2. Newton's laws of motion
Given :
The acceleration due to gravity on or near the surface of Earth is 32 ft/s/s
To Find :
From what height must a stone be dropped on Earth to strike the ground with a velocity of 136 ft/s.
Solution :
Initial velocity of stone, u = 0 ft/s.
Now, by equation of motion :

Therefore, height from which stone is thrown is 289 ft.
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
The answer is d. i hope this helps :D
Answer:
84.6 J
Explanation:
The work done by the force is given by

where
W is the work done
F = 15 N is the force applied
d = 6.0 m is the displacement
is the angle between the force's direction and the displacement
Substituting the numbers into the equation, we find
