The power delivered by the elevator is 34496 W.
<h3>What is power?</h3>
Power is the rate at which energy is used up.
To calculate the power delivered by the elevator, we use the formula below.
Formula:
- P = mgh/t.............. Equation 1
Where:
- P = Power delivered by the elevator
- m = Total mass on the elevator
- g = acceleration due to gravity
- h = Height
From the question,
Given:
- m = 1.1×10³ kg
- h = 40 m
- t = 12.5 s
- g = 9.8 m/s²
Substitute these values into equation 1
- P = 1.1×10³×40×9.8/12.5
- P = 34496 W.
Hence, the power delivered by the elevator is 34496 W.
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Answer:
h≅ 58 m
Explanation:
GIVEN:
mass of rocket M= 62,000 kg
fuel consumption rate = 150 kg/s
velocity of exhaust gases v= 6000 m/s
Now thrust = rate of fuel consumption×velocity of exhaust gases
=6000 × 150 = 900000 N
now to need calculate time t = amount of fuel consumed÷ rate
= 744/150= 4.96 sec
applying newton's law
M×a= thrust - Mg
62000 a=900000- 62000×9.8
acceleration a= 4.71 m/s^2
its height after 744 kg of its total fuel load has been consumed
h= 58.012 m
h≅ 58 m
Answer:
2.82 s
Explanation:
The ball will be subject to the acceleration of gravity which can be considered constant. Therefore we can use the equation for uniformly accelerated movement:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
Y0 is the starting position, 2.3 m in this case.
Vy0 is the starting speed, 13 m/s.
a will be the acceleration of gravity, -9.81 m/s^2, negative because it points down.
Y(t) = 2.3 + 13 * t - 1/2 * 9.81 * t^2
It will reach the ground when Y(t) = 0
0 = 2.3 + 13 * t - 1/2 * 9.81 * t^2
-4.9 * t^2 + 13 * t + 2.3 = 0
Solving this equation electronically gives two results:
t1 = 2.82 s
t2 = -0.17 s
We disregard the negative solution. The ball spends 2.82 seconds in the air.
Explanation:
Given that,
Mass of the particle, m = 4 kg
Speed of the particle, v = 2.5 m/s
The radius of the circle, r = 2 m
We need to find the angular momentum about the center of the circle. The formula for the angular momentum is given by :
Substitute all the values,
So, the angular momentum of the particle is 20 kg-m² s.
<span>Orbital radius would decrease because the gravitational pull would increase.</span>
