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3 years ago

A book has been lifted 1.5 meters into the air giving it 30J of potential energy. How much force was used to lift it

Physics

1 answer:

joja [24]3 years ago

6 0

Answer:

Force on the object is 20 N

Explanation:

As we know that work done to raise the book from initial position to final position is known as potential energy stored in it

So here we know that

$U = F.s$

here we know that

U = 30 J

s = displacement = 1.5 m

so we have

$30 = F(1.5)$

$F = 20 N$

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Playing soccer on a beach will require more effort because the sand causes a great deal of _______ to the ball.

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Answer:

I believe the answer to be B

Explanation:

If you were playing on grass, the ball would be able to roll around much easier rather than it to be on sand. If it's wrong I am so sorry

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2 years ago

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2 years ago

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A soccer player takes a corner kick, lofting a stationary ball 30.0° above the horizon at 18.0 m/s. If the soccer ball has a mas

Radda [10]

Answer:

change in momentum, $\Delta p=7.65 \,kg.m.s^{-1}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

Average Force, $F=144.3396\,N$

$F_x=125.0018\,N$

$F_y=72.1698\,N$

Explanation:

Given:

angle of kicking from the horizon, $\theta= 30^{\circ}$

velocity of the ball after being kicked, $v=18 m.s^{-1}$

mass of the ball, $m=0.425\, kg$

time of application of force, $t=5.3\times 10^{-2}\,s$

We know, since body is starting from the rest

$\Delta p=m.v$ .....................(1)

$\Delta p=0.425\times 18$

$\Delta p=7.65 \,kg.m.s^{-1}$

Now the components:

$\Delta p_x= 7.65\times cos 30^{\circ}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

similarly

$\Delta p_y= 7.65\times sin 30^{\circ}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

also, impulse

$I=F\times t$ .........................(2)

where F is the force applied for t time.

Then from eq. (1) & (2)

$F\times t=m.v$

$F\times 5.3\times 10^{-2}= 7.65$

$F=144.3396\,N$

Now, the components

$F_x=144.3396\times cos 30^{\circ}$

$F_x=125.0018\,N$

$F_y=144.3396\times sin 30^{\circ}$

$F_y=72.1698\,N$

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2 years ago

What does a calorimeter measure?

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2 years ago

Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.23 2.23 times a second. A tack is stuck in the t

elena-14-01-66 [18.8K]

Explanation:

The given data is as follows.

Angular velocity ( $\omega$ ) = 2.23 rps

Distance from the center (R) = 0.379 m

First, we will convert revolutions per second into radian per second as follows.

= 2.23 revolutions per second

= $2.23 \times 2 \times 3.14 rad/s$

= 14.01 rad/s

Now, tangential speed will be calculated as follows.

Tangential speed, v = $R \times \omega$

= 0.379 x 14.01

= 5.31 m/s

Thus, we can conclude that the tack's tangential speed is 5.31 m/s.

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2 years ago