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Tatiana [17]
3 years ago
14

When light strikes an object, what happens to some of the light?

Physics
1 answer:
svp [43]3 years ago
5 0

Answer:

The light wave could be reflected by the object

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Can someone help me
Illusion [34]
Thomas Edison is the answer im 100% sure of it.
5 0
3 years ago
A) an electron has an initial speed of 226000 m/s. if it undergoes an acceleration of 4.0 x 1014 m/s2, how long will it take to
KIM [24]

initial speed of 226000 m/s

acceleration of 4.0 x 1014 m/s2,

speed of 781000 m/s

What is Acceleration?

  • Acceleration is a rate of change of velocity with respect to time with respect to direction and speed.
  • A point or an object moving in a straight line is accelerated if it speeds up or slows down.
  • Acceleration formula can be written as,

                    a = (v - u ) / t m/s²

As we have to find the time taken, the formula can be altered as,

t = \frac{v-u}{a}

where, t - time taken to reach a final speed

v - final velocity

u - initial velocity

a - acceleration.

Substituting all the given values,

t =\frac{781000 - 226000} {4* 1014}

= 1.3875 × 10⁻⁹ seconds.

So, taken to reach the final speed is found to be 1.3 × 10⁻⁹ 8iH..

7 0
2 years ago
Two particles, each of mass m, are initially at rest very far apart.Obtain an expression for their relative speed of approach at
PSYCHO15rus [73]

Answer:

|\Delta v |=\sqrt{\frac{4Gm}{d} }

Explanation:

Consider two particles are initially at rest.

Therefore,

the kinetic energy of the particles is zero.

That initial K.E. = 0

The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are

\mu= \frac{m_1m_2}{m_1+m_2}

now, since both the masses have mass m

therefore,

\mu= \frac{m^2}{2m}

= m/2

The final K.E. of the particles is

KE_{final}=\frac{1}{2}\times \mu\times \Delta v^2

Distance between two particles is d and the gravitational potential energy between them is given by

PE_{Gravitational}= \frac{Gmm}{d}

By law of conservation of energy we have

KE_{initial}+KE_{final}= PE_{gravitaional}

Now plugging the values we get

0+\frac{1}{2}\frac{m}{2}\Delta v^2= -\frac{Gmm}{d}

|\Delta v |=\sqrt{\frac{4Gm}{d} }

=\sqrt{\frac{Gm}{d} }

This the required relation between G,m and d

5 0
3 years ago
Your backpack has a mass of 8 kg. You drop it from a height of 1.3m. How much work is done by gravity as the backpack falls?
olga_2 [115]

Answer:

The answer is C.

Explanation:

I guessed and it was right

6 0
3 years ago
Read 2 more answers
what is the name of the area around a charged object where the object can exert a force on other charged objects?
shusha [124]

Answer:

An electric field is a region around a charged object where the object's electric force is exerted on other charged objects. Electric fields get weaker the farther away they are from the charge. An electric field is invisible. You can use the field line to represent it.

Explanation:

4 0
2 years ago
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