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Assoli18 [71]
3 years ago
7

If the mass of the Earth somehow increased with no change in radius, your weight would

Physics
2 answers:
Margaret [11]3 years ago
5 0

Answer:

increase also

Explanation:

The weight of a person is equal to the gravitational pull exerted by the Earth on the person:

F=G\frac{mM}{R^2}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2} is the gravitational constant

M is the mass of the Earth

m is the mass of the person

R is the Earth's radius

We notice that the weight is directly proportional to the mass of the Earth. Therefore, if the mass of the Earth M increases, and the radius R does not change, the weight of the person increases as well.

Sever21 [200]3 years ago
5 0

<u>Answer</u>

Increases also.

The force due to gravity is given by,

F = GM.m/r²

Where G is a constant of proportionality

∴ F ∝ M.m/r²

When r remains constant, force due to gravity, F, will be;

F ∝ M.m

Where M is the mass of the earth and m is your mass

Since your mass does not change, we are going to have;

F ∝ M.

This means the weight F is directly proportional to the mass of the earth. when it increases the your weight also increases.

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A motorcycle starts from rest and has a constant acceleration. In a time interval t, it undergoes a displacement x and attains a
iren [92.7K]

Answer:

√(6ax)

Explanation:

Hi!

The question states that during a time t the motorcyle underwent a displacement x at constant acceleration a starting from rest, mathematically we can express it as:

x=(1/2)at^2

Then the we need to find the time t' for which the displacement is 3x

3x=(1/2)a(t')^2

Solving for t':

t'=√(6x/a)

Now, the velocity of the motorcycle as a function of time is:

v(t)=a*t

Evaluating at t=t'

v(t')=a*√(6x/a)=√(6*x*a)

Which is the final velocity

Have a nice day!

3 0
3 years ago
An electroscope is a fork-shaped device commonly used to detect the presence of charge. The tin leaves of an electroscope will s
velikii [3]

Answer:

All these is caused by the repulsion force.

Explanation:

The electroscope produces a series of electric charges that produce a repulsion force when is putted in contact with a electric charged object.

As the physics law mentions, two different forces are repealed, the electrocospe is charged negatively and the object positively, causing a repulsion force that avoids that both objects touch the other.

7 0
3 years ago
If the magnitude of a charge is half as much as another charge, but the force experienced is the same, then the electric field s
Kazeer [188]

Answer:

the electric field strength of this charge is two times the strength of the other charge

Explanation:

Using the relationship between electric field and the charge, which is inversely proportionality. Let the the magnitude of the first charge be Q and the respective electric field be E. It implies that;

E1/E2 = Q2/Q1

E2 = E1 x Q1/Q2

      = E x Q/ (Q/2)

       = 2E

8 0
2 years ago
Can i get help for the parts for the two questions? MATHPHYSSSS
Anestetic [448]

Explanation:

003 (part 1 of 2)

Pressure is force divided by area.

P = F / A

P = (117 kg × 9.8 m/s²) / (2 × (0.05 m)²)

P = 229,320 Pa

003 (part 2 of 2)

There are approximately 6895 Pa in 1 psi.

P = 229,320 Pa × (1 psi / 6895 Pa)

P = 33.3 psi

004 (part 1 of 2)

Since the collisions are elastic, the angle of reflection is the same as the angle of incidence (it bounces off at the same angle).

Impulse = change in momentum

F Δt = m Δv

F (36 s) = (300 × 0.003 kg) (5.2 sin 57° m/s − (-5.2 sin 57° m/s))

F = 0.218 N

004 (part 2 of 2)

Pressure is force over area.

P = F / A

P = 0.218 N / 0.712 m²

P = 0.306 N/m²

7 0
3 years ago
What is gravity’s role in tides?
kotegsom [21]

The relationship between the masses of the Earth, moon and sun and their distances to each other play critical roles in affecting tides

6 0
3 years ago
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