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denis-greek [22]
3 years ago
13

What is the only thing that can pull a beam of light towards itself ? ​

Physics
2 answers:
Anarel [89]3 years ago
6 0
Mirror trust me hjhgguujgffr
Elan Coil [88]3 years ago
4 0

Answer:

<h2><em>The </em><em>only </em><em>thing</em><em> </em><em>that</em><em> </em><em>can</em><em> </em><em>pull</em><em> </em><em>a </em><em>beam </em><em>of</em><em> light</em><em> </em><em>towards</em><em> </em><em>it</em><em> </em><em>self</em><em> </em><em>is </em><em>a </em><em>black </em><em>hole.</em></h2>

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The upward force on an airplane's wing is thrust.
kompoz [17]

No.  'Thrust' is what most people in aviation call the force
that pushes the aircraft forward. 

The same people generally call the upward force on the wing "lift".

3 0
3 years ago
Can u Anwser this Plzzzz
nata0808 [166]

Answer:

-0. 75m/s^2

Explanation:

use formula of acceleration

5 0
3 years ago
Without using a micrometer screw gauge, how do I find the average diameter of a long piece of thin wire using a metre rule and a
Mice21 [21]

Answer:

Wind the long piece of thin wire around the uniform glass rod multiple times, find the length of the total diameters using the metre ruler, and divide by the number of times you wound it around the rod.

Explanation:

Since the diameter of one long piece of thin wire is too thin to be measured by a metre ruler, you can wind it multiple times and push it side by side to get a length you can measure.

For example, if you wound it around 20 times and the total length of 20 diameters of the wire side-by-side is 2.0 cm, one winding, which is the diameter would be 2.0cm ÷ 20 = 0.10cm or 1mm.

5 0
2 years ago
A car travels along a straight line at a constant speed of 53.0 mi/h for a distance d and then another distance d in the same di
Shalnov [3]

A distance of d is covered with 53 mile/hr initially. Time taken to cover this distance t1 = d/53 hour Next distance of d is covered with x mile hours. Time taken to cover this distance t2 = d/x hours. We have average speed = 26.5 mile / hour          

                                         = Total distance traveled/ total time taken                  

                                         = \frac{2d}{\frac{d}{53}+\frac{d}{x}} = \frac{2}{\frac{1}{53}+\frac{1}{x} }  = \frac{106x}{x+53}

                              26.5 = \frac{106x}{x+53} \\ \\ 79.5 x = 1404.5\\ \\ x = 17.67 miles/hour

5 0
3 years ago
An object with a mass of 10 kg is rolled down a frictionless ramp from a height of 3 meters. If a factory worker at the bottom o
ruslelena [56]

Answer:

The amount of work the factory worker must to stop the rolling ramp is 294 joules

Explanation:

The object rolling down the frictionless ramp has the following parameters;

The mass of the object = 10 kg

The height from which the object is rolled = 3 meters

The work done by the factory worker to stop the rolling ramp = The initial potential energy, P.E., of the ramp

Where;

The potential energy P.E. = m × g × h

m = The mass of the ramp = 10 kg

g = The acceleration due to gravity = 9.8 m/s²

h = The height from which the object rolls down = 3 m

Therefore, we have;

P.E. = 10 kg × 9.8 m/s² × 3 m = 294 Joules

The work done by the factory worker to stop the rolling ramp = P.E. = 294 joules

8 0
2 years ago
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