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professor190 [17]
3 years ago
10

The mass of a certain neutron star is 2.5x10^30kg and the radius 7000m. what is the force of gravity on a 1kg object of the surf

ace of the neutron star
Physics
1 answer:
Makovka662 [10]3 years ago
8 0

Answer:

3.42N

Explanation:

*not too sure bc i left my physics notes at school so it might not be 100% accurate :p*

Use the equation: F = (GMm)/(r^2)

F = force of gravity

G = gravitational constant (6.7x10^-11)

M = mass1 (2.5x10^30kg)

m = mass2 (1kg)

r = radius (7000m)

Plug it in: F = ((6.7x10^-11)(2.5x10^30)(1)) / (7000^2)

F = (1.675x10^20) / (4.9x10^7)

F = 3.4183673x10^12

F = 3.42N

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Outside a spherically symmetric charge distribution of net charge Q, Gauss's law can be used to show that the electric field at
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Q at the center of the distribution.

Explanation:

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What injuries could occur due to incorrect movement and handling of equipment with any sport or activity? (Select ALL that apply
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3 years ago
At a height of ten meters above the surface of a freshwater lake, a sound pulse is generated. The echo from the bottom of the la
konstantin123 [22]

Answer:

Explanation:

Velocity of sound in air at 20 degree = 343 m/s

Velocity of sound in water at 20 degree = 1470 m/s

Time taken in to and fro movement in air

=( 2 x 10) / 343 = 0.0583 s

Rest of the time  is

.171 - .0583 = .1127 s

This time is taken to cover distance in water. If d be the depth of lake

2d / velocity = time taken

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5 0
3 years ago
A 20 kg wagon is rolling to the right across a floor. A person attempts to catch and stop the crate and applies a force of 70 N,
Serggg [28]

Answer:

1.736m/s²

Explanation:

According to Newton's second law;

\sum F_x = ma_x\\

Fm - Ff = ma_x\\ where;

Fm is the moving force = 70.0N

Ff is the frictional force acting on the body

Ff = \mu R\\Ff = \mu mg\\

\mu is the coefficient of friction

m is the mass of the object

g is the acceleration due to gravity

a is the acceleration/deceleration

The equation becomes;

Fm - Ff = ma_x\\Fm - \mu mg = ma\\

Substitute the given parameters

Fm - \mu mg = ma\\70 - 0.18(20)(9.8) = 20a\\70-35.28 = 20a\\34.72 = 20a\\a = \frac{34.72}{20}\\a =  1.736m/s^2\\

Hence the deceleration rate of the wagon as it is caught is 1.736m/s²

7 0
3 years ago
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