Answer:
-58.876 kJ
Explanation:
m = mass of air = 1 kg
T₁ = Initial temperature = 15°C
T₂ = Final temperature = 97°C
Cp = Specific heat at constant pressure = 1.005 kJ/kgk
Cv = Specific heat at constant volume = 0.718 kJ/kgk
W = Work done
Q = Heat = 0 (since it is not mentioned we are considering adiabatic condition)
ΔU = Change in internal energy
Q = W+ΔU
⇒Q = W+mCvΔT
⇒0 = W+mCvΔT
⇒W = -mCvΔT
⇒Q = -1×0.718×(97-15)
⇒Q = -58.716 kJ
The velocity of the stuntman, once he has left the cannon is 5 m/s.
The right option is O A. 5 m/s
The Kinetic energy of the stuntman is equal to the elastic potential energy of the spring.
<h3 /><h3>Velocity: </h3>
This is the ratio of displacement to time. The S.I unit of Velocity is m/s. The velocity of the stuntman can be calculated using the formula below.
⇒ Formula:
- mv²/2 = ke²/2
- mv² = ke².................. Equation 1
⇒ Where:
- m = mass of the stuntman
- v = velocity of the stuntman
- k = force constant of the spring
- e = compression of the spring
⇒ Make v the subject of the equation
- v = √(ke²/m)................. Equation 2
From the question,
⇒ Given:
- m = 48 kg
- k = 75 N/m
- e = 4 m
⇒ Substitute these values into equation 2
- v = √[(75×4²)/48]
- v = √25
- v = 5 m/s.
Hence, The velocity of the stuntman, once he has left the cannon is 5 m/s.
The right option is O A. 5 m/s
Learn more about velocity here: brainly.com/question/10962624
Present. NOT SURE IF THIS IS EVEN RIGHT!! I JUST GUESSED AND THOUGHT
Answer:
the answer is 2000Nm
Explanation:
wprk done = force × distance moved
w.d = 200N × 10m
w.d = 2000Nm
mark me as brainliest plyyzzz
You can use Vf^2-Vi^2 = 2ax
Vf^2 - 0 = 2(9.81)(25)
Or you can use energy
mgh = 1/2mv^2
2gh =v^2
Same thing
