<h3><u>Answer;</u></h3>
electric potential
<h3><u>Explanation;</u></h3>
Electric potential is the electric potential energy per unit charge.
Mathematically; V =PE/q
Where; PE is the electric potential energy, V is the electric potential and q is the charge.
Electric potential is more commonly known as voltage. If you know the potential at a point, and you then place a charge at that point, the potential energy associated with that charge in that potential is simply the charge multiplied by the potential.
To prevent the crate from slipping, the maximum force that the belt can exert on the crate must be equal to the static friction force.
Ff = 0.5 * 16 * 9.8 = 78.4 N
a = 4.9 m/s^2
If acceleration of the belt exceeds the value determined in the previous question, what is the acceleration of the crate?
In this situation, the kinetic friction force is causing the crate to decelerate. So the net force on the crate is 78.4 N minus the kinetic friction force.
Ff = 0.28 * 16 * 9.8 = 43.904 N
Net force = 78.4 – 43.904 = 34.496 N
To determine the acceleration, divide by the mass of the crate.
a = 34.496 ÷ 16 = 2.156 m/s^2
Answer:
The required pressure is 6.4866 atm.
Explanation:
The given data : -
In the afternoon.
Initial pressure of tire ( p₁ ) = 7 atm = 7 * 101.325 Kpa = 709.275 Kpa
Initial temperature ( T₁ ) = 27°C = (27 + 273) K = 300 K
In the morning .
Final temperature ( T₂ ) = 5°C = ( 5 + 273 ) K = 278 K
Given that volume remains constant.
To find final pressure ( p₂ ).
Applying the ideal gas equation.
p * v = m * R * T


= 657.2615 Kpa = 6.486 atm
1. A Medium
2. One of these 4, the refractive index of the medium <span>Choose whichever are on the test.
3. </span>properties and nature of medium, Hope this helps.
The force exerted on the car during this stop is 6975N
<u>Explanation:</u>
Given-
Mass, m = 930kg
Speed, s = 56km/hr = 56 X 5/18 m/s = 15m/s
Time, t = 2s
Force, F = ?
F = m X a
F = m X s/t
F = 930 X 15/2
F = 6975N
Therefore, the force exerted on the car during this stop is 6975N