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3 years ago

2. Place the pairs in order from weakest to strongest gravitational force. (Brainpop Gravity Challenge)

Physics

1 answer:

FromTheMoon [43]3 years ago

3 0

Answer:

B, A, C, D

Explanation:

i hope it helps. im just kinda guessing.

good luck!°°°°°⁰⁰⁰⁰⁰₀₀₀₀₀oo00OO

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You must exert a force of 4.5 N on a book to slide it across a table. If you do 2.7J

hoa [83]

Answer: 0.6m

Explanation:

Given that:

force = 4.5 N

Work done = 2.7J

Distance moved by the book = ?

Since work is done when force is applied on an object over a distance, apply the formula:

work = force x distance

2.7J = 4.5N x distance

Distance = (2.7J / 4.5N)

Distance = 0.6 m

Thus, the book was moved 0.6 metres far

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3 years ago

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A) it occurs when earth is between the sun and the moon

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3 years ago

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3 years ago

DEFINE UNIFORM AND NON UNIFORM VELOCITY

Inessa05 [86]

Explanation:

Uniform velocity is when an object goes an equal amount of space in an equal amount of time whereas non uniform velocity is when the object covers an unequal amount of distance in an equal amount of time.

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3 years ago

A stone with a mass of 0.100kg rests on a frictionless, horizontal surface. A bullet of mass 2.50g traveling horizontally at 500

jolli1 [7]

Answer:

Explanation:

Given that:

mass of stone (M) = 0.100 kg

mass of bullet (m) = 2.50 g = 2.5 ×10 ⁻³ kg

initial velocity of stone ( $u_{stone}$ ) = 0 m/s

Initial velocity of bullet ( $u_{bullet}$ ) = (500 m/s)i

Speed of the bullet after collision ( $v_{bullet}$ ) = (300 m/s) j

Suppose we represent $(v_{stone})$ to be the velocity of the stone after the truck, then:

From linear momentum, the law of conservation can be applied which is expressed as:

$m*u_{bullet} + M*{u_{stone}}= mv_{bullet}+Mv_{stone}$

$(2.50*10^{-3} \ kg) (500)i+0 = (2.50*10^{-3} \ kg)(300 \ m/s)j + (0.100 \ kg)v_{stone}$

$(2.50*10^{-3} \ kg) (500)i- (2.50*10^{-3} \ kg)(300 \ m/s)j= (0.100 \ kg)v_{stone}$

$v_{stone}= (1.25\ kg.m/s)i-(0.75\ kg m/s)j$

$v_{stone}= (12.5\ m/s)i-(7.5\ m/s)j$

∴

The magnitude now is:

$v_{stone}=\sqrt{ (12.5\ m/s)^2-(7.5\ m/s)^2}$

$\mathbf{v_{stone}= 14.6 \ m/s}$

Using the tangent of an angle to determine the direction of the velocity after the struck;

Let θ represent the direction:

$\theta = tan^{-1} (\dfrac{-7.5}{12.5})$

$\mathbf{\theta = 30.96^0 \ below \ the \ horizontal\ level}$

5 0

3 years ago