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3 years ago

How much work is done if you push a 200 N box across a floor with a force of 50 N for a distance of 20m

Physics

1 answer:

vovikov84 [41]3 years ago

8 0

<h2>Answer: 1000 J</h2>

The Work $W$ done by a Force $F$ refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.

It should be noted that it is a scalar magnitude, and its unit in the International System of Units is the Joule (like energy). Therefore, 1 Joule is the work done by a force of 1 Newton when moving an object, in the direction of the force, along 1 meter:

$1J=(1N)(1m)=Nm$

Now, when the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:

$W=(F)(d)$ (1)

When they are not parallel, both directions form an angle, let's call it $\alpha$ . In that case the expression to calculate the Work is:

$W=Fdcos{\alpha}$ (2)

For example, in order to push the 200 N box across the floor, you have to apply a force along the distance $d$ to overcome the resistance of the weight of the box (its 200 N).

In this case both <u>(the force and the distance in the path) are parallel</u>, so the work $W$ performed is the product of the force exerted to push the box $F=50N$ by the distance traveled $d$ . as shown in equation (1).

Hence:

$W=(50N)(20m)$

$W=1000Nm=1000J$ >>>>This is the work

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A satellite, moving in an elliptical orbit, is 368 km above Earth's surface at its farthest point and 164 km above at its closes

Marat540 [252]

Answer:

a) 6636 km

b) 0.0154

Explanation:

The height above the earth at its furthest point is 368 km

The height above the earth at its closest point is 164 km

Radius of the Earth is 6370 km

The distance of the satellite from the center of the earth to the furthest point is 6370 + 368 km = 6738 km

The distance of the satellite from the center of the earth to the closest point is 6370 + 164 = 6534 km

If we add together the sum of the distance of the satellite from the furthest and its closest distance, it is equal to the 2 major semi axis.

Basically,

2a = R + r

a = (R + r) / 2

a = (6738 + 6534) / 2

a = 13272 / 2

a = 6636 km

Eccentricity, e = (a - r) / a

Eccentricity, e = (6636 - 6534) / 6636

Eccentricity, e = 102 / 6636

Eccentricity, e = 0.0154

3 0

2 years ago

Ron fills a beaker with glycerin (n = 1.473) to a depth of 5.0 cm. if he looks straight down through the glycerin surface, he wi

Tju [1.3M]

By law of refraction we know that image position and object positions are related to each other by following relation

$\frac{\mu_1}{h_o} = \frac{\mu_2}{h_i}$

here we know that

$\mu_1 = 1.473$

$h_o = 5 cm$

$\mu_2 = 1$

now by above formula

$\frac{1.473}{5} = \frac{1}{h_i}$

$h_i = 3.39 cm$

so apparent depth of the bottom is seen by the observer as h = 3.39 cm

7 0

3 years ago

The sound waves travel along this

Anna35 [415]

Explanation:

sound waves go through almost every thing exept solid walls

4 0

2 years ago

A whistle you use to call your hunting dog has a frequency of 21 kHz, but your dog is ignoring it. You suspect the whistle may n

uranmaximum [27]

The Doppler effect is the right concept to solve this problem. The Doppler effect is understood as the change in apparent frequency of a wave produced by the relative movement of the source with respect to its observer. Mathematically it can be described as,

$f = (1-\frac{v_0}{v})f_0$

Here,

$f_0$ = Frequency of the sound from the Whistle

f = Frequency of sound heard

v = Speed of the sound in the Air

Replacing we have that

$1- \frac{v_0}{343} = \frac{20kHz}{21kHz}$

$\frac{v_0}{343} = 1-\frac{20}{21}$

$\frac{v_0}{343} = \frac{1}{21}$

$v_0 = \frac{1}{21}(343)$

$v_0 = 16.33m/s$

Therefore the minimum speed to know if the whistle is working is 16.33m/s

3 0

3 years ago

Rutherford tracked the motion of tiny, positively charged particles shot through a thin sheet of gold foil. Some particles

Anna007 [38]

Question

Rutherford tracked the motion of tiny, positively charged particles shot through a thin sheet of gold foil. Some particles travelled in a straight line and some were deflected at different angles.

Which statement best describes what Rutherford concluded from the motion of the particles?

A) Some particles travelled through empty spaces between atoms and some particles were deflected by electrons.

B) Some particles travelled through empty parts of the atom and some particles were deflected by electrons.

C) Some particles travelled through empty spaces between atoms and some particles were deflected by small areas of high-density positive charge in atoms.

D) Some particles travelled through empty parts of the atom and some particles were deflected by small areas of high-density positive charge in atoms.

Answer:

The right answer is C)

Explanation:

In the experiment described above, a piece of gold foil was hit with alpha particles, which have a positive charge. Alpha particles <em>α</em> were used because, if the nucleus was positive, then it would deflect the positive particles. The principles of physics posit that electric charges of the same orientation repel.

So most as expected some of the alpha particles went right through meaning that the gold atoms comprised mostly empty space except the areas that were with a dense population of positive charges. This area became known as the "nucleus".

Due to the presence of the positive charges in the nucleus, some particles had their paths bent at large angles others were deflected backwards.

Cheers!

6 0

3 years ago

Read 2 more answers