Ask question

Ask question

All categories

3 years ago

14

A cart is pushed on a frictionless floor with 29.6 Newtons of force along a distance of 5.9 meters in 8.1 seconds. How much powe

r is exerted
during this time period?

1 answer:

marta [7]3 years ago

8 0

Answer:

Power = 21.6 Watts

Explanation:

Power = $\frac{W}{t}$

Where W = work and t = time.

W = Fs

Where F is force and s is displacement.

29.6N(5.9m) = 174.64J

W = 174.64J

Power = $\frac{174.64J}{8.1s}$

Power = 21.6 Watts

You might be interested in

If you were creating a lamp by hand what type of material would you chose for the wiring and what material would you chose to co

anzhelika [568]

Copper because it contains alot of electricity

6 0

4 years ago

A 51-g rubber ball is released from rest and falls vertically onto a steel plate. The ball strikes the plate and is in contact w

Simora [160]

Answer:

Last option 3000N

Explanation:

4 0

3 years ago

What is projectile motion

MakcuM [25]

<h2><u>Projectile</u><u> </u><u>motion</u><u>:</u></h2>

<em>If</em><em> </em><em>an</em><em> </em><em>object is given an initial velocity</em><em> </em><em>in any direction and then allowed</em><em> </em><em>to travel freely under gravity</em><em>, </em><em>it</em><em> </em><em>is</em><em> </em><em>called a projectile motion</em><em>. </em>

It is basically 3 types.

horizontally projectile motion
oblique projectile motion
included plane projectile motion

7 0

3 years ago

Read 2 more answers

Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}$

$A=0.0198$

From equation (II)

$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

$A=0.0203$

From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

$A=0.0217$

From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

5 0

3 years ago

As light from a star spreads out and weakens, do gaps form between the photons?<br>

Zolol [24]

Answer:

There are no gaps in space between the photons as they travel. If you were to look at a wave then you'd come to a conclusion that indeed that there aren't any gaps unless they are specifically placed.The light from a distance star indeed spreads out and weakens as it travels, but this just reduces the wave strength and does not introduce gaps.

3 0

3 years ago