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 .
Answer:
a)10.28 Nm
b)9.93 Nm
Explanation:
Let g = 9.81m/s2. First we can calculate the weight of the trophy
W = mg = 1.6 * 9.81 = 15.696 N
(a) The torque is product of force and its moment arm
T = WL = 15.696 * 0.655 = 10.28 Nm
(b) Suppose his arm makes an angle of 15 degree with respect to the horizontal line. We can still calculate the arm length, or the horizontal distance from the trophy to the champion:
Again, torque is product of force and its moment arm
Answer:
(a) 83475 MW
(b) 85.8 %
Explanation:
Output power = 716 MW = 716 x 10^6 W
Amount of water flows, V = 1.35 x 10^8 L = 1.35 x 10^8 x 10^-3 m^3
mass of water, m = Volume x density = 1.35 x 10^8 x 10^-3 x 1000
= 1.35 x 10^8 kg
Time, t = 1 hr = 3600 second
T1 = 25.4° C, T2 = 30.7° C
Specific heat of water, c = 4200 J/kg°C
(a) Total energy, Q = m x c x ΔT
Q = 1.35 x 10^8 x 4200 x (30.7 - 25.4) = 3 x 10^12 J
Power = Energy / time
Power input =
Power input = 83475 MW
(b) The efficiency of the plant is defined as the ratio of output power to the input power.
Thus, the efficiency is 85.8 %.