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3 years ago

How do you think the gravitational force between two objects will change if the distance between the objects is doubled

1 answer:

4 0

We Know, F = Gm₁m₂ / r²
When, distance is doubled, it would be:
F' = Gm₁m₂ / (2r)²
F' = Gm₁m₂ / 4r

So, we can see, Force has decreased by 4 times.

Hope this helps!

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Momentum of the 2 kg mass moving with velocity 10 m/s is *

gulaghasi [49]

20 kg*m/s because there is 2 kg mass and 10 m/s so you can multiply.

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4 years ago

The first Leyden jar was probably discovered by a German clerk named E. Georg von Kleist. Because von Kleist was not a scientist

skelet666 [1.2K]

Answer:

a) q = 4.47 10⁻⁵ C

b) ΔV = 4.47 10⁴ V

Explanation:

A Leyden bottle works as a condenser that accumulates electrical charge, so we can use the formula of the energy stored in a capacitor

U = Q² / 2C

Q = √ (2UC)

let's reduce the magnitudes to the SI system

c = 1 nF = 1 10⁻⁹ F

let's calculate

q = √ (2 1 10⁻⁹-9)

q = 0.447 10⁻⁴ C

q = 4.47 10⁻⁵ C

b) for the potential difference we use

C = Q / ΔV

ΔV = Q / C

ΔV = 4.47 10⁻⁵ / 1 10⁻⁹

ΔV = 4.47 10⁴ V

8 0

3 years ago

A) What is the meaning of specific heat capacity ?

dem82 [27]

Explanation:

the heat required to raise the temperature of the unit mass of a given substance by a given amount (usually one degree).

3 0

3 years ago

Jeremy walked at an average rate of speed of 3 km/h. How far had Jeremy

makvit [3.9K]

D is the answer and also I love u

8 0

3 years ago

Read 2 more answers

A cyclist traveling at 30.0m/s along a straight road comes uniformly to stop in 5.00s. Determine the stopping acceleration, the

Amiraneli [1.4K]

Answer:

I. Stopping acceleration = 6 m/s²

II. Stopping distance, S = 75 meters

Explanation:

Given the following data;

Final velocity = 30 m/s

Time = 5 seconds

To find the stopping acceleration;

Mathematically, acceleration is given by the equation;

$Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}$

Substituting into the equation;

$a = \frac{30 - 0}{5}$

$a = \frac{30}{5}$

Acceleration = 6 m/s²

II. To find the stopping distance, we would use the third equation of motion;

$V^{2} = U^{2} + 2aS$

Where;

V represents the final velocity measured in meter per seconds.

U represents the initial velocity measured in meter per seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

30² = 0² + 2*6*S

900 = 12S

S = 900/12

S = 75 meters

7 0

3 years ago