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gizmo_the_mogwai [7]
2 years ago
13

What are the factor that affect gravity​

Physics
1 answer:
valentinak56 [21]2 years ago
6 0

Explanation:

The factors that affect gravity are as follows:

1. mass of body

2. acceleration

<em>Keep</em><em> </em><em>smiling </em><em>and</em><em> </em><em>hope</em><em> </em><em>u</em><em> </em><em>are</em><em> </em><em>satisfied</em><em> </em><em>with</em><em> </em><em>my</em><em> </em><em>answer</em><em>.</em><em>Have</em><em> </em><em>a</em><em> </em><em>good</em><em> </em><em>day</em><em> </em><em>:</em><em>)</em>

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According to the article "Nuclear shapes" by Renee Lucas the nucleus's shape is mainly modified by vibrational and rotational features happening within the cell. According to the article if i read correctly "near closed shells spherical shapes prevail, while between closed shells the large number of valence nucleons in orbit with large particle angular momentum leads to nuclei with large deformations leading them to not only maintain its shape but also alloying it to work.
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Two nuclei join to form a larger nucleus during which process
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A 42.0-kg parachutist is moving straight downward with a speed of 3.85 m/s. (a) If the parachutist comes to rest with constant a
RideAnS [48]

Answer:

-414.96 N

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-3.85^2}{2\times 0.75}\\\Rightarrow a=-9.88\ m/s^2

F=ma\\\Rightarrow F=42\times -9.88\\\Rightarrow F=-414.96\ N

The force the ground exerts on the parachutist is -414.96 N

If the distance is shorter than 0.75 m then the acceleration will increase causing the force to increase

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3 years ago
A physics student spends part of her day walking between classes or for recreation, during which time she expends energy at an a
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2 years ago
An isolated charged point particle produces an electric field with magnitude E at a point 2 m away. At a point 1 m from the part
guajiro [1.7K]

Explanation:

The electric field at a distance r from the charged particle is given by :

E=\dfrac{kq}{r^2}

k is electrostatic constant

if r = 2 m, electric field is given by :

E_1=\dfrac{kq}{(2)^2}\\\\=\dfrac{kq}{4}\ .....(1)

If r = 1 m, electric field is given by :

E_2=\dfrac{kq}{r_2^2}\\\\=\dfrac{kq}{1}\ ....(2)

Dividing equation (1) and (2) we get :

\dfrac{E_1}{E_2}=\dfrac{\dfrac{kq}{4}}{kq}\\\\\dfrac{E_1}{E_2}=\dfrac{1}{4}\\\\E_2=4\times E_1

So, at a point 1 m from the particle, the electric field is 4 times of the electric field at a point 2 m.

4 0
2 years ago
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