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3 years ago

Write down formula of power

Physics

2 answers:

r-ruslan [8.4K]3 years ago

8 0

There u go -> P=f/a

zzz [600]3 years ago

5 0

Answer:

Power is a measure of the amount of work that can be done in a given amount of time. So, the formula is the energy or work divided by time:

P = W ÷ T, where p is power , w is work and t is time.

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Why is the perfect mechine is not possible in the real life?

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Because the mechanical advantage of the machine is affected by friction and weight but velocity ratio is not. So, mechanical advantage is less than velocity rate. Thus, the machine's efficiency is less than 100% and can't be a perfect machine

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2 years ago

Two cars go through 2 different crashes. Car 1 experiences a 500 Ns impulse for a duration of 15s, while car 2 experiences that

tatuchka [14]

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3 years ago

A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo

Serggg [28]

Before the engines fail $(0\le t\le3.00\,\rm s)$ , the rocket's horizontal and vertical position in the air are

$x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2$

$y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t^2$

and its velocity vector has components

$v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t$

$v_y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t$

After $t=3.00\,\rm s$ , its position is

$x=273\,\rm m$

$y=362\,\rm m$

and the rocket's velocity vector has horizontal and vertical components

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}$

After the engine failure $(t>3.00\,\rm s)$ , the rocket is in freefall and its position is given by

$x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t$

$y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

and its velocity vector's components are

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}-gt$

where we take $g=9.80\,\frac{\rm m}{\mathrm s^2}$ .

a. The maximum altitude occurs at the point during which $v_y=0$ :

$159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s$

At this point, the rocket has an altitude of

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m$

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve $y=0$ for $t$ , then add 3 seconds to this time:

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s$

So the rocket stays in the air for a total of $37.6\,\rm s$ .

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute $x$ for this time $t$ :

$273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m$

5 0

3 years ago

Helpp pleaseeeeee …..

Lilit [14]

Answer:

300 cos 30 = 40 a + 40 * .2 * 10

Total force = mass * acceleration + frictional force

260 = 40 a + 80

a = 180 / 40 = 4.5 m/s^2

Check:

15 a + 15 * 10 * .2 = T acceleration of 15 kg block (assuming a = 4.5)

T = 15 (4.5) + 30 = 97.5 force required to accelerate 15 kg block

260 - 97.5 = 162.5 net force on 25 kg block

162.5 = 4.5 (25) + 25 * 10 * .2

162.5 = 112.5 + 50 = 162.5

4.5 m/s^2 checks out as correct

8 0

1 year ago

4025.05m +20.0m +0.050004m

PIT_PIT [208]

The answer is 4,045.1 meters

5 0

3 years ago

Write down formula of power​

Write down formula of power