Work Done = Force x distance
Since she exerted a horizontal force of 20N over a distance of 5m, the work done is 20N x 5m which is equals to 100 joules
Answer:
![\Delta K=0.07969 J - 0.0849 J = -0.00521 J](https://tex.z-dn.net/?f=%5CDelta%20K%3D0.07969%20J%20-%200.0849%20J%20%3D%20-0.00521%20J)
Explanation:
According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.
Being
and
the masses of pucks a and b respectively, the initial momentum of the system is
![M_1=m_av_a+m_bv_b](https://tex.z-dn.net/?f=M_1%3Dm_av_a%2Bm_bv_b)
Since b is initially at rest
![M_1=m_av_a](https://tex.z-dn.net/?f=M_1%3Dm_av_a)
After the collision and being
and
the respective velocities, the total momentum is
![M_2=m_av'_a+m_bv'_b](https://tex.z-dn.net/?f=M_2%3Dm_av%27_a%2Bm_bv%27_b)
Both momentums are equal, thus
Solving for ![v_a](https://tex.z-dn.net/?f=v_a)
![v_a=\frac{m_av'_a+m_bv'_b}{m_a}](https://tex.z-dn.net/?f=v_a%3D%5Cfrac%7Bm_av%27_a%2Bm_bv%27_b%7D%7Bm_a%7D)
![v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}](https://tex.z-dn.net/?f=v_a%3D%5Cfrac%7B0.254Kg%5Ctimes%20%28-0.123%20m%2Fs%29%2B0.367Kg%20%280.651m%2Fs%29%7D%7B0.254Kg%7D)
The initial kinetic energy can be found as (provided puck b is at rest)
![K_1=\frac{1}{2}m_av_a^2](https://tex.z-dn.net/?f=K_1%3D%5Cfrac%7B1%7D%7B2%7Dm_av_a%5E2)
![K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J](https://tex.z-dn.net/?f=K_1%3D%5Cfrac%7B1%7D%7B2%7D%280.254Kg%29%20%280.8176m%2Fs%29%5E2%3D0.0849%20J)
The final kinetic energy is
![K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B1%7D%7B2%7Dm_av_a%27%5E2%2B%5Cfrac%7B1%7D%7B2%7Dm_bv_b%27%5E2)
![K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B1%7D%7B2%7D0.254Kg%20%28-0.123m%2Fs%29%5E2%2B%5Cfrac%7B1%7D%7B2%7D0.367Kg%20%280.651m%2Fs%29%5E2%3D0.07969%20J)
The change of kinetic energy is
![\Delta K=0.07969 J - 0.0849 J = -0.00521 J](https://tex.z-dn.net/?f=%5CDelta%20K%3D0.07969%20J%20-%200.0849%20J%20%3D%20-0.00521%20J)
Complete question:
What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)
Answer:
The peak emf generated by the coil is 15.721 kV
Explanation:
Given;
Radius of coil, r = 0.250 m
Number of turns, N = 500-turn
time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s
magnetic field strength, B = 0.425 T
Induced peak emf = NABω
where;
A is the area of the coil
A = πr²
ω is angular velocity
ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s = 60 rev/s
Induced peak emf = NABω
= 500 x (π x 0.25²) x 0.425 x 376.738
= 15721.16 V
= 15.721 kV
Therefore, the peak emf generated by the coil is 15.721 kV
Answer: c. the molecules with the highest energy evaporate first, lowering the temperature of the sample
Explanation:
The process by which liquid starts to change into vapor phase at any temperature is known as evaporation.
During evaporation , the molecules which possess higher energies escape from the upper layer into vapor phase. the molecules which escape draw energy from surroundings and thus decrease the energy of the surroundings and hence lead to decrease in temperature.
As temperature of the system is directly proportional to the energy of the system , thus decrease in energy leads to decrease in temperature.
![K.E=\frac{3RT}{2}](https://tex.z-dn.net/?f=K.E%3D%5Cfrac%7B3RT%7D%7B2%7D)
K.E. = Kinetic energy
T = temperature
R= gas constant
Explanation:
The strength of the gravitational force between two objects depends on two factors, mass and distance. the force of gravity the masses exert on each other. ... increases, the force of gravity decreases. If the distance is doubled, the force of gravity is one-fourth as strong as before.