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3 years ago

7

Identify whether the following statements describe a change in acceleration. Explain your response answer them all plz

2 answers:

weqwewe [10]3 years ago

5 0

Answer:c

Explanation:it picking up speed

stepladder [879]3 years ago

3 0

I got C
Hoped it was help byee:)

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Someone please help its a simple power problem.

SOVA2 [1]

Well 200 doubled or (x2)=400 if that’s what it means

7 0

2 years ago

In order to change power, what else must be changed?

Genrish500 [490]

<span>In order to change power, current or voltage should also be changed. Voltage is an electromotive force, and also the quantitative expression that shows the potential difference of the two points charged in an electrical field. So, before power will take place, it would always be best to change also the voltage.</span>

6 0

3 years ago

Read 2 more answers

If the radius of the sun is 7.001×105 km, what is the average density of the sun in units of grams per cubic centimeter? The vol

xenn [34]

Answer:

Average density of Sun is 1.3927 $\frac{g}{cm}$ .

Given:

Radius of Sun = 7.001 × $10^{5}$ km = 7.001 × $10^{10}$ cm

Mass of Sun = 2 × $10^{30}$ kg = 2 × $10^{33}$ g

To find:

Average density of Sun = ?

Formula used:

Density of Sun = $\frac{Mass of Sun}{Volume of Sun}$

Solution:

Density of Sun is given by,

Density of Sun = $\frac{Mass of Sun}{Volume of Sun}$

Volume of Sun = $\frac{4}{3} \pi r^{3}$

Volume of Sun = $\frac{4}{3} \times 3.14 \times [7.001 \times 10^{10}]^{3}$

Volume of Sun = 1.436 × $10^{33}$ $cm^{3}$

Density of Sun = $\frac{ 2\times 10^{33} }{1.436 \times 10^{33} }$

Density of Sun = 1.3927 $\frac{g}{cm}$

Thus, Average density of Sun is 1.3927 $\frac{g}{cm}$ .

4 0

3 years ago

In a pool game, the cue ball, which has an initial speed of 3.0 m/s, make an elastic collision with the eight ball, which is ini

zhenek [66]

Explanation:

Given

initial speed(u)=3 m/s

mass of each ball is m

Let the cue ball is moving in x direction initially

In elastic collision Energy and momentum is conserved

Let u be the initial velocity and $v_1 , v_2$ be the final velocity of 8 ball and cue ball respectively

$\frac{mu^2}{2}+0=\frac{mv^2_1}{2}+\frac{mv^2_2}{2}$

The angle after which cue ball is deflected is given by

$\theta _1=90-40=50^{\circ}$

Conserving momentum in x direction

$mu=mv_1cos40+mv_2cos50$

$3=v_1cos40+v_2cos50$

Along Y axis

$0+0=v_1sin40-v_2sin50$

$v_1sin40=v_2sin50$

substitute the value of $v_1$

we get $v_2=1.912 m/s$

$v_1=2.27 m/s$

5 0

3 years ago

What is the weight on mass

Gre4nikov [31]

Answer:

6.39×10^23 kg is the weight on mass

3 0

2 years ago