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cluponka [151]
3 years ago
11

WILL MARK BRAINIEST!!!!!! (Fill in the blanks with the possible answers at the bottom)

Physics
1 answer:
drek231 [11]3 years ago
5 0
Number 3 I think there was so much going In ckcoc v K o n&+\
You might be interested in
If you exert a force of 10.0 N to lift a box a distance of 0.75 m, how much work do you do?
Bas_tet [7]

Answer:

<h2>7.5 J</h2>

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 10 × 0.75

We have the final answer as

<h3>7.5 J</h3>

Hope this helps you

7 0
3 years ago
Please help!!
Ronch [10]

Answer:

Explanation:

The distance will be the total distannce covered during the journey.

If you move 3 meters East and move 4 meters north, then the distance will be calculated as;

Distance = distnace through East+distance through north

Distance = 3m + 4m

<em>Distance = 7m</em>

Displacement is the distance covered in a specified direction. It is the shortest distance covered by me. This can be gotten using the Pythagoras theorem.

d² = 3²+4²

d² = 9+16

d² = 25

d = √25

d = 5m

<em>Hence the displacement of the object is 5metres</em>

5 0
4 years ago
8. The copper rod expands when its temperature
olchik [2.2K]

Answer:

c

Explanation:

4 0
3 years ago
monochromatic light from a distant source is incident on a slit 0.75 mm wide. on screen 2 m away, the distance from the central
hjlf

Displacement from the center line for minimum intensity is 1.35 mm , width of the slit  is 0.75 so  Wavelength of the light  is 506.25.

<h3>How to find Wavelength of the light?</h3>

When a wave is bent by an obstruction whose dimensions are similar to the wavelength, diffraction is observed. We can disregard the effects of extremes because the Fraunhofer diffraction is the most straightforward scenario and the obstacle is a long, narrow slit.

This is a straightforward situation in which we can apply the

Fraunhofer single slit diffraction equation:

y = mλD/a

Where:

y = Displacement from the center line for minimum intensity =  1.35 mm

λ =  wavelength of the light.

D = distance

a = width of the slit = 0.75

m = order number = 1

Solving for λ

λ = y + a/ mD

Changing the information that the issue has provided:

λ = 1.35 * 10^-3 + 0.75 * 10^-3 / 1*2  

=5.0625 *10^-7 = 506.25

so

Wavelength of the light 506.25.

To learn more about Wavelength of the light refer to:

brainly.com/question/15413360

#SPJ4

5 0
1 year ago
a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres
shusha [124]

Answer:

Part a) When collision is perfectly inelastic

v_m = \frac{m + M}{m} \sqrt{5Rg}

Part b) When collision is perfectly elastic

v_m = \frac{m + M}{2m}\sqrt{5Rg}

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

mv_m = (m + M)v

so we have

v = \frac{mv_m}{m + M}

now we know that in order to complete the circle we will have

v = \sqrt{5Rg}

\frac{mv_m}{m + M} = \sqrt{5Rg}

now we have

v_m = \frac{m + M}{m} \sqrt{5Rg}

Part b)

Now we know that collision is perfectly elastic

so we will have

v = \frac{2mv_m}{m + M}

now we have

\sqrt{5Rg} = \frac{2mv_m}{m + M}

v_m = \frac{m + M}{2m}\sqrt{5Rg}

6 0
4 years ago
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