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gtnhenbr [62]
3 years ago
12

If your mass is 63.7kg and standing 7.5m away from a boulder with a mass of 9750.6kg what is the gravitational force?

Physics
1 answer:
Anna11 [10]3 years ago
3 0

Answer:

7.37 x 10^--7 N

Explanation:

Did it on A P E X

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A 20 cm square frame that can rotate about the 00' axis is placed in a homogeneous magnetic field with 0.5T induction directed v
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Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon’
Yuki888 [10]

Answer:

0.2 m

Explanation:

PHASE 1

First, we calculate the distance the tongue moved in the first 20 ms (0.02 secs). We use one of Newton's equations of linear motion:

s = ut + \frac{1}{2}at^2

where u = initial velocity = 0 m/s

a = acceleration = 250 m/s^2

t = time = 0.02 s

Therefore:

s = 0 + \frac{1}{2} * 250 * (0.02)^2\\\\\\s = 0.05 m

PHASE 2

Then, for the next 30 ms (0.03 secs), we use the formula:

distance = speed * time

This speed is the same as the final velocity of the tongue after the first 20 ms.

This can be obtained by using the formula:

v = u + at\\\\=> v = 0 + (250 * 0.02)\\\\\\v = 5 m/s

Therefore:

distance = 5 * 0.03 = 0.15 m

Therefore, the total distance moved by the tongue in the 50 ms interval is:

0.05 + 0.15 = 0.2 m

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Is motion on earth different from space
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Why would it be cooler if you climb to a higher elevation in a desert
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Read 2 more answers
Erica is participating in a road race. The first part of the race is on a 5.2-mile-long straight road oriented at an angle of 25
Sveta_85 [38]

Answer:

8.6 miles

Explanation:

We need to calculate the components of the total displacement along the east-west and north-south directions first.

In the first part, Erica moves 5.2 miles at 25∘ north of east. So the components of this displacement along the two directions are:

East: d_{1x} = 5.2 cos 25^{\circ}=4.7 mi

North: d_{1y} = 5.2 sin 25^{\circ}=2.2 mi

In the second part, Erica moves 5.0 miles north. So, the components of this displacement are:

East: d_{2x}=0

North: d_{2y} = 5.0 mi

So the components of the total displacement are

East: d_x = d_{1x}+d_{2x}=4.7 + 0 = 4.7 mi

North: d_y = d_{1y}+d_{2y}= 2.2 + 5.0 = 7.2 mi

Therefore the magnitude of the displacement, which is the straight-line distance from the starting point to the end of the race, is

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