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3 years ago

The mass of the sun is 2*10^30 kg and its radius is

Physics

1 answer:

Marta_Voda [28]3 years ago

4 0

Explanation:

Distance d=1.5×108 km=1.5×1011 m

Mass of the sun, m=2×1030 kg

Mass of the earth, M=6×1024 kg

Force of gravitation, F=G×d2m×M

F=6.7×10−11×(1.5×1011)22×1030×6×1024=3.57×1022 N

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A 37 N block rests on a horizontal surface. The coefficients of static and kinetic friction between the surface and the block ar

Ne4ueva [31]

Answer:

The frictional force acting on the block is 14.8 N.

Explanation:

Given that,

Weight of block = 37 N

Coefficients of static = 0.8

Kinetic friction = 0.4

Tension = 24 N

We need to calculate the maximum friction force

Using formula of friction force

$f=\mu mg$

Put the value into the formula

$f=0.8\times37$

$f=29.6\ N$

So, the tension must exceeds 29.6 N for the block to move

We need to calculate the frictional force acting on the block

Using formula of frictional force

$f = \mu N$

Put the value in to the formula

$f=0.4\times37$

$f=14.8\ N$

Hence, The frictional force acting on the block is 14.8 N.

6 0

3 years ago

A force of 20 N is exerted on a box with a mass of 15 kg. if friction exerts a force of 4 N on the box, at what rate does the bo

MakcuM [25]

Answer:

1.06 metres per second squared

Explanation:

since friction acts against foward force

20 N - 4 N = 16 N

use Newtons 2nd law F=ma Solve for a:

a= F÷m

= 16 ÷ 15

= 1.06 metres per second squared

3 0

3 years ago

If a reaction starts with 30 grams how many should it end with?

GrogVix [38]

30 grams because of conservation

6 0

3 years ago

Read 2 more answers

What is the process called when the moon begins to fade from a full moon to the new moon?

satela [25.4K]

Answer: Waning

Explanation: Not much explanation for this

8 0

3 years ago

A singly charged ion of 7Li (an isotope of lithium which lost only one electron) has a mass of 1.16 ×10^-26 kg. It is accelerate

MaRussiya [10]

Explanation:

It is given that,

Mass of lithium, $m=1.16\times 10^{-26}\ kg$

It is accelerated through a potential difference, V = 224 V

Uniform magnetic field, B = 0.724 T

Applying the conservation of energy as :

$\dfrac{1}{2}mv^2=qV$

$v=\sqrt{\dfrac{2qV}{m}}$

q is the charge on an electron

$v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\ C\times 224\ V}{1.16\times 10^{-26}\ kg}}$

v = 78608.58 m/s

$v=7.86\times 10^4\ m/s$

To find the radius of the ion's path in the magnetic field. The centripetal force is balanced by the magnetic force as :

$qvB=\dfrac{mv^2}{r}$

$r=\dfrac{mv}{qB}$

$r=\dfrac{1.16\times 10^{-26}\ kg\times 7.86\times 10^4\ m/s}{1.6\times 10^{-19}\ C\times 0.724\ T}$

r = 0.0078 meters

So, the radius of the path of the ion is 0.0078 meters. Hence, this is the required solution.

4 0

3 years ago