I would have to say that it is Y
Answer:

Explanation:
During the exchange of applied force, thermal energy is generated by the friction that exists between the ground and the tire.
Said force according to the statement is the reaction of half the force on the rear tire. In this way the normal force acted is,

The work done is given by the friction force and the distance traveled,

Where ![\mu_k [/ tex] is the coefficient of kinetic frictionN is the normal force previously found d is the distance traveled,Replacing,[tex]W_f = (0.80)(441)(0.42)](https://tex.z-dn.net/?f=%20%5Cmu_k%20%5B%2F%20tex%5D%20is%20the%20coefficient%20of%20kinetic%20friction%3C%2Fp%3E%3Cp%3EN%20is%20the%20normal%20force%20previously%20found%20d%20is%20the%20distance%20traveled%2C%3C%2Fp%3E%3Cp%3EReplacing%2C%3C%2Fp%3E%3Cp%3E%5Btex%5DW_f%20%3D%20%280.80%29%28441%29%280.42%29)
The thermal energy released through the work done is,

Answer:
D. "The net force is zero, so the acceleration is zero"
Explanation:
edge 2020
Answer:
The magnitudes of the net magnetic fields at points A and B is 2.66 x
T
Explanation:
Given information :
The current of each wires, I = 4.7 A
dH = 0.19 m
dV = 0.41 m
The magnetic of straight-current wire :
B= μ
I/2πr
where
B = magnetic field (T)
μ
= 1.26 x
(N/
)
I = Current (A)
r = radius (m)
the magnetic field at points A and B is the same because both of wires have the same distance. Based on the right-hand rule, the net magnetic field of A and B is canceled each other (or substracted). Thus,
BH = μ
I/2πr
= (1.26 x
)(4.7)/(2π)(0.19)
= 4.96 x
T
BV = μ
I/2πr
= (1.26 x
)(4.7)/(2π)(0.41)
= 2.3 x
T
hence,
the net magnetic field = BH - BV
= 4.96 x
- 2.3 x 
= 2.66 x
T
Answer:
v = 54.2 m / s
Explanation:
Let's use energy conservation for this problem.
Starting point Higher
Em₀ = U = m g h
Final point. Lower
= K = ½ m v²
Em₀ = Em_{f}
m g h = ½ m v²
v² = 2gh
v = √ 2gh
Let's calculate
v = √ (2 9.8 150)
v = 54.2 m / s