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:
The diagram shows a heater above a thermometer. The thermometer bulb is in the position shown. How the heat
Answer:
Check Explanation
Explanation:
Let the speed of the drone relative to earth be u = ?
Let the speed of the drone relative to the starship be u' = 0.5c
Let the speed of the starship relative to earth be v = 0.8c
In the theory of relativity,
The 3 speeds are related thus
u = (u' + v)/[1 + (u'v/c²)]
u = (0.5c + 0.8c)[1 + ((0.8c × 0.5c)/c²)]
u = 1.3c/[1 + (0.4c²/c²)]
u = 1.3c/1.4 = 0.9286 c = 0.93 c = 93% c
Answer:
Explanation:
Given that,
Height of the bridge is 20m
Initial before he throws the rock
The height is hi = 20 m
Then, final height hitting the water
hf = 0 m
Initial speed the rock is throw
Vi = 15m/s
The final speed at which the rock hits the water
Vf = 24.8 m/s
Using conservation of energy given by the question hint
Ki + Ui = Kf + Uf
Where
Ki is initial kinetic energy
Ui is initial potential energy
Kf is final kinetic energy
Uf is final potential energy
Then,
Ki + Ui = Kf + Uf
Where
Ei = Ki + Ui
Where Ei is initial energy
Ei = ½mVi² + m•g•hi
Ei = ½m × 15² + m × 9.8 × 20
Ei = 112.5m + 196m
Ei = 308.5m J
Now,
Ef = Kf + Uf
Ef = ½mVf² + m•g•hf
Ef = ½m × 24.8² + m × 9.8 × 0
Ef = 307.52m + 0
Ef = 307.52m J
Since Ef ≈ Ei, then the rock thrown from the tip of a bridge is independent of the direction of throw