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tatuchka [14]
2 years ago
5

What are the properties of gravitational force

Physics
2 answers:
lys-0071 [83]2 years ago
7 0

Answer:

Gravitational force−Properties:

(1) It is a universal attractive force. It is directly proportional to the product of the masses of the two bodies.

(2) It obey inverse square law.

(3) It is the weakest force known in nature.

Irina18 [472]2 years ago
6 0

Explanation:

It obeys the inverse square law.

It is always attractive in nature.

It is a long range force. ...

The graviton is the field particle of gravitational force.

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Tcecarenko [31]
The answer of this question is B.
5 0
3 years ago
He generation of a magnetic field by an electric current is
USPshnik [31]

Answer: electromagnetism.

Explanation:The use of coils of wires produces a relationship between electricity and magnetism that gives us another magetism called electromagnetism.

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3 years ago
7. An 8 kg ball is travelling to the east at 10 ms', collides with a 2 kg ball travelling to the
Ymorist [56]

Answer:

The final velocity of the ball is 7m/s

Explanation:

M1=8kg,  V1 =10m/s , M2=2kg , V2=-5m/s

initial momentum before collison

m1v1+m2v2

=8×10 +2×(-5)  =80-10  = 70kg m/s

final momentum after collison

=(m1+m2)×v

=(8+2)×v

=10v

According to the law of conversion of momentum

initial momentum =final momentum

70=10v

10v=70

v=70/10

v=7m/s

3 0
3 years ago
To decrease the angle between the anterior surface of the foot and anterior surface of the lower leg is described as:
Westkost [7]

Answer:

dorsiflexion

Explanation:

To decrease the angle between the anterior surface of the foot and anterior surface of the lower leg is described as: dorsiflexion

7 0
3 years ago
Read 2 more answers
A girl rolls a ball up an incline and allows it to re- turn to her. For the angle and ball involved, the acceleration of the bal
zalisa [80]

Answer:

3.28 m

3.28 s

Explanation:

We can adopt a system of reference with an axis along the incline, the origin being at the position of the girl and the positive X axis going up slope.

Then we know that the ball is subject to a constant acceleration of 0.25*g (2.45 m/s^2) pointing down slope. Since the acceleration is constant we can use the equation for constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

X0 = 0

V0 = 4 m/s

a = -2.45 m/s^2 (because the acceleration is down slope)

Then:

X(t) = 4*t - 1.22*t^2

And the equation for speed is:

V(t) = V0 + a * t

V(t) = 4 - 2.45 * t

If we equate this to zero we can find the moment where it stops and begins rolling down, that will be the highest point:

0 = 4 - 2.45 * t

4 = 2.45 * t

t = 1.63 s

Replacing that time on the position equation:

X(1.63) = 4 * 1.63 - 1.22 * 1.63^2 = 3.28 m

To find the time it will take to return we equate the position equation to zero:

0 = 4 * t - 1.22 * t^2

Since this is a quadratic equation it will have to answers, one will be the moment the ball was released (t = 0), the other will eb the moment when it returns:

0 = t * (4 - 1.22*t)

t1 = 0

0 = 4 - 1.22*t2

1.22 * t2 = 4

t2 = 3.28 s

7 0
2 years ago
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