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3 years ago

If 35 J of work was performed in 70 seconds, how much power was used to do this task?

1 answer:

5 0

$P = w \times delta \: T$
P = Power
W = Work
Delta T = Change of Time

So:
P = 35 Joules
Delta T = 70 secs
W = ?

$P = 35 \div 70 \\ P = 0.5 \: watts$

* Joule/Sec = Watt

Answer: The power used to do this task was 0.5 Watts.

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A uniform solid disk of mass 5.00 kg and diameter 47.0 cm starts from rest and rolls without slipping down a 40.0 ∘ incline that

saveliy_v [14]

Answer:

The speed it reaches the bottom is

$v=6.51m/s$

Explanation:

Given: $m=5.0kg$ , $r=47cm\frac{1m}{100cm}=0.47m$

Using the conservation of energy theorem

$U_i=K_E+K_{ER}$

$m*g*h=\frac{1}{2}*m*v^2+\frac{1}{2}*I*w^2$

$v=r*w$ , $I=\frac{1}{2}*m*r^2$

$m*g*h=\frac{1}{2}*m*(r*w)^2 +\frac{1}{2}*[\frac{1}{2} *m*r^2]*w^2$

$m*g*h=\frac{3}{4}*m*r^2*w^2$

$g*h=\frac{3}{4}*r^2*w^2$

Solve to w'

$w^2=\frac{4*g*h}{3*r^2}$

$h=x*sin(30)=6.5m*sin(30)=3.25m$

$w=\sqrt{\frac{4*9.8m/s^2*3.25m}{3*(0.235m)^2}}$

$w=27.74rad/s$

$v=27.74rad/s*0.235m=6.51m/s$

7 0

3 years ago

The efficiency of a screw is low because of _____.

AfilCa [17]

The correct answer that would best complete the given statement above would be the second option. A screw is an inclined plane wrapped around a cylinder. <span>The efficiency of a screw is low because there is more input than output. In other words, it is because of friction. Hope this answer helps.</span>

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3 years ago

Read 2 more answers

I have a big test coming up and these two questions i don't understand: Choose the best answer: A device that transforms mechani

Jlenok [28]

1. B- electric generator

Although its really sometimes called an AC generator.

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3 years ago

When the biker is at the top of the ramp, he has a speed of 10 m/s at

oksian1 [2.3K]

Answer:

25000

Explanation:

100 x 25 x 10 = 25000

8 0

3 years ago

A positively charged particle is in the center of a parallel-plate capacitor that has charge ±Q on its plates. SUppose the dista

slamgirl [31]

Answer:

Stay the same

Explanation:

First of all, let's find how the capacitance of the capacitor changes.

Initially, it is given by

$C=\frac{\epsilon_0 A}{d}$

where

$\epsilon_0$ is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

From the formula, we see that the capacitance is inversely proportional to the separation between the plates. In this problem, the distance between the plates is doubled, so the capacitance will be halved:

$C' = \frac{1}{2}C$

The potential difference across the capacitor is given by

$V= \frac{Q}{C}$

where

Q is the charge on the plates

C is the capacitance

We see that the voltage is inversely proportional to the capacitance. We said that the capacitance has halved: therefore, the potential difference across the two plates will double:

$V' = 2 V$

Now we can analyze the electric field between the plates of the capacitor, which is given by

$E=\frac{V}{d}$

we said that:

- The voltage has doubled: V' = 2 V

- The distance between the plates has doubled: d' = 2 d

therefore, the new electric field will be

$E'=\frac{2V}{2d}=\frac{V}{d}=E$

So, the electric field is unchanged. And since the force on the particle at the center is directly proportional to the electric field:

F = qE

Then the force on the particle will stay the same.

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3 years ago