HELP PLSSS
1 answer:
Given :
A 13.3 kg box sliding across the ground decelerates at 2.42 m/s².
To Find :
The coefficient of kinetic friction.
Solution :
Frictional force applied to the box is :
....1)
Also, force of friction is given by :
....2)
Equating equation 1) and 2), we get :
Therefore, the coefficient of kinetic friction is 0.247 .
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Answer:
4Q
Explanation:
Case 1 :
= mass of the metal
= specific heat of the metal
= Change in temperature = 7 - 4 = 3 C
Amount of heat required for the above change of temperature is given as
Case 2 :
= mass of the metal
= specific heat of the metal
= Change in temperature = 19 - 7 = 12 C
Amount of heat required for the above change of temperature is given as
Hence the correct choice is
4Q
Answer:
A= 203 KJ
B= 54 Kg
Explanation:
The initial specific volumes and internal energies are obtained from A-12 for a given pressure and state. The enthalpy of the refrigerant in the supply line is determined using the saturated liquid approximation for the given temperature with data from A-11. The mass that has entered the tank is:
Δm = m₂ – m₁
= V(1/α₂ – 1/α₁)
= 0.05 (1/0.0008935 – 1/ 0.025645)Kg
= 54Kg
The heat transfer is obtained from the energy balance:
ΔU=
+
m₂u₂ – m₁u₂ = +
= m₂u₂ – m₁u₁ –
= V/α₂u₂ - V/α₁u₁ –
=(0.05/0.0008935 . 116.72 – 0.05/0.025645 . 246.82 – 54.108.28) Kj
= 203 KJ