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fgiga [73]
3 years ago
5

What is the relation between the gravitational force acting between any two bodies object with their masses and the distance bet

ween their centres?​
Physics
1 answer:
LenaWriter [7]3 years ago
7 0

Answer:

The strength of the gravitational force between two objects depends on two factors, mass and distance. the force of gravity the masses exert on each other. ... increases, the force of gravity decreases. If the distance is doubled, the force of gravity is one-fourth as strong as before.

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Why do metals have similar properties?
ra1l [238]
Core electrons are also referred as non-valence electrons. Two different elements have similar chemical properties when they have the same number of valence electrons in their outermost energy level. Elements in the same column of the Periodic Table have similar chemical propertie
5 0
3 years ago
Explain Bernoulli's Principle, air pressure, Down Wash, Angle of Attack, and vortices simply.
lord [1]

just listen I am also confused in this question if you get the answer so please just tell me also

3 0
3 years ago
2 kg<br> Use 10 m/s2 for g
Iteru [2.4K]

Answer:

See below

Explanation:

At point A    the PE = mgh = 2 * 10 * 1 = 20 J

  at point B, all of the PE , 20 J , is converted to Kinetic Energy

KE = 1/2 m v^2

20 = 1/2 (2)(v^2 )

20 = v^2     v = sqrt 20 = 4.47 m/s

for the friction part

 vf = vo t  + 1/2 a t^2      vf = final velocity = 0 (stopped)

                                       vo = original velocity = 4.47 m/s  

                                         a = -1 m/s^2

  0 = 4.47 t + 1/2 (-1) t^2

            - .5t^2  + 4.47 t = 0

                 t ( -.5t+ 4.47) = 0    shows t =  4.47/.5 = 8.9 seconds

6 0
1 year ago
1. What happens to the current in a series circuit as it moves through each component? a. The current stays the same throughout
Setler [38]

Answer: sorry here’s the answers, I didn’t feel like typing it all

Explanation:

3 0
2 years ago
A body is traveling at 5.0 m/s along the positive direction of an x axis; no net force acts on the body. An internal explosion s
loris [4]

To solve this problem it is necessary to apply the concepts related to the conservation of momentum and conservation of kinetic energy.

By definition kinetic energy is defined as

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

Where,

m = Mass

v = Velocity

On the other hand we have the conservation of the moment, which for this case would be defined as

m*V_i = m_1V_1+m_2V_2

Here,

m = Total mass (8Kg at this case)

m_1=m_2 = Mass each part

V_i = Initial velocity

V_2 = Final velocity particle 2

V_1 = Final velocity particle 1

The initial kinetic energy would be given by,

KE_i=\frac{1}{2}mv^2

KE_i = \frac{1}{2}8*5^2

KE_i = 100J

In the end the energy increased 100J, that is,

KE_f = KE_i KE_{increased}

KE_f = 100+100 = 200J

By conservation of the moment then,

m*V_i = m_1V_1+m_2V_2

Replacing we have,

(8)*5 = 4*V_1+4*V_2

40 = 4(V_1+V_2)

V_1+V_2 = 10

V_2 = 10-V_1(1)

In the final point the cinematic energy of EACH particle would be given by

KE_f = \frac{1}{2}mv^2

KE_f = \frac{1}{2}4*(V_1^2+V_2^2)

200J=\frac{1}{2}4*(V_1^2+V_2^2)(2)

So we have a system of 2x2 equations

V_2 = 10-V_1

200J=\frac{1}{2}4*(V_1^2+V_2^2)

Replacing (1) in (2) and solving we have to,

200J=\frac{1}{2}4*(V_1^2+(10-V_1)^2)

PART A: V_1 = 10m/s

Then replacing in (1) we have that

PART B: V_2 = 0m/s

8 0
3 years ago
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