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3 years ago

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A worker does 500 J of work on a 10 kg box.If the box transfers 375 J of heat to the floor through the friction between the box

and the floor,what is the velocity of the box after the work has been done on it?

1 answer:

kipiarov [429]3 years ago

3 0

The answer is in attachment

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A triangle can have at most how many right angle?

Anna [14]

Answer: 1 right angle

Explanation:

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3 years ago

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How can you make the moon go around in a bigger circle

zlopas [31]

You would have to give it more mechanical energy.

Like, strap a bunch of powerful rockets to one side of the moon, with all of them pointing in the direction that the moon is already moving in its orbit. Then blast away.

NOTE: There aren't enough rockets or rocket fuel on Earth to make a difference, even if you used ALL of them. The mass of the moon is about

<em>73,476,730,900,000,000,000,000 kilograms</em>

(rounded to the nearest hundred trillion kilograms.)

That's a lot.

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3 years ago

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A computer base unit of mass 7.5 kg is dragged along a smooth desk. If the normal contact force is 23N and the tension in the ar

kramer

<span>If there isn't any force then the normal contact force will be

N=m*g=7.5*9.81=73.58N

which is 73.58-23=50.58N less

so, there the person must pull at 23 degree upward

break down the tension in two components, vertical and horizontal.

vertical tension= 50.58=T*sin23

T=50.58/sin23=129.45N</span>

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2 years ago

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Two airplanes leave an airport at the same time. The velocity of the first airplane is 730 m/h at a heading of 65.3 ◦ . The velo

yanalaym [24]

Answer:

Plane will 741.6959 m apart after 1.7 hour

Explanation:

We have given time = 1.7 hr

So if we draw the vectors of a 2d graph we see that the difference in angles is = $102^{\circ}-65.3^{\circ}=36.7^{\circ}$

Speed of first plane = 730 m/h

So distance traveled by first plane = 730×1.7 = 1241 m

Speed of second plane = 590 m/hr

So distance traveled by second plane = 590×1.7 = 1003 m

We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 58.6.

Using the law of cosine, $r^2$ representing the distance between the planes, we see that:

$r^2=1241^2+1003^2-2\times 1003\times 1241cos(36.7)=550112.8295$

r = 741.6959 m

3 0

3 years ago

Two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg. If they are separated

Artemon [7]

Answer: Two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg. If they are separated by a distance of 1.43 km, then, the magnitude of the force of attraction (in newtons) between the objects will be 52.9kN

Explanation: To find the answer we need to know more about the Newton's law of gravitation.

<h3>What is Newton's law of gravitation?</h3>

Gravitation is the force of attraction between any two bodies.
Every body in the universe attracts every other body with a force.
This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between these two masses.
Mathematically we can expressed it as,

$F=\frac{GMm}{r^2} \\where, G=6.67*10^-11Nm^2kg^-2$

<h3>How to solve the problem?</h3>

Here, we have given with the data's,

$M=8.22*10^9kg\\m=1.38*10^8 kg\\r=1.43*10^3m$

Thus, the force of attraction between these two bodies will be,

$F=6.67*10^-11*\frac{8.22*10^9*1.38*10^8}{1.43*10^3} =52.9kN$

Thus, if two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg and, If they are separated by a distance of 1.43 km, then, the magnitude of the force of attraction (in newtons) between the objects will be 52.9kN.

Learn more about the Newton's law of gravitation here:

brainly.com/question/28045318

#SPJ4

6 0

1 year ago