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Ghella [55]
2 years ago
14

State the factor that affect the gravitational force​

Physics
2 answers:
Over [174]2 years ago
5 0

Answer:

<em>The force of gravity depends directly upon the masses of the two objects, and inversely on the square of the distance between them. This means that the force of gravity increases with mass, but decreases with increasing distance between objects.</em>

slamgirl [31]2 years ago
4 0

Answer:Mass and distance are the two factors that affect the gravitational force between any two objects.

mark me as brainliest please

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A pendulum oscillates 12 times in 4 seconds. what is the length of the pendulum?
seraphim [82]

Answer:

L = 2.8 cm

Explanation:

Period T = 4 / 12 = 1/3 s

T = 2π√(L/g)

L = (T/2π)²g

L = ((1/3)/2π)²9.8 = 0.02758... ≈ 2.8 cm

6 0
2 years ago
Is there a such thing as an elastic collision? Why do we study the case of the elastic collision in Physics?
fredd [130]
Yes there is an elastic collision in physics its when a collision occurs but no kinetic energy is loss. We study them in order to understand how to conserve momentum.
6 0
2 years ago
A square plate of edge length 9.0 cm and negligible thickness has a total charge of 6.90 10-6 C. Estimate the magnitude E of the
SVETLANKA909090 [29]

Answer:

E= 4.35*10^6 N/C

Explanation:

Let's find the area charge density of the plate

α= 6.9*10^-6/9*10^-2 = 7.7*10^-5C/m2

Now we can calculate the electric field just of the plate

E =α/2e =7.7*10^-5/2*8.85*10^-12 = 4.35*10^6 N/C

7 0
2 years ago
The mean time between collisions for electrons in room temperature copper is 2.5 x 10-14 s. What is the electron current in a 2
Darya [45]

Answer:

1.87 A

Explanation:

τ = mean time between collisions for electrons = 2.5 x 10⁻¹⁴ s

d = diameter of copper wire = 2 mm = 2 x 10⁻³ m

Area of cross-section of copper wire is given as

A = (0.25) πd²

A = (0.25) (3.14) (2 x 10⁻³)²

A = 3.14 x 10⁻⁶ m²

E = magnitude of electric field = 0.01 V/m

e = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

m = mass of electron = 9.1 x 10⁻³¹ kg

n = number density of free electrons in copper = 8.47 x 10²² cm⁻³ = 8.47 x 10²⁸ m⁻³

i = magnitude of current

magnitude of current is given as

i = \frac{Ane^{2}\tau E}{m}

i = \frac{(3.14\times 10^{-6})(8.47\times 10^{28})(1.6\times 10^{-19})^{2}(2.5\times 10^{-14}) (0.01)}{(9.1\times 10^{-31})}

i  = 1.87 A

4 0
2 years ago
A car travels initially at 24 m/s, until it enters the highway. If the car accelerates at 4 m/s^2 for a 96 meters, what is the c
marishachu [46]
  • initial velocity=u=24m/s
  • Acceleration=a=4m/s^2
  • Distance=s=96m
  • Final velocity=v

Using 3rd equation of kinematics

\boxed{\Large{\sf v^2-u^2=2as}}

\\ \Large\sf\longmapsto v^2=u^2+2as

\\ \Large\sf\longmapsto v^2=24^2+2(4)(96)

\\ \Large\sf\longmapsto v^2=576+768

\\ \Large\sf\longmapsto v^2=1344

\\ \Large\sf\longmapsto v=\sqrt{1344}

\\ \Large\sf\longmapsto v=36.6m/s

3 0
3 years ago
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