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liraira [26]
3 years ago
5

10) If the mass 2m, the left mass

Physics
1 answer:
Romashka-Z-Leto [24]3 years ago
7 0

Answer:

F = \frac{-Gm_{1}m_{2} }{r^{2} }.

Explanation:

Gravitational force between two objects of masses m_{1},  m_{2} kept at a distance r is given by the formula

F = \frac{-Gm_{1}m_{2} }{r^{2} }

Here ,m_{1} = 2m

         m_{2} = \frac{m}{2}

         

Thus , F = \frac{-G.2m.\frac{m}{2} }{r^{2} }

          F = \frac{-Gm_{1}m_{2} }{r^{2} }.

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The football player running toward the goal line has
soldi70 [24.7K]

Kinetic energy.

Kinetic energy is the type of energy observed in moving objects. In this case the football player is running, ie moving, so he/she must have kinetic energy.

5 0
3 years ago
What is the energy of moving electrical charges
Simora [160]
The energy of moving electrical charges is Electrical energy



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3 0
3 years ago
A ball is thrown vertically upward. What is its acceleration right before it hits the earth?
Bezzdna [24]
<span>The entire time the ball is in the air, its acceleration is 9.8 m/s2 down provided this occurs on the surface of the Earth. Note that the acceleration can be either 9.8 m/s2 or -9.8 m/s2.
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7 0
3 years ago
Read 2 more answers
The electron gun in a television tube is used to accelerate electrons with mass 9.109 × 10−31 kg from rest to 3 × 107 m/s within
zaharov [31]

Answer:

Electric field, E = 40608.75 N/C

Explanation:

It is given that,

Mass of electrons, m=9.1\times 10^{-31}\ kg

Initial speed of electron, u = 0

Final speed of electrons, v=3\times 10^7\ m/s

Distance traveled, s = 6.3 cm = 0.063 m

Firstly, we will find the acceleration of the electron using third equation of motion as :

a=\dfrac{v^2-u^2}{2s}

a=\dfrac{(3\times 10^7)^2}{2\times 0.063}

a=7.14\times 10^{15}\ m/s^2

Now we will find the electric field required in the tube as :

ma=qE

E=\dfrac{ma}{q}

E=\dfrac{9.1\times 10^{-31}\times 7.14\times 10^{15}}{1.6\times 10^{-19}}

E = 40608.75 N/C

So, the electric field required in the tube is 40608.75 N/C. Hence, this is the required solution.

3 0
3 years ago
You throw a stone horizontally at a speed of 5.0 m/s from the top of a cliff that is 78.4 m high.
Monica [59]

a)You throw a stone horizontally at a speed of 5.0 m/s from the top of a cliff that is 78.4 m high.

from above statement we got

height = 78.4 m

since the ball is thrown, so its vertical velocity would be zero

u = 0

taking g = 9.8m/s^2

now, using the equation of motion

h = ut + gt^2/2

now putting all the values in it

we got ,

78.4 = 9.8 * t^2/ 2

by solving we got,

t = 4 sec

b) now, since along the horizontal , no force acting and accelaration is zero so

R = ut , R is RANGE

R = 5 * 4

range =  20 m

c)  vertical components of the stone’s velocity just before it hits the ground = v sin θ =

horizontal   components of the stone’s velocity  just before it hits the ground = v cos θ

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4 0
1 year ago
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