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Yanka [14]
3 years ago
12

If the average mass of an adult is 86kg. Determine the mass and weight of an 86kg man in the moon where gravitational field is o

ne-sixth that of the earth.
Physics
1 answer:
gavmur [86]3 years ago
3 0

Answer:

See the explanation below, and the different answers

Explanation:

It is important to clarify that the mass is always preserved, that is, it does not vary regardless of the location of the person, whether the moon, Saturn, Mars. The value that changes is the weight, due to the gravity of each planet.

<u>On Earth</u>

<u />

m = 86[kg]

w = m * g

where:

g = gravity acceleration = 9.81[m/s^2]

w = weight = [N]

w = 86 * 9.81

w = 843.66 [N]

<u>On the Moon</u>

<u />

m = 86[kg]

w = m * g

where:

g = gravity acceleration * (1/6) = 9.81 * (1/6) = 1.635 [m/s^2]

w = weight = [N]

w = 86 * 1.635

w = 140.61 [N]

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AveGali [126]

Answer: 2000 J.

Explanation: Since work is force*displacement, we just have to multiply the force by the distance: w = f*d = 400 N*5.0 m = 2000 J.

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A police officer uses a radar gun to determine the speed of a car. A specialized radar gun uses ultraviolet light to determine t
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A transverse and D electromagnetic

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3 years ago
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During the first 50 s a truck traveled at constant speed of 25 m/s. Find the distance that it is traveled.
Allushta [10]
Time=50s
speed=25m/s

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6 0
2 years ago
What is the kinetic energy of an automobile with a mass of 1252 kg traveling at a speed of 12 m/s?
torisob [31]
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4 0
3 years ago
Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first interference
strojnjashka [21]

Answer:

1.7323

Explanation:

To develop this problem, it is necessary to apply the concepts related to refractive indices and Snell's law.

From the data given we have to:

n_{air}=1

\theta_{liquid} = 19.38\°

\theta_{air}35.09\°

Where n means the index of refraction.

We need to calculate the index of refraction of the liquid, then applying Snell's law we have:

n_1sin\theta_1 = n_2sin\theta_2

n_{air}sin\theta_{air} = n_{liquid}sin\theta_{liquid}

n_{liquid} = \frac{n_{air}sin\theta_{air}}{sin\theta_{liquid}}

Replacing the values we have:

n_{liquid}=\frac{(1)sin(35.09)}{sin(19.38)}

n_liquid = 1.7323

Therefore the refractive index for the liquid is 1.7323

6 0
3 years ago
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