V
V = 3.22 m/s to the left.
The question was incomplete. After searching it, it is likely that it was as given below:
In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s. Ignore friction between the cart and the floor. A 15.0-kg package slides down a chute that is inclined at 27∘ from the horizontal and leaves the end of the chute with a speed of 3.00 m/s. The package lands in the cart and they roll together. If the lower end of the chute is a vertical distance of 4.00 m above the bottom of the cart, what is (1) the speed of the package just before it lands in the cart and (2) the final speed of the cart?
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What is the density of an 100g object with volume of 50cm3?
Answer:
mass m= 100 g or .1 kg
volume V= 50 cm3 or 50×[tex]10^{-6[/tex] m3
density ρ= [tex]\frac{m}{V}[/tex]
So,
ρ=[tex]\frac{.1}{50*10^{-6} }[/tex]= 2000 kgm-3 or 2 gcm-3
Explanation:
The density is calculated as 2 g / cm³.
To calculate the density, mass = 100 g
volume = 50 cm ³
What is density?The principle of density was discovered by the Greek scientist Archimedes.
The mass of a unit volume of a material substance.
Then, formula for density
ρ = M/V,
where d is density, M is mass, and V is volume.
Unit of density is grams per cubic centimeter (g / cm³) .
Substituting all the values,
ρ = 100 / 50
= 2
Hence, the density of an object is ρ = 2 g / cm³.
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Cindy pours cool milk into a hot bowl of oatmeal and then stirs it with a room-temperature spoon. Which substance will have heat flow from it to the other substances?
Answer:
The answer is oatmeal.
Explanation:
Since the oatmeal is the hottest substance, it will spread heat to the other substances.
Answer:
C, oatmeal
Explanation:
To understand the formula representing a traveling electromagnetic wave. Light, radiant heat (infrared radiation), X rays, and radio waves are all examples of traveling electromagnetic waves. Electromagnetic waves comprise combinations of electric and magnetic fields that are mutually compatible in the sense that the changes in one generate the other. The simplest form of a traveling electromagnetic wave is a plane wave. For a wave traveling in the x direction whose electric field is in the y direction, the electric and magnetic fields are given by
Answer:
E(x,t) = Emaxcos(kx - ωt + φ),
B(x,t) = Bmaxcos(kx - ωt + φ).
Explanation:
E is the electric field vector, and B is the magnetic field vector of the EM wave. For electromagnetic waves the electric field E and the magnetic field B are always perpendicular to each other and perpendicular to the direction of propagation. The direction of propagation is the direction of E x B.
A bungee jumper jumps from a tall bridge that is 60m above a flat, dry creek bed. She is carrying an audio generator that emits sound at frequency 1200 Hz. The unstretched length of the bungee cord is 26m, and the spring constant is selected so that she stops just before hitting the creek bed. At what point in her fall is the beat frequency, determined by listening to the audio generator and the reflection from the creek bed, a maximum
Answer:
Explanation:
The original frequency of sound f₀
The apparent frequency of sound fa
For apparent frequency the formula is
fa = [tex]f_0\times\frac{V+v}{V-v }[/tex] , v is velocity of jumper which increases as he goes down .
Beat frequency
= fa - f₀
= [tex]f_0\times(\frac{V+v}{V-v }-1)[/tex]
= [tex]f_0\times(\frac{2v}{V-v })[/tex]
since v is very small in comparison to V , velocity of sound , in the denominator , v can be neglected.
beat frequency = [tex]f_0\times(\frac{2v}{V })[/tex]
v , the velocity of jumper will go on increasing as long as net force on the jumper is positive or
mg > kx where x is extension in the cord and k is its force constant . Below this point kx or restoring force becomes more than weight of the jumper and then net force on the jumper directs upwards. At this point beat frequency becomes maximum.
A plane circular loop of conducting wire of radius with possesses turn is placed in a uniform magnetic field. The direction of the magnetic field makes an angle of 30° with respect to the normal direction to the loop. The magnetic field-strength Bis increased at a constant rate from to in a time interval of 10s. (a) What is the emf generated around the loop? (b) If the electrical resistance of the loop is , what current flows around the loop as the magnetic field is increased?
Complete question;
A plane circular loop of conducting wire of radius r-10 cm which possesses 15 turns is placed in a uniform magnetic field. The direction of the magnetic field makes an angle of 30° with respect to the normal direction to the loop. The magnetic field-strength is increased at a constant rate from IT to 5T in a time interval of 10 s.
a) What is the emf generated around the loop?
b) If the electrical resistance of the loop is 15Ω, what current flows around the loop as the magnetic field is increased?
Answer:
A) E.M.F generated around loop = 0.163 V
B)Current in loop; I = 0.011 A
Explanation:
A) We are given;
Initial magnetic field strength;B1 = 1T
Final magnetic field strength;B2 = 5T
Number of turns;N = 15 turns
Radius; r = 10cm = 0.1m
Angle;θ = 30°
Time interval; Δt = 10 s
Now, the formula for magnetic flux is: Φ = NABcosθ
Where;
N is number of turns
A is area = πr²
B is magnetic field strength
θ is angle
So, initial magnetic flux is;
Φ1 = NA(B1)cosθ
Plugging in the relevant values to obtain;
Φ1 = 15*(π*0.1²)(1)cos30
Φ1 = 0.4081 Wb
Similarly, final magnetic flux is;
Φ2 = NA(B2)cosθ
Plugging in the relevant values to obtain;
Φ2 = 15*(π*0.1²)(5)cos30
Φ2 = 2.0405 Wb
The time rate of change of the flux is;
dΦ_B/dt = (Φ2 - Φ1)/Δt
So, dΦ_B/dt = (2.0405 - 0.4081)/10
dΦ_B/dt = 0.163 Wb/s
Thus, the emf generated around the loop is; E = dΦ_B/dt = 0.163 V
B) from Ohm's law, the current which flows around the loop in response to the emf is given as;
I = E/R
We are given R =15Ω
Thus; I = 0.1632/15
I = 0.011 A
A 0.060-kg ice hockey puck comes toward a player with a high speed. The player hits it directly back softly with an average force of 1.50 x 10^3 N. The hockey stick is in contact with the ball for 1.20 ms, and the ball leaves the stick with a velocity of 8.00 m/s. Let the direction of the force be the + x direction. Find the following (note: be careful with the sign/direction of the values):_______. 1. The final momentum of the ball 2. The impulse on the ball 3. The initial velocity of the ball
Answer:u=-22 m/s
Explanation:
Given
mass of puck [tex]m=0.06\ kg[/tex]
Average force [tex]f_{avg}=1.5\times 10^3\ N[/tex]
time of contact [tex]t=1.2ms=1.2\times 10^{-3}\ s[/tex]
puck leaves with a velocity of [tex]v=8\ m/s[/tex]
We know impulse is [tex]F_{avg}\Delta t[/tex][tex]=\text{change in momentum}[/tex]
therefore
[tex]1.5\times 10^3\times (1.2\times 10^{-3})=P_f-P_i[/tex]
[tex]P_i=0.06\times 8-1.8[/tex]
[tex]P_i=0.48-1.8=-1.32\ kg-m/s[/tex]
Final momentum [tex]P_f=m\times v_f[/tex]
[tex]P_f=0.06\times 8[/tex]
[tex]P_f=0.48\ kg-m/s[/tex]
Impulse on the ball [tex]=F_{avg}\Delta t[/tex]
Impulse[tex]=1.5\times 10^3\times 1.2\times 10^{-3}=1.8\ N-s[/tex]
Initial velocity is given by
[tex]u=\frac{P_i}{m}=\frac{-1.32}{0.06}[/tex]
[tex]u=-22\ m/s[/tex]
i.e. initially ball is moving towards -x-axis
The three small spheres are welded to the light rigid frame which is rotating in a horizontal plane about a vertical axis through O with an angular velocity w= 20 rad/s. If a couple MO= 30 N.m is applied to the frame for 5 seconds, compute the new angular velocity.
The new angular velocity is computed using the kinematic expression w² = wo² + 2a0, where 'wo' is the original angular velocity 'a' is the angular acceleration and 'w' is the new angular velocity. The calculation is guided by principles of physics primarily involving angular momentum and acceleration.
Explanation:The calculation for the new angular velocity can be gleaned from the kinematic expression w² = wo² + 2a0. The formula described represents the angular motion of the frame. The angular velocity originally is 20 rad/s and a couple MO of 30 N.m is applied to the frame for 5 seconds. The development of this concept involves principals of angular momentum as well as angular acceleration.
The angular acceleration can be calculated using the relation a = nett
Bearing in mind the initial angular velocity, the applied couple MO, and the duration of its application, the new angular velocity can be computed using the given formula. Note that the laws of physics, specifically the law of conservation of angular momentum, play an essential part in this calculation.
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You are packing for a trip to another star. During the journey, you will be traveling at 0.99c. You are trying to decide whether you should buy smaller sizes of your clothing because you will be thinner on your trip due to length contraction. You also plan to save money by reserving a smaller cabin to sleep in because you will be shorter when you lie down. Should you (a) buy smaller sizes of clothing, (b) reserve a smaller cabin, (c) do neither of these things, or (d) do both of these things?
Answer:
Do neither of these things ( c )
Explanation:
For length contraction : Is calculated considering the observer moving at a speed that is relative the object at rest applying this formula
L = (l) [tex]\sqrt{1 -\frac{v^{2} }{c^{2} } }[/tex]
where l = Measured distance from object at rest, L = contracted measured in relation to the observer , v = speed of clock , c = speed of light
you will do neither of these things because before you can make such decisions who have to view the object in this case yourself from a different frame from where you are currently are, if not your length and width will not change hence you can't make such conclusions/decisions .
A reddish layer of rust can form on old iron nails and bikes. In this chemical reaction, iron reacts slowly with oxygen in the air. The chemical name for rust is iron oxide. What are the reactant(s) and product(s) of this reaction?
Answer:
Reactants: Fe, [tex]H_2O[/tex] and [tex]O_2[/tex]
Product: [tex]Fe_2O_3.H_2O[/tex]
Explanation:
Two electrochemical reactions occur during the formation of rust;
Oxidation of iron (Fe) in water: [tex]2Fe --> 2Fe^{2+} + 4e^-[/tex] Reduction of oxygen in water: [tex]O_2 + 2H_2O + 4e^- --> 4OH^-[/tex]The two products from the two electrochemical equations then react together to form iron hydroxide.
[tex]2Fe^{2+} + 4OH^- --> 2Fe(OH)_2[/tex]
The iron hydroxide thus produced slowly reacts with oxygen to form rust which is [tex]Fe_2O_3.H_2O[/tex].
Hence, the reactants are iron (Fe), water ([tex]H_2O[/tex]) and oxygen ([tex]O_2[/tex]) while the product is rust ([tex]Fe_2O_3.H_2O[/tex]).
Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the equation PV = C, where C is a constant. Suppose that at a certain instant the volume is 600 cm3, the pressure is 150 kPa, and the pressure is increasing at a rate of 40 kPa/min. At what rate is the volume decreasing at this instant?
Answer:
The volume is decreasing at 160 cm³/min
Explanation:
Given;
Boyle's law, PV = C
where;
P is pressure of the gas
V is volume of the gas
C is constant
Differentiate this equation using product rule:
[tex]V\frac{dp}{dt} +P\frac{dv}{dt} = \frac{d(C)}{dt}[/tex]
Given;
[tex]\frac{dP}{dt}[/tex] (increasing pressure rate of the gas) = 40 kPa/min
V (volume of the gas) = 600 cm³
P (pressure of the gas) = 150 kPa
Substitute in these values in the differential equation above and calculate the rate at which the volume is decreasing ( [tex]\frac{dv}{dt}[/tex]);
(600 x 40) + (150 x [tex]\frac{dv}{dt}[/tex]) = 0
[tex]\frac{dv}{dt} = -\frac{(600*40)}{150} = -160 \ cm^3/min[/tex]
Therefore, the volume is decreasing at 160 cm³/min
Final answer:
Boyle's law states that the volume of a gas is inversely proportional to its pressure when temperature and the amount of gas are constant. Mathematically, this can be expressed as PV = C. To find the rate at which the volume is decreasing at a given instant, we differentiate the equation PV = C with respect to time and substitute the given values.
Explanation:
Boyle's law states that the volume of a gas is inversely proportional to its pressure when temperature and the amount of gas are constant. Mathematically, this can be expressed as PV = C, where P is the pressure, V is the volume, and C is a constant. In this scenario, the pressure is increasing at a rate of 40 kPa/min. To find the rate at which the volume is decreasing at this instant, we differentiate the equation PV = C with respect to time. Applying the chain rule, we get:
d(PV)/dt = d(C)/dt
P(dV/dt) + V(dP/dt) = 0
dV/dt = - (V/P)(dP/dt)
We substitute the given values where the volume V = 600 cm3 and pressure P = 150 kPa:
dV/dt = - (600 cm3 / 150 kPa) (40 kPa/min)
dV/dt = -160 cm3/min
Therefore, the volume is decreasing at a rate of 160 cm3/min at this instant.
Archimedes tells us the lifting power of a balloon (how much mass it can lift) is equal to the difference between the mass of the balloon and the mass of the air it displaces. That is, if the balloon occupies the same volume as of air, but the balloon only weighs , then the balloon can lift . Airships have sometimes been filled with hydrogen (), but hydrogen is very flammable, and after the Hindenburg caught fire and crashed in with serious loss of life airships have nearly always been filled with helium () instead.
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
the force that is exerted when a shopping cart is pushed. the forces that causes a metal ball to move toward a magnet
Answer:
The force that is exerted when a shopping cart is pushed is a type of push force, supplied by the muscles of the cart pusher's body.
The forces that causes a metal ball to move toward a magnet is a type of pull force that is as a result of the magnetic field forces.
Explanation:
Forces are divided into push forces that tends to accelerate a body away from the source of the force, and pull forces that accelerates the body towards the force source.
Examples of push forces includes pushing a cart, pushing a table, repulsion of two similar poles of a magnet etc. Examples of pull forces includes a attractive force between two dissimilar poles of a magnet, pulling a load by a rope, a dog pulling on a leash etc.
Answer:
The force that is exerted when a shopping cart is pushed:
-Contact
The force that causes a metal ball to move toward a magnet:
-Noncontact
a minute later the green car is traveling 60 miles per hour north, the blue car traveling 60 miles per hour south how is the green cars velocity similar to the blue cars velocity? what is the different about their velocities
Answer:
A) The green cars's velocity has the same magnitude as the blue car's velocity.
B) the difference about their velocity is the direction of the velocity which is 180° in the opposite directions.
Explanation:
Velocity is a vector quantity that has both magnitude and direction. The magnitude shows the size, and the direction shows the path the motion takes.
Both cars have the same magnitude but different directions. The difference in their direction is at a 180°angle.
The expression known as Froude's number is given as Upper F equals StartFraction v squared Over gl EndFraction . This number was used in shipbuilding and also in research on dinosaurs. In this example, l is given as well as Froude's number. Find the velocity, v (in meters per second) of a rhinoceros. It is known that gequals9.8 meters per second squared. Rhinoceros: lequals 1.2; Froude number equals 2.6
Answer:[tex]v=5.529\ m/s[/tex]
Explanation:
Given
number is given as [tex]=\frac{v^2}{gl}[/tex]
and [tex]number=2.6[/tex]
[tex]l=1.2\ m[/tex]
g=acceleration due to gravity[tex](9.8\ m/s^2)[/tex]
calculating for [tex]v[/tex]
[tex]v^2=gl(2.6)[/tex]
[tex]v=\sqrt{2.6\times 9.8\times 1.2}[/tex]
[tex]v=\sqrt{30.576}[/tex]
[tex]v=5.529\ m/s[/tex]
Two loudspeakers are placed on a wall 2 m apart. A listener stands directly in front of one of the speakers, 81.7 m from the wall. The speakers are being driven by the same electric signal generated by a harmonic oscillator of frequency 4450 Hz. What is the phase difference ∆Φ between the two waves (generated by each speaker) when they reach the listener? The speed of the sound in air is 343 m/s. Answer in units of rad.
Answer:
The phase difference is [tex]\Delta \phi = 1.9995 rad[/tex]
Explanation:
From the question we are told that
The distance between the loudspeakers is [tex]d = 2m[/tex]
The distance of the listener from the wall [tex]D = 81.7 \ m[/tex]
The frequency of the loudspeakers is [tex]f = 4450Hz[/tex]
The velocity of sound is [tex]v_s = 343 m/s[/tex]
The path difference of the sound wave that is getting to the listener is mathematically represented as
[tex]\Delta z =\sqrt{d^2 + D^2} -D[/tex]
Substituting values
[tex]\Delta z =\sqrt{2^2 + 81.7^2 } -81.7[/tex]
[tex]\Delta z =0.0245m[/tex]
The phase difference is mathematically represented as
[tex]\Delta \phi[/tex] = [tex]\frac{2 \pi}{\lambda } * \Delta z[/tex]
Where [tex]\lambda[/tex] is the wavelength which is mathematically represented as
[tex]\lambda = \frac{v_s }{f}[/tex]
substituting value
[tex]\lambda = \frac{343 }{4450}[/tex]
[tex]\lambda = 0.0770 m[/tex]
Substituting value into the equation for phase difference
[tex]\Delta \phi[/tex] = [tex]\frac{2 * 3.142 * 0.0245}{0.0770}[/tex]
[tex]\Delta \phi = 1.9995 rad[/tex]
A hydrogen atom can be in the 1S state, whose energy we'll call 0, the 2S state, or any of 3 2P states. The 2S and 2P states have energies of 10.2 eV. There are other states with higher energy but we'll ignore them for simplicity. The 2P states have distinctive optical properties, so we're interested in how many are present even when it's a small fraction of the total. 1) What fraction of the H is in 2P states at T=5900 K, a typical Sun surface temperature?
Final answer:
The fraction of hydrogen atoms in the 2P states at T=5900 K is calculated using the Boltzmann distribution. The energy of the 2S and 2P states is 10.2 eV, and their degeneracies are factored into the Boltzmann factor to determine the relative populations of these states.
Explanation:
The fraction of hydrogen atoms in the 2P states at a temperature of 5900 K can be calculated using the Boltzmann distribution. According to quantum mechanics, the energy levels of a hydrogen atom involve principal quantum numbers, with the ground state being 1s and higher energy excited states being designated by higher quantum numbers and corresponding letters (s, p, d, f, etc.) for their angular momentum quantum numbers. Since we are given that the energy of 2S and 2P states are the same at 10.2 eV and the ground state (1S) has an energy we'll call 0, we can use the Boltzmann factor to find the relative populations of these states.
To calculate the fraction of hydrogen atoms in the 2P state, use the following Boltzmann factor equation: fraction = (g2P / gTotal) * exp(-E2/kT), where g2P is the degeneracy of the 2P state, gTotal is the total degeneracy of all states considered, E2 is the energy of the 2P state, k is the Boltzmann constant, and T is the temperature. The degeneracy of the 2P state is 3 (since there are three 2P states) and the only other state considered here is the 2S state, which has degeneracy 1, making gTotal = 4. Plugging in the energy of 10.2 eV for E2 and converting it to joules (multiply by 1.602 x 10-19 J/eV), using k = 1.38 x 10-23 J/K, and T = 5900 K, we can calculate the fraction of hydrogen in the 2P state.
A circuit connected to a 12V battery has three 50 Ohm resistors with two connected in series, and parallel to the third. Calculate the total resistance and current.
Answer:
Total resistance=33.3 ohms
Current=0.36 amperes
Explanation:
Voltage=12v
R1=50 ohms
R2=50 ohms
R3=50 ohms
Two connected in series:
R1+R2=50+50=100 ohms
In parallel
1/R=1/50 + 1/100
1/R=(2+1)/100
1/R=3/100
Cross multiplying we get
100=3R
Divide both sides by3
100/3=3R/3
33.3=R
Total resistance=33.3 ohms
Current=voltage ➗ resistance
Current=12 ➗ 33.3
Current=0.36 amperes
A mother is helping her children, of unequal weight, to balance on a seesaw so that they will be able to make it tilt back and forth without the heavier child simply sinking to the ground. Given that her heavier child of weight WWW is sitting a distance LLL to the left of the pivot, at what distance L1L1L_1 must she place her second child of weight www on the right side of the pivot to balance the seesaw? Express your answer in terms of LLL, WWW, and www.
Answer:
L1L1L_1 = ( LLL × WWW)/ www.
Explanation:
From the question we are given that the children's weight are not equal, thus, there will be a need to balance the seesaw "so that they will be able to make it tilt back and forth without the heavier child simply sinking to the ground''.
Also, From the question, we are gven that the weight of the heavier child = WWW, the heavier child's is sitting a distance = LLL, the weight of the second child= www.
Therefore, in order to balance the seesaw, she will need to place her second child of weight www on the right side of the pivot to balance the seesaw at a distance of;
L1L1L_1 = ( LLL × WWW)/ www.
Calculate the induced electric field (in V/m) in a 52-turn coil with a diameter of 12 cm that is placed in a spatially uniform magnetic field of magnitude 0.30 T so that the face of the coil and the magnetic field are perpendicular. This magnetic field is reduced to zero in 0.10 seconds. Assume that the magnetic field is cylindrically symmetric with respect to the central axis of the coil. (Enter the magnitude.)
Given Information:
Number of turns = N = 52
Diameter of coil = d = 12 cm = 0.12 m
Time = t = 0.10 seconds
Magnetic field = B = 0.30 T
Required Information:
Induced electric field = E = ?
Answer:
Induced electric field = E = 4.68 V/m
Explanation:
The Maxwell's third equation can be used to find out the induced electric field,
∫E.dl = -dΦ/dt
Where E is the induced electric field, dl is the circumference of the loop and dΦ/dt is the rate of change of magnetic flux and is given by
Φ = NABcos(θ)
Where N is the number of turns, A is the area of coil and B is the magnetic field and cos(θ) = 1
Φ = NAB
∫E.dl = -dΦ/dt
E(2πr) = -d(NAB)/dt
E =1/(2πr)*-d(NAB)/dt
E =NA/(2πr)*-dB/dt
Area is given by
A = πr²
E =Nπr²/(2πr)*-dB/dt
E =Nr/2*-dB/dt
The magnetic field reduce from 0.30 to zero in 0.10 seconds
E =Nr/2*-(0.30 - 0)/(0 - 0.10)
E =Nr/2*-(0.30)/(-0.10)
E = Nr/2*-(-3)
The radius r is given by
r = d/2 = 0.12/2 = 0.06 m
E = (52*0.06)/2*(3)
E = 1.56*3
E = 1.56*3
E = 4.68 V/m
Therefore, the induced electric field in the coil is 4.68 V/m
Compared to the current in the 10.-ohm resistance in the circuit shown at the right, the current in the 5.0-ohm resistance is
a. one-half as great
b. one-fourth as great
c. the same
d. twice as great
Answer:
The answer is d. Twice as great.
Explanation:
The German physicist and mathematician Georg Simon Ohm says in his basic law of electrical circuits or Ohm's law that the potential difference V that is applied at the ends of a conductor is proportional to the intensity I of the current that circulates and that the electrical resistance R is the ratio factor between I and V.
The equation would be I = V / R
If by any chance R is reduced because it is inversely proportional, the current would increase.
So if the current goes from 10 to half 5, its current would double.