It can either be all of them or just 1 and 3
Answer:
the average force 11226 N
Explanation:
Let's analyze the problem we are asked for the average force, during the crash, we can find this from the impulse-momentum equation, but this equation needs the speeds and times of the crash that we could look for by kinematics.
Let's start looking for the stack speeds, it has a free fall, from rest (Vo=0)
Vf² = Vo² - 2gY
Vf² = 0 - 2 9.8 7.69 = 150.7
Vf = 12.3 m / s
This is the speed that the battery likes when it touches the beam. They also give us the distance it travels before stopping, let's calculate the time
Vf = Vo - g t
0 = Vo - g t
t = Vo / g
t = 12.3 / 9.8
t = 1.26 s
This is the time to stop
Now let's use the equation that relates the impulse to the amount of movement
I = Δp
F t = pf-po
The amount of final movement is zero because the system stops
F = - po / t
F = - mv / t
F = - 1150 12.3 / 1.26
F = -11226 N
This is the average force exerted by the stack on the vean
Wave power can be regarded as a reliable source of energy because the ocean currents are always moving.
<h3>What can be the challenges of wave power?</h3>
Wave power is a device that can be used to convert the mechanical energy of the ocean waves into electrical energy based on the principle of conservation of energy.
The major challenges that face the use of wave power in electricity generation is the unreliability of the waves which leads to uncertainty in the quantity of power generated Also, the wave direction and direction of ocean currents all limit the amount of power generated by this method. However, in spite of challenges, it can be regarded as a reliable source of energy because the ocean currents are always moving.
Learn more about wave power:brainly.com/question/1362067
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Answer:
η = 2.57%
Explanation:
Given:
Output power of the steam power plant, P(out) = 200 MW
Consumption of coal, m = 700 tons/h = (700 × 10³)/3600 kg/s= 194.44 kg/s
Heating capacity of the coal, Cv = 40000 kJ/kg
now,
Input power, P(in) = mCv
or
P(in) = 194.44 kg/s × 40000 = 7777.77 MW
thus,
efficiency, η = P(out)/P(in) = (200 MW) / (7777.77 MW) = 0.02571
or
η = 2.57%
