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2 years ago

Select the statement that correctly describes how light travels.

Physics

1 answer:

Paladinen [302]2 years ago

5 0

The speed of light, traveling in a vacuum, will not change if the light source is moving.

The speed of light in vacuum does not depend on whether the light source is moving or not.

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A 2 kg object traveling at 5 m s on a frictionless horizontal surface collides head-on with and sticks to a 3 kg object initiall

Svetlanka [38]

Answer: (d)

Explanation:

Given

Mass of object $m=2\ kg$

Speed of object $u=5\ m/s$

Mass of object at rest $M=3\ kg$

Suppose after collision, speed is v

conserving momentum

$\Rightarrow mu+0=(m+M)v\\\\\Rightarrow v=\dfrac{2\times 5}{2+3}\\\\\Rightarrow v=2\ m/s$

Initial kinetic energy

$k_1=\dfrac{1}{2}\times 2\times 5^2\\\\k_1=25\ J$

Final kinetic energy

$k_2=\dfrac{1}{2}\times (2+3)\times 2^2\\\\k_2=10\ J$

So, it is clear there is decrease in kinetic energy . Thus, energy decreases and velocity becomes 2 m/s.

4 0

2 years ago

The half-life of the radioactive isotope Carbon-14 is about 5730 years. Find N in terms of t. The amount of a radioactive elemen

Fudgin [204]

The complete queston is The amount of a radioactive element A at time t is given by the formula

A(t) = A₀e^kt

Answer: A(t) =N e^( -1.2 X 10^-4t)

Explanation:

Given

Half life = 5730 years.

A(t) =A₀e ^kt

such that

A₀/ 2 =A₀e ^kt

Dividing both sides by A₀

1/2 = e ^kt

1/2 = e ^k(5730)

1/2 = e^5730K

In 1/2 = 5730K

k = 1n1/2 / 5730

k = 1n0.5 / 5730

K= -0.00012 = 1.2 X 10^-4

So that expressing N in terms of t, we have

A(t) =A₀e ^kt

A₀ = N

A(t) =N e^ -1.2 X 10^-4t

7 0

3 years ago

A singly charged ion of 7Li (an isotope of lithium which lost only one electron) has a mass of 1.16 ×10^-26 kg. It is accelerate

MaRussiya [10]

Explanation:

It is given that,

Mass of lithium, $m=1.16\times 10^{-26}\ kg$

It is accelerated through a potential difference, V = 224 V

Uniform magnetic field, B = 0.724 T

Applying the conservation of energy as :

$\dfrac{1}{2}mv^2=qV$

$v=\sqrt{\dfrac{2qV}{m}}$

q is the charge on an electron

$v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\ C\times 224\ V}{1.16\times 10^{-26}\ kg}}$

v = 78608.58 m/s

$v=7.86\times 10^4\ m/s$

To find the radius of the ion's path in the magnetic field. The centripetal force is balanced by the magnetic force as :

$qvB=\dfrac{mv^2}{r}$

$r=\dfrac{mv}{qB}$

$r=\dfrac{1.16\times 10^{-26}\ kg\times 7.86\times 10^4\ m/s}{1.6\times 10^{-19}\ C\times 0.724\ T}$

r = 0.0078 meters

So, the radius of the path of the ion is 0.0078 meters. Hence, this is the required solution.

4 0

3 years ago

Plz help 20 points!!

vazorg [7]

Answer:

1-d

2- a

3-e

4-b

5-c

This is the correct sequence

5 0

2 years ago

How can parents help children to gain friends?

Andre45 [30]

Answer:

You could try finding a familiar peer to join the activity with your child. Or ask your child who their friends are at school, or what they look for in a friend at school.

7 0

3 years ago

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