Answer:
8 m/s
Explanation:
KE Formula: KE = 1/2mv^2
Kinetic Energy is equal to energy, J.
6400 = 1/2 * 200 * V^2
1/2 * 200 = 100
6400 = 100 * V^2
6400/100 = V^2
64 = V^2
Square root both sides
Speed is 8 m/s
Friction, wind resistance, gravity, tension force
Thermal potential energy is converted into electric energy.
Answer:
The the quality of the refrigerant at the exit of the expansion valve is 0.179.
Explanation:
Given that,
Initial pressure = 10 bar
Temperature = 22°C
Final pressure = 2.0 bar
We using the value of h

The refrigerant during expansion undergoes a throttling process
Therefore, 
We need to calculate the quality of the refrigerant at the exit of the expansion valve
At 2.0 bar,
The property of ammonia


Using formula

Put the value into the formula



Hence, The the quality of the refrigerant at the exit of the expansion valve is 0.179.
Answer:
9.89 m/s.
Explanation:
Given that,
The radius of the circular arc, r = 25 m
The acceleration of the vehicle is 0.40 times the free-fall acceleration i.e.,a = 0.4(9.8) = 3.92 m/s²
Let v is the maximum speed at which you should drive through this turn. It can be solved as follows :

So, the maximum speed of the car should be 9.89 m/s.