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Nadya [2.5K]
2 years ago
8

When you turn on a lightbulb in a room, the entire room appears to flood with light at the same time. Your eyes cannot perceive

light originating at the bulb and then moving outward from it. This either means that light has infinite speed (appears everywhere instantaneously), or it moves so fast that it is difficult to tell the difference. How might you improve your chances of detecting the motion of light if it moves at finite speed?
Physics
1 answer:
Evgen [1.6K]2 years ago
7 0

Explanation:

Assuming we can turn on the lightbulb from any distance with a device. We can gradually increase the distance that separates us from lightbulb, in this way, if the speed of light is finite  we can see a temporary delay between the moment we turn on the lightbulb and the moment in which we observe its light.

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Equal masses are suspended from two separate wires made of the same material. The wires have identical lengths. The first wire h
choli [55]
D. I think is the correct answer
6 0
3 years ago
An electron is released from rest in a weak electric field given by E =-2.30 x 10-10 N/Cj. After the electron has traveled a ver
Goshia [24]

Explanation:

It is given that,

An electron is released from rest in a weak electric field of, E=2.3\times 10^{-10}\ N/C

Vertical distance covered, s=1\ \mu m=10^{-6}\ m

We need to find the speed of the electron. Let its speed is v. Using third equation of motion as :

v^2-u^2=2as

v^2=2as.............(1)

Electric force is F_e and force of gravity is F_g. As both forces are acting in downward direction. So, total force is:

F=mg+qE

F=9.1\times 10^{-31}\times 9.8+1.6\times 10^{-19}\times 2.3\times 10^{-10}

F=4.57\times 10^{-29}\ N

Acceleration of the electron, a=\dfrac{F}{m}

a=\dfrac{4.57\times 10^{-29}\ N}{9.1\times 10^{-31}\ kg}

a=50.21\ m/s^2

Put the value of a in equation (1) as :

v=\sqrt{2as}

v=\sqrt{2\times 50.21\times 10^{-6}}

v = 0.010 m/s

So, the speed of the electron is 0.010 m/s. Hence, this is the required solution.

6 0
3 years ago
1 2 3 4 5 6 7 8 9 10
7nadin3 [17]

Answer:

<h2>C. <u>0.55 m/s towards the right</u></h2>

Explanation:

Using the conservation of law of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.

Momentum = Mass (M) * Velocity(V)

BEFORE COLLISION

Momentum of 0.25kg body moving at 1.0m/s = 0.25*1 = 0.25kgm/s

Momentum of 0.15kg body moving at 0.0m/s(body at rest) = 0kgm/s

AFTER COLLISION

Momentum of 0.25kg body moving at x m/s = 0.25* x= 0.25x kgm/s

<u>x is the final velocity of the 0.25kg ball</u>

Momentum of 0.15kg body moving at 0.75m/s(body at rest) =

0.15 * 0.75kgm/s = 0.1125 kgm/s

Using the law of conservation of momentum;

0.25+0 = 0.25x + 0.1125

0.25x = 0.25-0.1125

0.25x = 0.1375

x = 0.1375/0.25

x = 0.55m/s

Since the 0.15 kg ball moves off to the right after collision, the 0.25 kg ball will move at <u>0.55 m/s towards the right</u>

<u></u>

4 0
2 years ago
A planet exerts a gravitational force of magnitude 9e22 N on a star. If the planet were 2 times closer to the star (that is, if
Dmitrij [34]

To solve this problem we will use the related concepts in Newtonian laws that describe the force of gravitational attraction. We will use the given value and then we will obtain the proportion of the new force depending on the Radius. From there we will observe how much the force of attraction increases in the new distance.

Planet gravitational force

F_p = 6*10^{22}N

F_p = \frac{GMm}{R^2}

F_p = 9*10^{22}N

Distance between planet and star

r = \frac{R}{2}

Gravitational force is

F = \frac{GMm}{r^2}

Applying the new distance,

F = \frac{GMm}{(\frac{R}{2})^2}

F =  4\frac{GMm}{R^2}

Replacing with the previous force,

F = 4F_p

Replacing our values

F= 4(9*10^{22}N)

F = 36*10^{22}N

Therefore the magnitude of the force on the star due to the planet is  36*10^{22}N

5 0
3 years ago
A cyclist travels 36km in 3 hours. What is the cyclist’s average speed?
dybincka [34]
To determine what the cyclists average speed is, simply divide the distance the cyclist has travelled by the time the cyclist has traveled for.

Assuming that this is the average rate the cyclist is moving at it would be 12 km/hr.
4 0
3 years ago
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