All categories

3 years ago

What is physics mean and branch

1 answer:

5 0

Physics is a branch of science that deals with the properties of matter and energy and relationship between them it also tries to explain the material world of the natural phenomenon of the universe the scope of physics is very wide and vast

You might be interested in

During energy transformation, all energy systems: are

Tom [10]

B: Energy lose

i say this because in order to change they lose energy.

3 0

3 years ago

Read 2 more answers

The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9

kirill [66]

Answer:

Speed of the car 1 = $V_1=8.98m/s$

Speed of the car 2 = $V_2=17.96m/s$

Explanation:

Given:

Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)

mathematically,

M₁ = 2M₂

Kinetic Energy of the car 1 = Half the kinetic energy of the car 2

KE₁ = 0.5 KE₂

Now, the kinetic energy for a body is given as

$KE =\frac{1}{2}mv^2$

where,

m = mass of the body

v = velocity of the body

thus,

$\frac{1}{2}M_1V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

$\frac{1}{2}2M_2V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

$2M_2V_1^2=0.5\times M_2V_2^2$

$2V_1^2=0.5\times V_2^2$

$4V_1^2= V_2^2$

$2V_1= V_2$ .................(1)

also,

$\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2$

$\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2$

$2(V_1+9.0)^2=(2V_1+9.0)^2$

$\sqrt{2}(V_1+9.0)=(2V_1+9.0)$

$(\sqrt{2}V_1+ \sqrt{2}\times 9.0)=(2V_1+9.0)$

$(\sqrt{2}V_1+ 12.72)=(2V_1+9.0)$

$(2V_1-\sqrt{2}V_1)=(12.72-9.0)$

$(0.404V_1)=(3.72)$

$V_1=8.98m/s$

and, from equation (1)

$V_2=2\times 8.98m/s = 17.96m/s$

Hence,

Speed of car 1 = $V_1=8.98m/s$

Speed of car 2 = $V_2=17.96m/s$

8 0

2 years ago

Read 2 more answers

24 How long does it take for the Earth to orbit the Sun?

Zarrin [17]

Answer:

365 days

So The Final Answer is a Year

D is the answer

Explanation:

7 0

2 years ago

Read 2 more answers

50POINTS! Find the orbital speed of a satellite in a circular orbit 1700km above the surface of the Earth. M_earth 5.97e24kg, r_

ivolga24 [154]

<h2>Answer: 7020.117 m/s</h2>

Explanation:

The velocity of a satellite describing a circular orbit is<u> constant</u> and defined by the following expression:

$V=\sqrt{G\frac{M}{R}}$ (1)

Where:

$G=6.674({10}^{-11})\frac{N{m}^{2}}{{kg}^{2}}$ is the gravity constant

$M_{Earth}=5.97{10}^{24}kg$ the mass of the massive body around which the satellite is orbiting, in this case, the Earth .

$R=r_{Earth}+h=8080000m$ the radius of the orbit (measured from the center of the planet to the satellite).

This means the radius of the orbit is equal to <u>the sum</u> of the average radius of the Earth $r_{Earth}$ and the altitude of the satellite above the Earth's surface $h$ .

Note this orbital speed, as well as orbital period, does not depend on the mass of the satellite. It depends on the mass of the massive body (the Earth).

Now, rewriting equation (1) with the known values:

$V=\sqrt{(6.674({10}^{-11})\frac{N{m}^{2}}{{kg}^{2}})\frac{5.97{10}^{24}kg}{8080000m}}$