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Talja [164]
2 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]2 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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Two groups of students were tested to compare their speed working math problems,Each group was given the same problems.One group
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A house brick has a volume of 1900 cm³ and a weight in air of 80N.What is its apparent weight in water?The density of water is 1
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Answer:

61 N

Explanation:

We'll begin by calculating the mass of the brick when placed in water. This can be obtained as follow:

Volume of brick = 1900 cm³

Density of water = 1 g/cm³

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Density = mass / volume

1 = mass of brick in water / 1900

Cross multiply

Mass of brick in water = 1 × 1900

Mass of brick in water = 1900 g

Next, we shall convert 1900 g to Kg.

1000 g = 1 Kg

Therefore,

1900 g = 1900 g × 1 Kg / 1000 g

1900 g = 1.9 Kg

Next, we shall determine the weight in water. This can be obtained as follow:

Mass (m) = 1.9 Kg

Acceleration due to gravity (g) = 10 m/s²

Weight (W) =?

W = m × g

W = 1.9 × 10

W = 19 N

Thus, the weight of the brick in water is 19 N.

Finally, we shall determine the apparent weight of the brick in water. This can be obtained as follow:

Weight in air = 80 N

Weight in water = 19 N

Apparent weight =?

Apparent weight = weight in air – weight in water

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2 years ago
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All these resistors are in series so we can take the sum of them by:

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So...

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I = 3 amps (A)

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3 years ago
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