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4 years ago

HURRY!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

1 answer:

5 0

<span>Kinetic energy doesn't depend on a object's relative height.</span>

In short, Your Answer would be Option D

Hope this helps!

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A 4.0-kilogram block, initially at rest, is acted upon by an 8.0 newtons force for 10 seconds. How far, in meters, will the mass

Shtirlitz [24]

Answer:

Explanation:

5 0

3 years ago

During aerobic activity, if your heart rate is lower than the lower limit, you are ___________________.

soldier1979 [14.2K]

Not pushing yourself hard enough is the answer since your heart rate doesn't even hit your lower minimum.
Pushing yourself to the limit is at your max heart rate.
Just at the right spot is at your max heart rate.
Pushing yourself too hard is above your max heart rate.

4 0

4 years ago

A cyclist travels 20km in 4hrs.What speed did the cyclist cycle at?

Tanya [424]

Answer: 5 km/hr

Explanation:

speed= distance divided by time

20/4

= 5 km/hr

3 0

3 years ago

Read 2 more answers

Derive the following equations of motion

xz_007 [3.2K]

Answer:

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<h3>a. Let us assume a body has initial velocity 'u' and it is subjected to a uniform acceleration 'a' so that the final velocity 'v' after a time interval 't'. Now, By the definition of acceleration, we have:</h3>

$a = \frac{v - u}{t} \\ or \: at = v - u \\ v = u + at \:$

It is first equation of motion.

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<h3>b. Let us assume a body moving with an initial velocity 'u'. Let it's final body 'v' after a time interval 't' and the distance travelled by the body becomes 's' then we already have,</h3>

$v = u + at...........(i) \\ s = \frac{u + v}{2} \times t.........(ii)$

Putting the value of v from the equation (i) in equation (ii), we have,

$s= \frac{u + (u + at)}{2} \times t \: \: \\ or \: s = \frac{(2u + at)t}{2} \\ or \: s = \frac{2ut + a {t}^{2} }{2} \\ s = ut + \frac{1}{2} a {t}^{2}$

It is third equation of motion.

________________________________

<h3>c. Let us assume a body moving with an initial velocity 'u'. Let it's final velocity be 'v' after a time and the distance travelled by the body be 's'. We already have,</h3>

$v = u + at.....(i) \\ s = \frac{u + v}{2} \times t......(ii) \\$

$v = u + at \\ or \: at = v - u \\ t = \frac{v - u}{a}$

Putting the value of t from (i) in the equation (ii)

$s = \frac{u + v}{2} \times \frac{v - u}{a} \\ or \: s = \frac{ {v}^{2} - {u}^{2} }{2a} \\ or \: 2as = {v}^{2} - {u}^{2} \\ {v}^{2} = {u}^{2} + 2as$

It is forth equation of motion.

________________________________

Hope this helps...

Good luck on your assignment..

3 0

3 years ago

Which of the following measurements involve a direction

insens350 [35]

Answer:

I think it is Velocity idk but does it hurt to try

8 0

3 years ago