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Oksi-84 [34.3K]
3 years ago
10

A container of rocks is brought back from the Moon's surface where the acceleration due to gravity is 162 meters per second if t

he total mass of the container of rocks is 650 kilograms, what is its weight on
Earth's surface
Physics
1 answer:
spin [16.1K]3 years ago
6 0

Answer:

6318 N

Explanation:

From the question given above, the following data were obtained:

Acceleration due to gravity of the moon (gₘ) = 1.62 m/s²

Mass (m) of container = 650 kg

Weight (W) of container on the earth =.?

Next, we shall determine the acceleration due to gravity of the earth. This can be obtained as follow:

Acceleration due to gravity of the moon (gₘ) = 1.62 m/s²

Acceleration due to gravity of the earth (gₑ) =.?

gₘ = 1/6 × gₑ

1.62 = 1/6 × gₑ

1.62 = gₑ /6

Cross multiply

gₑ = 1.62 × 6

gₑ = 9.72 m/s²

Finally, we shall determine the weight of the container on the earth as follow:

Mass (m) of container = 650 kg

Acceleration due to gravity of the earth (gₑ) = 9.72 m/s²

Weight (W) of container on the earth =.?

W = m × gₑ

W = 650 × 9.72

W = 6318 N

Therefore, the weight of the container on earth is 6318 N

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A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s.
Mama L [17]

Answer: f_{r} = 16.49N

Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,

W_{x} = horizontal component of the weight = mgsinФ

W_{y} = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - f_{r} = ma

where m = mass of object=5kg

a = acceleration= unknown

Ф = angle of inclination = 37°

g = acceleration due to gravity = 9.8m/s^{2}

f_{r} = frictional force = unknown

we need to first get the acceleration before the frictional force which is gotten by using the equation below

v^{2} = u^{2} + 2aS

where v = final velocity = 2m/s

u = initial velocity = 0m/s (because the object started from rest)

a= unknown

S= distance covered = length of plane = 5m

2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}

we slot in a into the equation below to get frictional force

mgsinФ - f_{r} = ma

3 * 9.8 * sin 37 - f_{r} = 3* 0.4

17.9633 - f_{r} =  1.2

f_{r} = 17.9633 - 1.2

f_{r} = 16.49N

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Find the equilibrant of two 10.0-N forces acting upon a body when the angle between the forces is 90° Solve graphically using a
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The equilibrant force of the two given forces is 14.14 N.

<h3 /><h3 /><h3>What is equilibrant force?</h3>
  • This is a single force that balances other given forces.

The given parameters:

  • First force, F₁ = 10 N
  • Second force, F₂ = 10 N
  • Angle between the forces, θ = 90⁰

The equilibrant force of the two given forces is calculated as follows;

F = \sqrt{F_1 ^2 + F_2 ^2} \\\\F = \sqrt{(10)^2 + (10)^2} \\\\F = 14.14 \ N

Thus, the equilibrant force of the two given forces is 14.14 N.

Learn more about equilibrant force here: brainly.com/question/8045102

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