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8_murik_8 [283]

3 years ago

14

How to know if a position time graph table is balanced or unbalanced

1 answer:

Goshia [24]3 years ago

3 0

When balanced forces act on an object at rest, the object will not move. If you push against a wall, the wall pushes back with an equal but opposite force. Neither you nor the wall will move. Forces that cause a change in the motion of an object are unbalanced forces.

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What is the distance from one crest to the next crest?

ella [17]

<span>The distance between wave crests is called wavelength. It is a characteristic shared by waves of all kinds, including ocean waves and sound waves. Wavelength is measured from the highest point, or summit, of one wave's crest to the summit of the next wave's <span>crest</span></span>

<span><span>hope this helps</span></span>

3 0

3 years ago

A student is creating a model of a concave lens. The diagram shows her incomplete model.

valentinak56 [21]

Answer:

D

Explanation:

I’m pretty sure it’s correct but I don’t really know. Just trying to pass science

8 0

3 years ago

A sports car accelerates from rest for 5 seconds reaching a velocity of 23.0 m/s.

denis-greek [22]

Answer:

<h2>4.6 m/s²</h2>

Explanation:

The acceleration of an object given it's velocity and time taken can be found by using the formula

<h3> $a = \frac{v - u}{t} \\$ </h3>

where

v is the final velocity

u is the initial velocity

t is the time taken

a is the acceleration

Since the body is from rest u = 0

From the question we have

$a = \frac{23 - 0}{5} = \frac{23}{5} \\$

We have the final answer as

<h3>4.6 m/s²</h3>

Hope this helps you

4 0

3 years ago

12. AABC is a right triangle. If AB = 3 and AC = 7, find BC. Leave your answer in simplest radical form.

Dennis_Churaev [7]

Answer: A 2 square root 3

Explanation:

4 0

3 years ago

The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles

belka [17]

Answer:

$F_a=5.67\times 10^{-5}\ N$

<u /> $F_b=3.49\times 10^{-5}\ N$

$F_c=9.16\times 10^{-5}\ N$

Explanation:

Given:

mass of particle A, $m_a=363\ kg$

mass of particle B, $m_b=517\ kg$

mass of particle C, $m_c=154\ kg$

All the three particles lie on a straight line.

Distance between particle A and B, $x_{ab}=0.5\ m$

Distance between particle B and C, $x_{bc}=0.25\ m$

Since the gravitational force is attractive in nature it will add up when enacted from the same direction.

<u>Force on particle A due to particles B & C:</u>

$F_a=G. \frac{m_a.m_b}{x_{ab}^2} +G. \frac{m_a.m_c}{(x_{ab}+x_{bc})^2}$

$F_a=6.67\times 10^{-11}\times (\frac{363\times 517}{0.5^2}+\frac{363\times 154}{(0.5+0.25)^2})$

$F_a=5.67\times 10^{-5}\ N$

<u>Force on particle C due to particles B & A:</u>

<u /> $F_c=G.\frac{m_c.m_b}{x_{bc}^2} +G.\frac{m_c.m_a}{(x_{ab}+x_{bc})^2}$ <u />

$F_c=6.67\times 10^{-11}\times (\frac{154\times 517}{0.25^2}+\frac{154\times 363}{(0.25+0.5)^2} )$

$F_c=9.16\times 10^{-5}\ N$

<u>Force on particle B due to particles C & A:</u>

<u /> $F_b=G.\frac{m_b.m_c}{x_{bc}^2} -G.\frac{m_b.m_a}{x_{ab}^2}$ <u />

<u /> $F_b=6.67\times 10^{-11}\times (\frac{517\times 154}{0.25^2}-\frac{517\times 363}{0.5^2} )$ <u />

<u /> $F_b=3.49\times 10^{-5}\ N$ <u />

3 0

3 years ago