Answer:
The specific heat capacity is q_{L}=126.12kJ/kg
The efficiency of the temperature is n_{TH}=0.67
Explanation:
The p-v diagram illustration is in the attachment
T_{H} means high temperature
T_{L} means low temperature
The energy equation :
= R* 
in(
/
)
The specific heat capacity:
=q_{h}*(T_{L}/T_{H})
q_{L}=378.36 * (400/1200)
q_{L}=378.36 * 0.333
q_{L}=126.12kJ/kg
The efficiency of the temperature will be:
=1 - (
/
)
n_{TH}=1-(400/1200)
n_{TH}=1-0.333
n_{TH}=0.67
From the geometry of the problem, the 20 m-long cable creates
the hypotenuse of a right triangle, with the extended of the other two sides of
size 20 m * cos(30 deg), which is around 17.3 m. Therefore, the ball has increased
by 20 m - 17.3 m = 2.7 m.
The potential energy will have altered by m*g*h, which is 1400 kg * 9.8 m/s^2 *
1.6 m , or about 37044 joules.
A child height can be very tall or short
Answer:
D. Friction slow down or stop objects in motion.
