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Veseljchak [2.6K]

2 years ago

9

how does gravity affect objects of different mass close to earth, and how does that effect change as an object moves farther fro

m earth?

1 answer:

Luda [366]2 years ago

3 0

Answer:

It makes it lighter when its closer and heavier when its farther way.

Explanation:

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What does this image reveal about gravityon the moon compared to Earth?

elena-14-01-66 [18.8K]

ANSWER:

The Moon's gravity is less than Earth's.

STEP-BY-STEP EXPLANATION:

When you jump, you fall back to the ground. Apples or leaves also fall: we are all attracted to the Earth. It is the terrestrial attraction due to the force of gravity.

The force of gravity also exists on the Moon. But since the Moon is smaller than the Earth, the attraction felt on the Moon is smaller than the Earth's attraction.

As the gravity is less, you can do things such as the one shown in the image.

As the force of attraction is less, the weight is less on the Moon, which can cause things that would be impossible on Earth.

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9 months ago

Match the situation with the energy type.

anastassius [24]

Answer:

snowplow would be m, sun r , hb n , FB e ,and tiaa c

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2 years ago

What should you do if you see lightning while you are outside exercising? Head inside Stay under a tree Lie in the grass Move qu

Arada [10]

If you are under a tree, they kind of act like lightning rods, so stay away from trees, so roll in dat grass. Fun Fact: Lightning comes from the ground more than it comes from the sky, its like when you rub a blanket on your head, some lil' lightnings come from your head while a little come from the blanket, its the same with grass and clouds only, 10000 volts stronger and deadly.

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2 years ago

Read 2 more answers

This phenomenon is known as the bends. If a scuba diver rises quickly from a depth of 25.0 m in Lake Michigan (which is fresh wa

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Explanation:

Attachments of the solution are below.

6 0

2 years ago

A 4.00 m long, massless beam rests horizontally on a support 3.00 m from the left

Rus_ich [418]

If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Given the data in the question;

Length of the massless beam; $L = 4.00m$

Distance of support from the left end; $x = 3.00m$

First mass; $m1 = 31.3 kg$

Distance of beam from the left end( m₁ is attached to ); $x_1 = ?$

Second mass; $m_2 = 61.7 kg$

Distance of beam from the right of the support( m₂ is attached to ); $x_1 = 0.273m$

Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.

Hence, $m_1g( x-x_1) = m_2gx_2$

we divide both sides by $g$

$m_1( x-x_1) = m_2x_2$

Next, we make $x_1$ , the subject of the formula

$x_1 = x - [ \frac{m_2x_2}{m_1} ]$

We substitute in our given values

$x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]$

$x_1 = 3.00m - 0.538m$

$x_1 = 2.46m$

Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Learn more; brainly.com/question/3882839

6 0

2 years ago