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Talja [164]
3 years ago
14

A cheerleader lifts his 79.4 kg partner straight up off the ground a distance of 0.945 m before releasing her. the acceleration

of gravity is 9.8 m/s 2 . if he does this 33 times, how much work has he done? answer in units of j
Physics
1 answer:
Oksi-84 [34.3K]3 years ago
5 0
To find out how much work he has done, we must first calculate force using the force formula (F= Mass*Acceleration). In this case, mass is 79.4 and acceleration is the gravitational constant of 9.8m/s, plugging this into the formula we find that force is 778.12Newtons. Next, we need to multiply force by the distance to get the amount of energy used to lift his partner once. Which is 778.12 * .945 = 735.32. Finally, we need to multiply 735.32 by the number of times he lifts his partner, 33, to get 735.32 * 33 to find that the energy he has expended 24,265.56 Joules of energy.
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3 years ago
a fan acquires a speed of 180 rpm in 4s, starting from rest. calculate the speed of the fan at the end of the 5th second startin
KengaRu [80]

Answer:

225 rpm

Explanation:

The angular acceleration of the fan is given by:

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

where

\omega_f is the final angular speed

\omega_i is the initial angular speed

\Delta t is the time interval

For the fan in this problem,

\omega_i = 0\\\omega_f = 180 rpm\\\Delta t=4 s

Substituting,

\alpha = \frac{180-0}{4}=45 rpm/s

Now we can find the angular speed of the fan at the end of the 5th second, so after t = 5 s. It is given by:

\omega' = \omega_i + \alpha t

where

\omega_i = 0\\\alpha = 45 rpm/s\\t = 5 s

Substituting,

\omega' = 0 + (45)(5)=225 rpm

7 0
3 years ago
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I think it’s A pretty sure...
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3 0
3 years ago
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