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3 years ago

Write any two importance of gravitational force

Physics

1 answer:

saul85 [17]3 years ago

5 0

Answer:

plz mark me as brainliest plz

Explanation:

The gravitational force of the earth keeps us bound to the earth. Gravitational force between earth and sun makes the earth move around the sun. Gravitational force between moon and earth makes the moon go around the earth.

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An object is thrown vertically up and attains an upward velocity of 9.6 m/s when it reaches one fourth of its maximum height abo

ki77a [65]

Answer:

11.09 m/s

Explanation:

Given that an object is thrown vertically up and attains an upward velocity of 9.6 m/s when it reaches one fourth of its maximum height above its launch point.

The parameters given are:

Initial velocity U = ?

Final velocity V = 9.6 m/s

Acceleration due to gravity g = 9.8m/s^2

Let first assume that the object is thrown from rest with the velocity U, at maximum height V = 0

Using third equation of motion

V^2 = U^2 - 2gH

0 = U^2 - 2 × 9.8H

U^2 = 19.6H ........ (1)

Using the formula again for one fourth of its maximum height

9.6^2 = U^2 - 2 × 9.8 × H/4

92.16 = U^2 - 19.6/4H

92.16 = U^2 - 4.9H

U^2 = 92.16 + 4.9H ...... (2)

Substitute U^2 in equation (1) into equation (2)

19.6H = 92.16 + 4.9H

Collect the like terms

19.6H - 4.9H = 92.16

14.7H = 92.16

H = 92.16/14.7

H = 6.269

Substitute H into equation 2

U^2 = 92.16 + 4.9( 6.269)

U^2 = 92.16 + 30.72

U^2 = 122.88

U = 11.09 m/s

Therefore, the initial velocity of the object is 11.09 m/s

3 0

4 years ago

How long does it take (in minutes) for light to reach mercury from the sun, a distance of 6.02 × 107 km?

Ksju [112]

Light travels at a speed of:
$c=3 \cdot 10^8 m/s$
The distance between Mercury and Sun is $S=6.02 \cdot 10^7 km=6.02 \cdot 10^{10} m$ , so the time it takes is
$t= \frac{S}{c}= \frac{6.02 \cdot 10^{10} m}{3 \cdot 10^8 m/s} =200.7 s$
if we want to convert this into minutes, keeping in mind that 1 min = 60 seconds, we should divide this value by 60:
$t= \frac{200.7 s}{60 s/min}=3.34 min$

6 0

4 years ago

Read 2 more answers

A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co

Tems11 [23]

(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>

The wheels under the wind do not pass through the center line.
The position of the front wheel is constant and it is zero mark (origin).
The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

$X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}$

$18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg$

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

$W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN$

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

Learn more about center mass here: brainly.com/question/13499822

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2 years ago

After three half-lives have elapsed, the amount of an 8.0 g sample of a radionuclide that

almond37 [142]

After three half-lives have elapsed, the amount of an 8.0 g sample of a radionuclide that remains undecayed is 1.0 g.

Half-Life refers to the time it takes for half the amount of a substance to disappear or change.

The nucleus of the atoms of radioactive elements disintegrate to half their starting amounts after every Half-Life.

After three half-lives one-eight of the original atoms remain.

Therefore, after three half-lives have elapsed, the amount of an 8.0 g sample of a radionuclide that remains undecayed is 1.0 g.

Learn more about Half-Life at: brainly.com/question/26689704

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7 0

2 years ago

A(n ___ is used to determine where an implanted electrode is to be placed in a brain in 3-dimensions.

Fynjy0 [20]

A stereotaxic atlas is used to determine where an implanted electrode is to be placed in a brain in 3-dimensions. A stereotaxic atlas is used for doing a stereotactic surgery in order to cure the disorder occurred in the brain. This tool increase the accuracy, reliability, and the measurement of the brain because it gives the accurate brain map<span>.</span>

5 0

3 years ago

Write any two importance of gravitational force​

Write any two importance of gravitational force