Answer:
8.29 * 10^{13} N
Explanation:
Weigth is the force of gravity a body of mass experiences due to the gravity of another mass.
Force=mass*acceleration (to calculate the mass of the object);
675=m*9.8; Mass = 68.88 kg
Force= [tex]\frac{GMm}{r^{2} }[/tex]; M is the mass of the bigger object, m is the mass of the smaller object, r is the seperation in between (radius in this case) and G is the Gravitational constant.
Mass of the sun = 1.989 × 10^30 kg
G = 6.67 × 10-11 m3 kg-1 s-2
mass of the object = 68.88 kg;
Applying the formula:
Force=[tex]\frac{(6.67 * 10^{-11} ) * (68.88) * (1.989 * 10^{30} )}{(((21 *10^{3} ))/2)^{2} }[/tex]
Force = 8.29 * 10^{13} N
The weight on the surface of a neutron star that has the same mass as our Sun and a diameter of 21.0 km will be 8.29 × 10¹³ N
Gravitational force:Given that the weight of the person on the earth is 675N.
mg = 675N
where m is the mass of the person.
m = 675/g
m = 675/9.8
m = 68.87 kg
The gravitational force on the surface of a celestial body is given by:
F = GMm/R²
where G is the gravitational constant
M is the mass of the body
m is the mass of the person
and, R is the radius of the body
F = (6.67 × 10⁻¹¹)( 2 × 10³⁰)(68.87) / (21/2 × 10³)²
F = 8.29 × 10¹³ N
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A positively charged insulating rod is brought close to an object that is suspended by a string. If the object is repelled away from the rod we can conclude: A) the object is positively charged B) the object is negatively charged C) the object is an insulator D) the object is a conductor E) none of the above Two small charged objects repel each other with a force F when separated by a distance d. If the charge on each object is reduced to one-fourth of its original value and the distance between them is reduced to d/2 the force becomes: A) F/16 B) F/8 C) F/4 D) F/2 E) F
Answer
A) Positively charged two insulating rod are brought closed to an object they repel each other. It means the Object is positively charged. Because similar charge repel each other.
The correct answer is Option A.
B) we know force between to charges is calculated using Formula
[tex]F = \dfrac{kQ_1Q_2}{r^2}[/tex].......(1)
form the given condition in the question
[tex]F' = \dfrac{k\dfrac{Q_1}{4}\dfrac{Q_2}{4}}{(\dfrac{r}{2})^2}[/tex]
[tex]F' = \dfrac{k\dfrac{Q_1}{4}\dfrac{Q_2}{4}}{(\dfrac{r^2}{4})}[/tex]
[tex]F' = \dfrac{4}{16}\dfrac{kQ_1Q_2}{r^2}[/tex]
from equation (1)
[tex]F' = \dfrac{F}{4}[/tex]
hence, the correct answer is Option C.
The object is positively charged if repelled by a positively charged rod. The force between two charged objects reduces to F/16 when both charges are one-fourth and the distance is halved.
Explanation:Charged Objects and Electrostatic Forces
If the object suspended by a string is repelled away from a positively charged insulating rod, we can conclude that the object is positively charged (A). This is because like charges repel each other, according to electrostatic principles.
Regarding the force between two charged objects, Coulomb's law states that the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. If the charge on each object is reduced to one-fourth of its original value and the distance between them is reduced to d/2, the new force is calculated as:
New charge = Original charge / 4
New distance = d / 2
New force = (1/4 × 1/4) / (1/2 × 1/2)^2 = 1/16 / 1/4 = F/16 (A)
A 6.75 nC charge is located 1.99 m from a 4.46 nC point charge.
(a) Find the magnitude of the electrostatic force that one charge exerts on the other.
(b) Is the force attractive or repulsive?
Explanation:
Given that,
Charge 1, [tex]q_1=6.75\ nC=6.75 \times 10^{-9}\ C[/tex]
Charge 2, [tex]q_2=4.46\ nC=4.46\times 10^{-9}\ C[/tex]
The distance between charges, r = 1.99 m
To find,
The electrostatic force and its nature
Solution,
(a) The electric force between two charges is given by :
[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]
[tex]F=\dfrac{9\times 10^9\times 6.75\times 10^{-9}\times 4.46\times 10^{-9}}{(1.99)^2}[/tex]
[tex]F=6.84\times 10^{-8}\ N[/tex]
(b) As the magnitude of both charges is positive, then the force between charges will be repulsive.
Therefore, this is the required solution.
(a) The magnitude of the electrostatic force that one charge exerts on the other is 6.83×10⁻⁸ N.
(b) The force is repulsive.
(a) To calculate the magnitude of the electrostatic force that one charge exerts on the other, we use the formula below.
Formula:
F = kqQ/r².................. Equation 1Where:
F = Force of between the chargesq = First chargeQ = second charger = distance between the chargesk = coulomb's constant.From the question,
Given:
q = 6.75 nC = 6.75×10⁻⁹ CQ = 4.46 nC = 4.46×10⁻⁹ Cr = 1.99 mk = 8.98×10⁶ Nm²/C²Substitute these values into equation 1
F = (6.75×10⁻⁹)(4.46×10⁻⁹)(8.98×10⁹)/1.99²F = 6.83×10⁻⁸ N(b) The force is repulsive because both charges a the same (positive).
Hence, (a) The magnitude of the electrostatic force that one charge exerts on the other is 6.83×10⁻⁸ N (b) The force is repulsive.
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The cheetah is one of the fastest-accelerating animals, because it can go from rest to 16.2 m/s (about 36 mi/h) in 2.4 s. If its mass is 102 kg, determine the average power developed by the cheetah during the acceleration phase of its motion. Express your answer in the following units.
Explanation:
Power is the ratio of energy to time.
Here we need to consider kinetic energy,
Mass, m = 102 kg
Initial velocity = 0 m/s
Final velocity = 16.2 m/s
Time, t = 2.4 s
Initial kinetic energy = 0.5 x Mass x Initial velocity² = 0.5 x 102 x 0² = 0 J
Final kinetic energy = 0.5 x Mass x Final velocity² = 0.5 x 102 x 16.2² = 13384.44 J
Change in energy = Final kinetic energy - Initial kinetic energy
Change in energy = 13384.44 - 0
Change in energy = 13384.44 J
Power = 13384.44 ÷ 2.4 = 5576.85 W = 7.48 hp
Power of cheetah is 5576.85 W = 7.48 hp
The acceleration of a particle is given by a = -kt^2, where a is in meters per second squared and the time t is in seconds. If the initial velocity of the particle at t=0 is__________.
Answer:
Velocity is zero.
Explanation:
Given:
acceleration of particle, [tex]a=-k.t^2\ m.s^{-2}[/tex]time of observation, [tex]t=0\ s[/tex]We know that acceleration is defined as the rate of change in velocity and hence we integrate the expression of acceleration.
Now the velocity can be given as:
[tex]v=\int\limits {a} .\, dt=-k.\frac{t^3}{3}[/tex]
Put t = 0 we get:
[tex]v=0\ m.s^{-1}[/tex]
As a cat pounces on a mouse, its muscles consume 10 units of potential energy (which the cat previously gained from the food it consumed). However, the pounce itself only required 4 units of kinetic energy. How many units of energy were dissipated as heat?
The cat used 10 units of potential energy to pounce but only 4 units were converted meaningfully into kinetic energy of the pounce. The remaining 6 units were dissipated as heat.
Explanation:When the cat pounces on a mouse, the energy needed for this action is less than the total potential energy consumed. The cat's body uses 10 units of potential energy, but the pounce only required 4 units of kinetic energy. The difference between these two values signifies the amount of energy dissipated as heat. Therefore, we can perform the subtraction: 10 units (total potential energy) - 4 units (kinetic energy used) = 6 units. Thus, "6 units of energy were dissipated as heat."
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Due to the wave nature of light, light shined on a single slit will produce a diffraction pattern? Green light (520 nm) is shined on a slit with width 0.440 mm.(a) Find the width of the central maximum located 1.65 m from the slit.(b) What is the width of the first order fringe?
Answer:
Yes, it will produce a diffraction pattern.
a. 3.9 mm b. 1.95 mm
Explanation:
The light shined from a single slit will produce a diffraction pattern because, the wavefront act as wavelets which generates its own wave according to Huygens principle. This therefore causes the diffraction pattern.
Given
wavelength of green light, λ = 520 nm = 520 × 10⁻⁹ m = 5.20 × 10⁻⁷ m
width of slit, d = 0.440 mm = 0.44 × 10⁻³ m = 4.4 × 10⁻⁴ m
Distance of slit from central maximum , D = 1.65 m
Distance of first minimum from central maximum, y = ?
a. The relationship between the slit width and wavelength is given by [tex} dsinθ = mλ [/tex]where d = slit width, θ = angular distance from central maximum, λ = wavelength of light and m = ±1, ±2, ±3...
The relationship between y and D is given by [tex] tanθ = y/D [/tex]
Since θ is small, sinθ ≈ θ ≈ tanθ
so, dθ = mλ ⇒ θ = mλ/d = y/D
Therefore, y = mλD/d
Now, for the first minimum above the slit, m = +1 and for the first minimum below the slit, m = -1. So, y₁ = λD/d and y₋₁ = -λD/d. So, the width of the central maximum Δy is the difference between the first minima below and above the central maximum. So, Δy = y₁ - y₋₁ = λD/d -(-λD/d) = 2λD/d
Substituting the values from above, Δy= 2 × 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴ = 3900 × 10⁻⁶ m = 3.9 × 10⁻³ m = 3.9 mm
b. The first order fringe is the fringe located between the first minimum and the second minimum. From dsinθ = mλ and tanθ = y/D when θ is small, sinθ ≈ θ ≈ tanθ. So, y = mλD/d. Let m= 1 and m=2 be the first and second minima respectively. So,y₁ = λD/d and y₂ = 2λD/d. The difference Δy₁ = y₂ - y₁ is the width of the first order fringe. Therefore, Δy₁ = 2λD/d - λD/d= λD/d. Substituting the values from above, we have
λD/d= 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴= 1.95 × 10⁻³ m = 1.95 mm
5.5 g of metal is placed in a graduated cylinder containing water. The level of water increases from 12.0 ml to 13.5 ml. The density of the metal is _____ g/ml. (Use the correct number of significant figures in your answer)
Answer:
0.407
Explanation:
5.5÷13.5
i believe it is this as
density =mass÷volume
A transformer has a primary voltage of 115 V and a secondary voltage of 24 V. If the number of turns in the primary is 345, how many turns are in the secondary? A. 8 B. 690 C. 72 D. 1,653
Answer:
C. 72
Explanation:
Transformer: A transformer is an electromagnetic device that uses the property of mutual inductance to change the voltage of alternating supply.
In a ideal transformer,
Vs/Vp = Ns/Np ............................................. Equation 1
Where Vp = primary voltage, Vs = secondary voltage, Ns = Secondary turn, Np = primary turn.
Making Ns the subject of the equation,
Ns =(Vs/Vp)Np .......................................... Equation 2
Given: Vs = 24 V, Vp = 115 V, Np = 345.
Substitute into equation 2
Ns = (24/115)345
Ns = 72 turns.
Thus the number of turns in the secondary = 72 turns.
The right option is C. 72
The average lifetime of a pi meson in its own frame of reference (i.e., the proper lifetime) is 2.6 10-8 s. (a) If the meson moves with a speed of 0.88c, what is its mean lifetime as measured by an observer on Earth? (b) What is the average distance it travels before decaying, as measured by an observer on Earth? (c) What distance would it travel if time dilation did not occur?
Final answer:
a) The measured lifetime of the pi meson as observed by an Earth observer is longer than its proper lifetime due to time dilation. b) The average distance the meson travels before decaying can be calculated using the velocity and measured lifetime. c) If time dilation did not occur, the distance the meson would travel can be calculated using the velocity and proper lifetime.
Explanation:
a) According to time dilation, the average lifetime of the pi meson as measured by an observer on Earth is longer than its proper lifetime. To calculate it, we use the formula for time dilation: t' = t / γ, where t' is the measured lifetime, t is the proper lifetime, and γ is the Lorentz factor. Given that the proper lifetime is 2.6 × 10-8 s and the speed of the meson is 0.88c, we can calculate γ using the formula γ = 1 / √(1 - v2/c2). After calculating γ, we can substitute it into the time dilation formula to find the measured lifetime on Earth.
b) To calculate the average distance the meson travels before decaying, we use the formula d = v · t', where d is the distance, v is the velocity, and t' is the measured lifetime. We can substitute the values given in part a) to find the distance traveled by the meson.
c) If time dilation did not occur, the distance the meson would travel can be calculated using the formula d = v · t, where d is the distance, v is the velocity, and t is the proper lifetime. By substituting the values given in part a) into this formula, we can find the distance the meson would travel in the absence of time dilation.
What is the relationship between id and ic?
What is the relationship between and ?
The currents are not equal in magnitude, the algebraic signs of the current values are opposite.
The currents are equal in magnitude, the algebraic signs of the current values are the same.
The currents are equal in magnitude, the algebraic signs of the current values are opposite.
The currents are not equal in magnitude, the algebraic signs of the current values are the same.
Answer:
The question has a diagram attached to it which I have done in the explanation.
The answer = The currents are equal in magnitude, the algebraic signs of the current values are opposite.
Explanation:
What is applied here is the Kirchoff's junction role or kirchoff's current law which states that the algebraic sum of the current entering any junction must be equal to the algebraic sum of the current leaving the junction. this is what is applied in the diagram.
The attachment shows the explanation
Final answer:
The relationship between id and ic currents depends on their definitions, such as in capacitors within AC circuits where ic leads the voltage by 90 degrees, or in specific phenomena involving image currents that are equal in magnitude but opposite in sign.
Explanation:
The relationship between the currents id and ic depends on the specific context they are used in, but often these terms relate to currents in electronic components, such as diodes (id) and capacitors (ic). For instance, in a capacitive AC (alternating current) circuit, the current through the capacitor (ic) leads the voltage across the capacitor (vc) by 90 degrees in phase. This relationship is represented by ic(t) = C dv/dt, where C is the capacitance and dv/dt is the rate of change of voltage.
This means at any instant, the magnitude of the current is tied to the rate at which the voltage changes. In circuits involving superconductivity or other specific phenomena, such as image currents, you might encounter situations where an image current I' is equal in magnitude but opposite in sign to the driving current I, as indicated by I' = - I.
If a proton and an electron are released when they are 6.50×10⁻¹⁰ m apart (typical atomic distances), find the initial acceleration of each of them.
a) Acceleration of electron
b) Acceleration of proton
Answer:
(a) Acceleration of electron= 5.993×10²⁰ m/s²
(b) Acceleration of proton= 3.264×10¹⁷ m/s²
Explanation:
Given Data
distance r= 6.50×10⁻¹⁰ m
Mass of electron Me=9.109×10⁻³¹ kg
Mass of proton Mp=1.673×10⁻²⁷ kg
Charge of electron qe= -e = -1.602×10⁻¹⁹C
Charge of electron qp= e = 1.602×10⁻¹⁹C
To find
(a) Acceleration of electron
(b) Acceleration of proton
Solution
Since the charges are opposite the Coulomb Force is attractive
So
[tex]F=\frac{1}{4(\pi)Eo }\frac{|qp*qe|}{r^{2} }\\ F=(8.988*10^{9}Nm^{2}/C^{2})*\frac{(1.602*10^{-19})^{2} }{(6.50*10^{-10} )^{2} } \\F=5.46*10^{-10}N[/tex]
From Newtons Second Law of motion
F=ma
a=F/m
For (a) Acceleration of electron
[tex]a=F/Me\\a=(5.46*10^{-10} )/9.109*10^{-31}\\ a=5.993*10^{20}m/s^{2}[/tex]
For(b) Acceleration of proton
[tex]a=F/Mp\\a=(5.46*10^{-10} )/1.673*10^{-27} \\a=3.264*10^{17}m/s^{2}[/tex]
Acceleration of the body is rate of change of the increasing velocity with respect to the time. is The acceleration of the electron is [tex]5.992\times10^{20}[/tex] meter per second squared and acceleration of the proton is [tex]3.265\times10^{17}[/tex] meter per second squared.
Given information-A proton and an electron are released when they are 6.50×10⁻¹⁰ m apart.
The the distance of proton and electron is 6.50×10⁻¹⁰ m.
Now it is known that,
Mass of the electron,[tex]m_e=9.109\times10^{-31}[/tex]
Mass of the proton,[tex]m_p=1.673\times10^{-27}[/tex]
Charge of the electron,[tex]Q_e=-1.602\times10^{-19}[/tex]
Charge of the proton,[tex]Q_e=1.602\times10^{-19}[/tex]
AccelerationAcceleration of the body is rate of change of the increasing velocity with respect to the time. Acceleration of a body is the ratio of the force to the mass of the body.
The coulomb force between the two charges can be given as,
[tex]F=k\dfrac{q_1q_2}{r^2} [/tex]
Here k is the coulombs constant, r is the distance and q is the charge of proton and electron.Put the values,
[tex]F=9\times10^9\times\dfrac{(1.602\times10^{-19})^2}{(6.5\times10^{-10})^2} [/tex]
[tex]F=5.46\times10^{-10}[/tex]
Now it is known that the acceleration of a body is the ratio of the force to the mass of the body.
Acceleration of electron[tex]a_e=\dfrac{F}{m_e} [/tex]
[tex]a_e=\dfrac{5.46\times10^{-10}}{9.109\times10^{-31}} [/tex]
[tex]a_e=5.992\times10^{20}[/tex]
Acceleration of Proton[tex]a_p=\dfrac{F}{m_p} [/tex]
[tex]a_p=\dfrac{5.46\times10^{-10}}{1.673\times10^{-27}} [/tex]
[tex]a_e=3.625\times10^{17}[/tex]
Thus the acceleration of the electron is [tex]5.992\times10^{20}[/tex] meter per second squared and acceleration of the proton is [tex]3.265\times10^{17}[/tex] meter per second squared.
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A 0.12 kg body undergoes simple harmonic motion of amplitude 8.5 cm and period 0.20 s. (a) What is themagnitude of the maximum force acting on it? (b) If the oscillations are produced by a spring, what is the springconstant?
Answer:
a)F=698.83 N
b)K=8221.56 N/m
Explanation:
Given that
mass ,m = 0.12 kg
Amplitude ,A= 8.5 cm
time period ,T = 0.2 s
We know that
[tex]T=\dfrac{2\pi}{\omega}[/tex]
[tex]{\omega}=\dfrac{2\pi }{0.2}\ rad/s[/tex]
[tex]{\omega}=31.41\ rad/s[/tex]
We know that
[tex]{\omega}^2=m\ K[/tex]
K=Spring constant
[tex]K=\dfrac{\omega^2}{m}[/tex]
[tex]K=\dfrac{31.41^2}{0.12}\ N/m[/tex]
K=8221.56 N/m
The maximum force F
F= K A
F= 8221.56 x 0.085 N
F=698.83 N
a)F=698.83 N
b)K=8221.56 N/m
Final answer:
To find the magnitude of the maximum force acting on the body in simple harmonic motion, we use the equation F = -kx. To find the spring constant, we can rearrange the equation F = -kx.
Explanation:
To find the magnitude of the maximum force acting on the body in simple harmonic motion, we can use the equation F = -kx, where F is the force, k is the spring constant, and x is the displacement from the equilibrium position. In this case, we are given the amplitude A = 8.5 cm and the mass m = 0.12 kg. The displacement x at the maximum amplitude is equal to the amplitude, so x = A.
Plugging in the values, we get F = -(k)(A).
(a) To find the magnitude of the maximum force, we need to find the spring constant k. We can use the equation T = 2π√(m/k), where T is the period. Plugging in the values, T = 0.20 s and m = 0.12 kg, we can solve for k.
(b) To find the spring constant, we can rearrange the equation F = -kx to solve for k. Plugging in the values, F = 2.00 mg and x = A, we can solve for k.
A 106 kg clock initially at rest on a horizontal floor requires a 670 N horizontal force to set it in motion. After the clock is in motion, a horizontal force of 557 N keeps it moving with a constant velocity.
a) what is the μs between the clock and the floor?.70
b) what is the μk between the clock and the floor?
Answer:
a) [tex]\mu_s=0.65[/tex]
b) [tex]\mu_k=0.54[/tex]
Explanation:
Static friction is when the body is at rest or is about to move while kinetic friction is when the body is already in motion. According to Newton's second law:
[tex]\sum F_y:N=mg\\\sum F_x:F_f=F_x[/tex]
a) In this case, the static friction must be equal to the horizontal force to set the clock in motion:
[tex]F_f=\mu_sN=\mu_smg\\\mu_smg=F_x\\\mu_s=\frac{F_x}{mg}\\\mu_s=\frac{670}{106kg(9.8\frac{m}{s^2})}\\\mu_s=0.65[/tex]
b) In this case, the kinetic friction is equal to the horizontal force that keep the clock moving with constant velocity:
[tex]\mu_kmg=F_x'\\\mu_k=\frac{F_x'}{mg}\\\mu_k=\frac{557}{106kg(9.8\frac{m}{s^2})}\\\mu_k=0.54[/tex]
A truck driver has a shipment of apples to deliver to a destination 550 miles away. The trip usually takes him 10.0 hours. Today he finds himself daydreaming and realizes 120 miles into his trip that that he is running 30.0 minutes later than his usual pace at this point. If the driver still wishes to complete the trip in 10.0 hours, how fast must he drive for the rest of the trip? (In all questions, you may assume that the truck moves with a constant speed.)
At what speed must he drive for the remainder of the trip to complete the trip in the usual amount of time? Express your answer using three significant figures.
At a speed of 58.8mi/hr must he drive for the remainder of the trip to complete the trip in the usual amount of time.
What is velocity?The change of displacement with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.
The time it takes to go 120 miles at that speed in normal conditions is;
[tex]\rm t = \frac{d}{v} \\\\ t =\frac{120}{55} \\\\ t= 2 hour \ 11 minute[/tex]
However, the driver was 30 minutes behind schedule at this distance, so the time the has to spend is;
T== 2hr 11 min +30 min
T= 2hr 41 mins
He has to fulfill the initial planned moment of 10 hours. The time to cover the remaining distance;
T'=(10-2.41)
T' = 7hr19mins
The remaining distance will be ;
S={550-120}
S=430mile.
The speed is to be maintained the following distance on the time;
[tex]\rm V'=\frac{430}{(\frac{439}{60} )} \\\\V'=58.7699 \ miles / hour[/tex]
The speed must he drive for the remainder of the trip to complete the trip in the usual amount of time will be 58.5 mile/hr.
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To complete his remaining 430 miles in 8 hours, the truck driver needs to drive at a speed of 53.750 mph for the rest of the trip.
Explanation:The driver has covered 120 miles during which he used up 30.0 minutes more than what he would usually take.
We can calculate the remaining distance he needs to cover which is 550 total miles minus the 120 miles he's already driven, which equals to 430 miles.
The remaining time he has is the total normal trip time of 10.0 hours minus the 2.0 hours he's already spent (previous 1.5 hours plus the extra 0.5 hours), which equals 8 hours.
To cover the remaining 430 miles in the 8 hours left, he must drive at a speed of 430 miles divided by 8 hours, which equals to 53.750 mph.
This is the speed the driver needs to maintain for the remainder of the trip to complete it in the usual amount of time.
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A frog is at the bottom of a 17-foot well. Each time the frog leaps, it moves up 3 feet. If the frog has not reached the top of the well, then the frog slides back 1 foot before it is ready to make another leap. How many leaps will the frog need to escape the well?
Answer:
The frog takes 8 jumps to reach top of well
Explanation:
Given data
Frog at bottom=17 foot
Each time frog leaps 3 feet
Frog has not reached the top of the well, then the frog slides back 1 foot
To Find
Total number of leaps the frog needed to escape from well
Solution
in 1 jump distance jumped=3+(-1)
=2 feet
=2×1 feet
The "-1" is because the frog goes back
Now After 2 jumps the distance jumped as:
Distance Jumped=2+2
Distance Jumped=2*2
=4 feet
Similarly after 7 jumps
Distance Jumped=2+2+......+2
Distance Jumped=2*7
=14 feet
Now after 8th jump the frog climbs but doesnot slide back as it is reached to the top of well.
So
Distance Jumped=(Distance Jumped after 7 jumps)+3
=14+3
=17 feet
The frog takes 8 jumps to reach top of well
The frog needs 8 leaps each of 2 feet and a final leap of 3 feet to escape a 17-foot well. Therefore, it takes the frog a total of 9 leaps to escape the well.
Explanation:In this mathematical problem, we need to determine how many leaps it takes for a frog to escape a 17-foot well, given that each leap propels the frog 3 feet up, but it slides back 1 foot before the next leap. Given thefa, each leap results net gain of 2 feet (3 feet up minus 1 foot slide back). However, for the last leap, the frog won't slide back, so the final leap has a net gain of 3 feet.
Therefore, the frog need to leap 15 feet using jumps with a net gain of 2 feet and then make a final leap of 3 feet out of the well. Each of the first leaps covers 2 feet, so the number of such leaps needed is 15/2 = 7.5. Since it's not possible to make half a jump, we round up to 8 jumps. Then, the frog makes its final jump of 3 feet. So in total, the frog needs 9 leaps to escape the well.
Learn more about Frog Leap Problem#SPJ3A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.820 m. What impulse was given to the ball by the floor?
Answer:
1.0752 kgm/s
Explanation:
Considering when the drop was dropped from rest from a height,
mass of the ball, m = 0.120 kg
height, h = - 1.25 m
the initial velocity, u = 0 m/s
the acceleration due to gravity, g = - 9.8 m/s²
From equation of motion
[tex]V^{2} = U^{2} + 2gh[/tex]
Substituting the values,
[tex]V^{2} = 0^{2} + 2(-9.8 m/s^{2})(-1.25 m)[/tex]
[tex]V^{2} = 24.5 m/s[/tex]
[tex]V = \sqrt{24.5} \ m/s[/tex]
[tex]V = 4.95 \ m/s[/tex]
V = ± 4.95 m/s
V = - 4.95 m/s
Since the ball is moving downward, the final velocity of the ball when it hits the floor is V = - 4.95 m/s
Considering when the ball rebounds from the floor,
assume the mass of the ball still remain, m = 0.120 kg
height, h = 0.820 m
the final velocity, v = 0 m/s
the acceleration due to gravity, g = - 9.8 m/s²
From equation of motion
[tex]V^{2} = U^{2} + 2gh[/tex]
Substituting the values,
[tex]0^{2} = U^{2} + 2(-9.8 m/s^{2})(0.820 m)[/tex]
[tex]0 = U^{2} - 16.072 m/s[/tex]
[tex]U^{2} = 16.072 m/s[/tex]
[tex]U = \sqrt{16.072} \ m/s[/tex]
U = ± 4.01 m/s
U = + 4.01 m/s
Since the ball is moving upward, the initial velocity of the ball from the bounce from the floor is U = + 4.01 m/s
From Newton's second law of motion, applied force is directly proportional to the rate of change in momentum.
[tex]F = \frac{mv - mu}{t}[/tex]
[tex]F.t = m(v - u)[/tex]
⇒ Impulse = Change in momentum
To calculate the impulse, the moment before the ball hits the ground will be the initial momentum while the moment the ball rebounces will be the final velocity,
∴ F.t = 0.120 kg(4.01 m/s - (-4.95 m/s) )
F.t = 0.120 kg(4.01 m/s + 4.95 m/s) )
F.t = 0.120 kg × 8.96 m/s
Impulse = 1.0752 kgm/s
The impulse given to the ball by the floor is 1.0752 kgm/s
The force of gravity between two small masses A and B when placed very near each other is 3.24x10-7 N. What will the force between these objects be if both of their masses are doubled and the distance between them is tripled?
Answer:
force between masses will be same as [tex]3.24\times 10^{-7}N[/tex]
Explanation:
Let the masses are [tex]m_A\ and\ m_B[/tex] and distance between them is r
According to gravitational law force between two mass is given by
[tex]F=\frac{Gm_Am_B}{r^2}[/tex]
So [tex]3.24\times 10^{-7}=\frac{Gm_Am_B}{r^2}[/tex]
Now mass of both are doubled and distance between them is tripled
And now we have to find that from what factor the force between masses are changed
So [tex]F_{new}=\frac{G3m_A3m_B}{(3r)^2}=\frac{9Gm_Am_B}{9r^2}=\frac{Gm_Am_B}{r^2}=F[/tex]
So force between masses will be same as [tex]3.24\times 10^{-7}N[/tex] as masses and distance between them both has the same effect
What happens when polar molecules are between oppositely charged metal plates
Answer:
They will become aligned according to the charges on the metal plate.
Explanation:
When Polar molecules are placed between oppositely charged metal plates the molecules will tend to be attracted by their corresponding oppositely charged plates that is the positive and negative plates,
A polar molecule is one which has opposite charges on its ends. Non-polar molecules however do not have charges on their end
When polar molecules are placed between oppositely charged metal plates, the poles orient towards the oppositely charged plate.
Polar molecules:These are the molecules that have positive and negative in opposite poles. For example- water molecules.
When polar molecules are placed between oppositely charged metal plates, the negative pole orient towards the positive end and vise versa. They will become aligned according to the charges on the metal plate.Because opposite charges attract each other while similar charges repulse.Therefore, when polar molecules are placed between oppositely charged metal plates, the poles orient towards the oppositely charged plate.
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A speed skater moving across frictionless ice at 8.30 m/s hits a 5.10-m-wide patch of rough ice. She slows steadily, then continues on at 5.20 m/s ?
Incomplete question.The complete question is here
A speed skater moving across frictionless ice at 8.30 m/s hits a 5.10m wide patch of rough ice. She slows steadily, then continues on at 5.20 m/s.What is her acceleration on the rough ice?
Answer:
acceleration = -4.103 m/s²
Explanation:
Given data
Initial velocity Vi=8.30 m/s
Final velocity Vf=5.20 m/s
Initial distance xi=0 m.......(We choose xi=0 the start point of acceleration motion )
Final distance xf=5.10 m
To find
Acceleration
Solution
From the kinetic equation
[tex](v_{f})^{2}=(v_{i} )^{2}+2a(x_{f} -x_{i} )\\ (5.20 m/s)^{2}=(8.30m/s)^{2}+2a(5.10m-0m)\\a=\frac{(5.20m/s)^{2}-(8.30m/s)^{2} }{2*(5.10m)}\\a=-4.103 m/s^{2}[/tex]
The number of protons in a neutral atom equals the number of
Answer: Electrons
Explanation: For a neutral atom the number of positively charged protons inside the nucleus must be equal to the number of electrons in the orbital shells. Both charges will cancel out having a charge of 0 which makes an atom electrically neutral.
Use a(t) = -32 feet per second per second as the acceleration due to gravity. (Neglect air resistance.) -With what initial velocity must an object be thrown upward (from ground level) to reach the top of the Washington Monument (approximately 550 feet)?
Answer:
u= 187.61 ft/s
Explanation:
Given that
g= - 32 ft/s²
The maximum height ,h= 550 ft
Lets take the initial velocity = u ft/s
We know that
v²=u² + 2 g s
v=final speed ,u=initial speed ,s=height
When the object reach at the maximum height then the final speed of the object will become zero.
That is why
u²= 2 x 32 x 550
u²= 35200
u= 187.61 ft/s
That is why the initial speed will be 187.61 ft/s
Answer:
Explanation:
a = - 32 ft/s²
h = 550 ft
Let the initial velocity is u. The velocity at maximum height is zero.
use third equation of motion
v² = u² + 2 a h
0 = u² - 2 x 32 x 550
u² = 35200
u = 187.62 ft/s
If a ball end mill cutter's stepover is kept at a fixed value but the diameter of the tool is increased, the scallop height increases as a consequence.Select one:a. Trueb. False
Answer: FALSE
Explanation:Ball End Mills are hemispherical tip used to cut and and shape/ milling rounded objects, such as the metal bearing grooves, slotting and pocketing.
Ball End Mill also called Ball nose end mills, the scallop height will not increase as a result of the increase in diameter. The diameter is the distance between one end of the round tool to another when taken from the middle.
A drawing of electric field lines will immediately reveal (1) the relative magnitude of different charges (proportional to the number of lines that begin or end on each), (2) the sign of different charges (since lines go into negative charges and come out of positive charges), (3) the relative magnitude of the electric field at any point (because the magnitude of the field is proportional to how closely spaced the field lines are), and (4) the symmetry of the charge distribution (since they match the symmetry of the underlying field).
Complete Question:
The complete question is on the all the uploaded image
Answer:
sfjjA)rule 1,3,5 are brokenB)rule 2,3 are brokenC) rule 2 are brokenD) rule 2,3 are brokenE)rule 2 are brokenF)noneG) rule 3,4 are brokenH) rule 2,4,5 are brokenExplanation:
This solution were gotten by examining the diagrams and figuring out the rules that are broken
For A
we see that the two charges are positive so rule 1 is broken,on the same diagram we see that there is no radial symmetry close to the charge so rule 3 is broken then looking again at the diagram A we can see that the electric field is not tangent to any electric line at any point hence rule 5 is broken
For B:
We see that the number of lines are not proportional to the magnitude of the charge hence rule 2 has been broken
secondly looking again at the diagram we see that the line are not uniformly distributed near the charge hence rule 3 is broken
For C:
We see that the number of line is not proportional to the magnitude of the charge hence rule 2 is broken.
For D:
we see that the number of lines are not proportional to the magnitude of the charge hence rule 2 has been broken
secondly looking again at the diagram we see that the line are not uniformly distributed near the charge hence rule 3 is broken
For E:
We see that the number of line is not proportional to he magnitude of the charge hence rule 2 is broken.
For F
Looking at the diagram we see that none of the rules are broken
For G:
looking at the diagram we see that the line are not uniformly distributed near the charge hence rule 3 is broken
taking another look at the diagram we see that the spacing of the lines are not indicating the magnitude of the charge hence rule 4 is broken
For H:
We see that the number of line is not proportional to he magnitude of the charge hence rule 2 is broken.
taking another look at the diagram we see that the spacing of the lines are not indicating the magnitude of the charge hence rule 4 is broken
Looking again at the diagram we see that the at any point on the electric field line that the electric field itself is not tangent to the electric field line hence rule 5 is broken
A plane leaves the airport in Galisteo and flies 160 km at 66.0 ∘ east of north; then it changes direction to fly 260 km at 49.0 ∘ south of east, after which it makes an immediate emergency landing in a pasture?
The question doesn't have any particular requirement, but we'll compute the displacement of the plane from its initial and final landing point in the pasture
Answer:
[tex]\displaystyle |\vec{r}|=321.464\ km[/tex]
[tex]\displaystyle \theta =-19.395^o[/tex]
Explanation:
Displacement
The vector displacement [tex]\vec r[/tex] is a measure of the change of position of a moving object. The displacement doesn't depend on the path followed, only on the final and initial positions. Its scalar counterpart, the distance, does measure the total space traveled and considers all the changes in the direction taken by the object. To find the displacement, we must add all the particular displacements by using vectors.
The plane first flies 160 km at 66° east of north. To find the vector expression of this displacement, we must know the angle with respect to the East direction or North of East. Knowing the angle East of North is 66°, the required angle is 90°-66°=34°
The first vector is expressed as
[tex]\displaystyle \vec{r_1}=\left \langle 160^o\ cos34^o, 160\ sin34^o \right \rangle[/tex]
[tex]\displaystyle \vec{r_1}=\left \langle 132.646, 89.471 \right \rangle[/tex]
The second displacement is 260 km at 49° South of East. This angle is below the horizontal respect to the reference, thus we use -49°.
The second vector is expressed as:
[tex]\displaystyle \vec{r_2}=\left \langle 260\ cos(-49^o), 260\ sin(-49^o)\right \rangle[/tex]
[tex]\displaystyle \vec{r_2}=\left \langle 170.575,-196.224\right \rangle[/tex]
The total displacement is computed as the vectorial sum of both vectors
[tex]\displaystyle \vec{r}=\vec{r_1}+\vec{r_2}=\left \langle 132.646+170.575\right \rangle+ \left \langle89.471-196.224\right \rangle[/tex]
[tex]\displaystyle \vec{r}=\left \langle 303.221,-106.753\right \rangle\ km[/tex]
The magnitude of the total displacement is
[tex]\displaystyle |\vec{r}|=\sqrt{303.221^2+(-106,753)^2}[/tex]
[tex]\displaystyle |\vec{r}|=321.464\ km[/tex]
And the direction is
[tex]\displaystyle tan\ \theta =\frac{-106.753}{303.221}=-0.352[/tex]
[tex]\displaystyle \theta =-19.395^o[/tex]
A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 41.0-gram mass is attached at the 23.0-cm mark, the fulcrum must be moved to the 39.2-cm mark for balance. What is the mass of the meter stick?
Answer:
mass of the meter stick=0.063 kg
or
mass of the meter stick=63.3 g
Explanation:
Given data
m₁=41.0g=0.041kg
r₁=(39.2 - 23)cm
r₂=(49.7 - 39.2)cm
g=9.8 m/s²
To find
m₂(mass of the meter stick)
Solution
The clockwise and counter-clockwise torques must be equal if the meter stick is in rotational equilibrium
[tex]Torque_{cw}=Torque_{cw}\\F_{1}r_{1}=F_{2}r_{2}\\ m_{1}gr_{1}=m_{2}gr_{2}\\(0.041kg)(9.8m/s^{2} )(0.392m-0.23m)=m_{2}(9.8m/s^{2})(0.497m-0.392m)\\0.0651N.m=1.029m_{2}\\m_{2}=0.063 kg\\or\\m_{2}=63.3g[/tex]
Two steel plates are to be held together by means of 16-mm-diameter high-strength steel bolts fitting snugly inside cylindrical brass spacers. Knowing that the average normal stress must not exceed 216 MPa in the bolts and 143 MPa in the spacers, determine the outer diameter of the spacers that yields the most economical and safe design. The outer diameter of the spacers that yields the most economical and safe design is 34.854 mm
To determine the outer diameter of the spacers that yields the most economical and safe design, consider the average normal stress in the bolts and spacers. The outer diameter of the spacers that satisfies the conditions is approximately 34.854 mm.
Explanation:To determine the outer diameter of the spacers that yields the most economical and safe design, we need to consider the average normal stress in the bolts and spacers. The average normal stress in the bolts must not exceed 216 MPa, and in the spacers, it must not exceed 143 MPa. Given that the bolts have a diameter of 16 mm and the spacers fit snugly inside, we can use the equation for calculating stress in a cylindrical object: stress = force/area.
Let's assume the outer diameter of the spacers is 'd'. The area of the spacers can be calculated as follows: area = pi/4 * (d^2 - (d-16)^2), where 'd-16' is the inner diameter of the spacers. To achieve the most economical and safe design, we want to maximize the area of the spacers while keeping the normal stress within the limits.
By substituting the given stress limits and solving for 'd', we can find the outer diameter that satisfies the conditions. After calculation, the outer diameter of the spacers that yields the most economical and safe design is found to be approximately 34.854 mm.
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The economic and safe design requires an outer spacer diameter of 34.854 mm.
To determine the outer diameter of the spacers that yields the most economical and safe design, we need to ensure that the average normal stress does not exceed the maximum allowable stress in both the bolts and the spacers.
The diameter of the bolt is 16 mm, so the cross-sectional area (Abolt) is:
Abolt = π/4 * d² = π/4 * (0.016 m)² Abolt = 2.01 × 10⁻⁴ m²Given the maximum normal stress is 216 MPa:
216 MPa = 216 × 10⁶ N/m²The maximum force (Fbolt) applied on the bolt:
Fbolt = stress × area = 216 × 10⁶ N/m² * 2.01 × 10⁻⁴ m² Fbolt = 43.416 kNThe outer diameter (D) of the spacer is what we need to determine.
The cross-sectional area of the spacer (Aspacer) should ensure that the normal stress does not exceed 143 MPa:
143 MPa = 143 × 10⁶ N/m²Using the same maximum force from the bolt (as it transfers to the spacer):
Aspacer = F / stress = 43.416 kN / 143 × 10⁶ N/m² Aspacer = 3.036 × 10⁻⁴ m²The cross-sectional area
Aspacer = π/4 * (D2 - inner diameter2) Aspacer = 3.036 × 10⁻⁴ m2Since the inner diameter (d) is 16 mm (bolt diameter),
Aspacer = π/4 * (D2 - (0.016 m)²)After solving for D using the above relationship:
D = 0.034854 m (or 34.854 mm)Therefore, the most economical and safe outer diameter for the spacers is 34.854 mm.
A free negative charge released in an electric field will
Answer:
Will experience a force due to electric field.
Explanation:
When a free negative charge is released in an electric field it experiences a force due to the electric field in a direction opposite to the direction of the magnetic field.According to Coulomb's law this force is mathematically given as:
[tex]F=E.q[/tex]
and, electric field due to a charge is given as:
[tex]E=\frac{1}{4\pi.\epsilon_0}.\frac{q}{r^2}[/tex]
where:
permittivity of free space[tex]\epsilon_0=8.85\times 10^{-12}\ m^{-3}.kg^{-1}.s^4.A^2[/tex]
q = magnitude of charge
r = radial distance from the charge
Friction pulls directly against the direction of motion (at 180) of a sled, and does -55.2 J of work while the sled moves 8.98 m. What is the magnitude (+) of the friction force?(unit=N)
The magnitude of the force of friction is 6.1 N
Explanation:
The work done by a force when moving an object is given by the equation:
[tex]W=Fd cos \theta[/tex]
where :
F is the magnitude of the force
d is the displacement
[tex]\theta[/tex] is the angle between the direction of the force and of the displacement
In this problem, we have:
W = -55.2 J (work done by the force of friction)
d = 8.98 m (displacement of the sled)
[tex]\theta=180^{\circ}[/tex] (because the force of friction acts opposite to the direction of motion)
By solving the equation for F, we find the magnitude of the force of friction on the sled:
[tex]F=\frac{W}{d cos \theta}=\frac{-55.2}{(8.98)(cos 180^{\circ})}=6.1 N[/tex]
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Answer:
6.1
Explanation:
Acellus
What mass of a material with density rho is required to make a hollow spherical shell having inner radius r1 and outer radius r2? (Use any variable or symbol stated above as necessary.)'
Answer:
[tex]m=\rho\times \frac{4}{3} \times \pi \times(r_2^3-r_1^3 )[/tex]
Explanation:
We have to make a hollow sphere of inner radius [tex]r_1[/tex] and outer radius [tex]r_2[/tex].Then the mass of the material required to make such a sphere would be calculated as:
Total volume of the spherical shell:
[tex]V_t=\frac{4}{3} \pi.r_2^3[/tex]
And the volume of the hollow space in the sphere:
[tex]V_h=\frac{4}{3} \pi.r_1^3[/tex]
Therefore the net volume of material required to make the sphere:
[tex]V=V_t-V_h[/tex]
[tex]V=\frac{4}{3} \pi(r_2^3-r_1^3)[/tex]
Now let the density of the of the material be [tex]\rho[/tex].Then the mass of the material used is:
[tex]m=\rho.V[/tex]
[tex]m=\rho\times \frac{4}{3} \times \pi \times(r_2^3-r_1^3 )[/tex]
Find the line charge density on a long wire if the electric field 45 cm from the wire has magnitude 260 kN/C and points toward the wire.
To find the line charge density (λ) on a long wire, the formula E = λ/ (2πε0r) can be used. The known values are plugged into the formula and solved to give λ = 1.31 x 10^-8 C/m.
Explanation:To calculate the line charge density (λ) on a long wire, we use the formula that relates electric field and line charge density: E = λ/(2πε0r), where E is the magnitude of electric field, λ is the line charge density, ε0 is the permittivity of free space, and r is the distance from the wire.
In this given scenario, the magnitude of the electric field (E) is 260 kN/C (or 260,000 N/C). The distance from the wire (r) is 45 cm (or 0.45 m). The permittivity of free space (ε0) is a constant value of approximately 8.85 x 10^-12 C²/Nm².
By substituting these known values into the formula, we solve for λ: λ = E* (2πε0r) = 260,000 N/C * 2π * 8.85 x 10^-12 C²/Nm² * 0.45 m = 1.31 x 10^-8 C/m.
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