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svetlana [45]
3 years ago
13

(b) A car of mass 3000 kg travels at a constant velocity of 5.0 m/s.

Physics
1 answer:
Tamiku [17]3 years ago
4 0

Answer:

16.7 s

Explanation:

T= <u>Vf - Vo</u>          a= <u>F</u>

         a                    m

4,500 / 3000 = 1.5 (a)

30 - 5 / 1.5(a) = 16.7 s      

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Zepler [3.9K]
I believe it would be weight. mass never changes.
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2 years ago
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A hockey ball accelerates from 0.m/s to 25m/s in 0.05 seconds what is the acceleration of the ball ?
S_A_V [24]

Answer:

500

Explanation:

25/0.05 .

I think this ?

3 0
3 years ago
Is there a such thing as an elastic collision? Why do we study the case of the elastic collision in Physics?
fredd [130]
Yes there is an elastic collision in physics its when a collision occurs but no kinetic energy is loss. We study them in order to understand how to conserve momentum.
6 0
3 years ago
A satellite that goes around the earth once every 24 hours (86,400 s) is called a geosynchronous satellite. If a geosynchronous
Olegator [25]

Answer:

42244138.951 m

Explanation:

G = Gravitational constant = 6.667 × 10⁻¹¹ m³/kgs²

r = Radius of orbit from center of earth

M = Mass of Earth = 5.98 × 10²⁴ kg

m = Mass of Satellite

The satellite revolves around the Earth at a constant speed

Speed = Distance / Time

The distance is the perimeter of the orbit

v=\frac{2\pi \times r}{24\times 3600}

The Centripetal force of the satellite is balanced by the universal gravitational force

m\frac{v^2}{r}=\frac{GMm}{r^2}\\\Rightarrow \frac{\left(\frac{2\pi \times r}{24\times 3600}\right)^2}{r}=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{r^2}\\\Rightarrow \left(\frac{2\pi \times r}{24\times 3600}\right)^2=6.667\times 10^{-11}\times 5.98\times 10^{24}\\\Rightarrow r^3=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\\\Rightarrow r=\left(\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\right)^{\frac{1}{3}}\\\Rightarrow r=42244138.951\ m

The radius as measured from the center of the Earth) of the orbit of a geosynchronous satellite that circles the earth is 42244138.951 m

6 0
3 years ago
"My distance to the center of the earth is about 4000 miles when I am on the surface. If I go to a height of 8000 miles above th
lord [1]

Given,

Distance from the surface to the center of the earth, d=4000 miles

Distance from the center to you at a height of 8000 miles, a= 8000+4000=12000 miles

The gravitational force acting on a person at the surface is equal to his weight.

From Newton's Universal Law of Gravitation, the gravitational force is

F=\frac{G\times M\times m}{r^2}

Where G is the gravitational constant, M is the mass of the earth, m is the mass of the object/person, r is the distance between the center of the earth and the object/person

At the surface, this force is equal to the weight of the person, W=mg

i.e.

F_s=\frac{G\times M\times m}{d^2}=W

On substituting the of d,

W=\frac{\text{GMm}}{4000^2}

At a height of 8000 miles from the surface, the gravitational force is equal to,

F_a=\frac{GMm}{12000^2}

On dividing the above two equations,

\frac{F_a}{W}=\frac{4000^2^{}}{12000^2}=\frac{1}{9}

Therefore,

F_a=\frac{1}{9}W

Therefore at a height of 8000 miles above the surface of the earth, the force of gravity becomes 1/9 time your weight.

5 0
1 year ago
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