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3 years ago

PLEASEEE HELPPP!!!

Physics

1 answer:

Alja [10]3 years ago

3 0

Given :

A mover slides a refrigerator weighing 650 N at a constant velocity across the floor a distance of 8.1 m.

The force of friction between the refrigerator and the floor is 230 N.

To Find :

How much work has been performed by the mover on the refrigerator.

Solution :

Since, refrigerator is moving with constant velocity.

So, force applied by the mover is also 230 N ( equal to force of friction ).

Now, work done in order to move the refrigerator is :

$W = Force\times distance\\\\W = 230 \times 8.1\ N\ m\\\\W = 1863\ N\ m$

Hence, this is the required solution.

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a lawn mower is pushed with a force of 50n. if the angle between the handle of the mower and the ground is 30°. why doesn't the

Mademuasel [1]

Answer:

IT doesnt push down because the lawnmower has wheels and the ground is hard

Explanation:

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3 years ago

How long have advances in astronomy been occurring? Only for the past few years

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I'd have to say that the list of choices doesn't go far enough.

Advances in Astronomy have been occurring for at least the past two millennia (2000 years). Maybe longer.

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3 years ago

1. While John is traveling along an interstate highway, he notices a 160-mile marker as he passes through town. Later John passe

pogonyaev

Number of miles that marker shows when passes through town= 160 miles.

Number of miles that marker shows currently to John = 115 miles.

We need to find the distance between town and John's current location.

For the problem, we can clearly see that Town is at 160 miles away but when John passes the marker shows 115 miles.

So, it's just the difference between 160 miles and 115 miles.

In order to find that difference, we need to subtract those two numbers.

160miles - 115miles = 45 miles.

So, we could say the distance between town and John's current location is 45 miles.

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3 years ago

(b)

Leona [35]

Answer:

i wish i knew your answer.

Explanation:

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2 years ago

A proton that has a mass m and is moving at 270 m/s in the i hat direction undergoes a head-on elastic collision with a stationa

Nataly_w [17]

Answer:

$V_p = 267.258 m/s$

$V_n = 38.375 m/s$

Explanation:

using the law of the conservation of the linear momentum:

$P_i = P_f$

where $P_i$ is the inicial momemtum and $P_f$ is the final momentum

the linear momentum is calculated by the next equation

P = MV

where M is the mass and V is the velocity.

so:

$P_i = m(270 m/s)$

$P_f = mV_P + M_nV_n$

where m is the mass of the proton and $V_p$ is the velocity of the proton after the collision, $M_n$ is the mass of the nucleus and $V_n$ is the velocity of the nucleus after the collision.

therefore, we can formulate the following equation:

m(270 m/s) = m $V_p$ + 14m $V_n$

then, m is cancelated and we have:

270 = $V_p$ + $14V_n$

This is a elastic collision, so the kinetic energy K is conservated. Then:

$K_i = \frac{1}{2}MV^2 = \frac{1}{2}m(270)^2$

and

Kf = $\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2$

then,

$\frac{1}{2}m(270)^2$ = $\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2$

here we can cancel the m and get:

$\frac{1}{2}(270)^2$ = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$

now, we have two equations and two incognites:

270 = $V_p$ + $14V_n$ (eq. 1)

$\frac{1}{2}(270)^2$ = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$

in the second equation, we have:

36450 = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$ (eq. 2)

from this last equation we solve for $V_n$ as:

$V_n = \sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }$

and replace in the other equation as:

270 = $V_p +$ 14 $\sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }$

so,

$V_p = -267.258 m/s$

Vp is negative because the proton go in the -i hat direction.

Finally, replacing this value on eq. 1 we get:

$V_n = \frac{270+267.258}{14}$

$V_n = 38.375 m/s$

3 0

3 years ago