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ddd [48]
2 years ago
12

What is the speed of light in a vacuume?

Physics
1 answer:
-Dominant- [34]2 years ago
5 0

The speed of light in vacuum is exactly

299,792,458 meters per second.

That's so exact that it's the official scientific definition of a "meter".

The number doesn't t change even if the flashlight or other light source is moving.

You might be interested in
Suppose a 49-N sled is resting on packed snow. The coefficient of kinetic friction is 0.11. If a person weighing 585 N sits on t
Andre45 [30]

Answer:

69.74 N

Explanation:

We are given that

Weight of sled=49 N

Coefficient of kinetic friction,\mu_k=0.11

Weight of person=585 N

Total weight==mg=49+585=634 N

We know that

Force needed to pull the sled across the snow at constant speed,F=Kinetic friction

F=\mu_k N

Where N= Normal=mg

F=0.11\times 634=69.74 N

Hence, the force is needed  to pull the sled across the snow at constant speed=69.74 N

7 0
2 years ago
In the long jump, an athlete launches herself at an angle above the ground and lands at the same height, trying to travel the gr
NikAS [45]

A) 2.64t

B) 2.64h

C) 2.64D

Explanation:

A)

The motion of the athlete is equivalent to the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion (constant acceleration) along the vertical direction

The time of flight of a projectile can be found from the equations of motion, and it is found to be

t=\frac{2u sin \theta}{g}

where

u is the initial speed

\theta is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the time of flight is t.

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. Therefore, the time of flight on Mars will be:

t'=\frac{2usin \theta}{g'}=\frac{2u sin \theta}{0.379g}=\frac{1}{0.379}t=2.64t

B)

The maximum height reached by a projectile can be also found using the equations of motion, and it is given by

h=\frac{u^2 sin^2\theta}{2g}

where

u is the initial speed

\theta is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the maximum height is h.

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. So, the maximum height reached on Mars will be:

h'=\frac{u^2 sin^2\theta}{2g'}=\frac{u^2 sin^2\theta}{(0.379)2g}=\frac{1}{0.379}h=2.64h

C)

The horizontal distance covered by a projectile is also found from the equations of motion, and it is given by

D=\frac{u^2 sin(2\theta)}{g}

where:

u is the initial speed

\theta is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the horizontal distance covered is D.

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. Therefore, the horizontal distance reached on Mars will be:

D'=\frac{u^2 sin(2\theta)}{g'}=\frac{u^2 sin(2\theta)}{(0.379)g}=\frac{1}{0.379}D=2.64D

7 0
3 years ago
You walk with a velocity of 2 m/s north. You see a man approaching you, and from your frame of
solong [7]

Answer:

The velocity of the man from the frame of  reference of a stationary observer is, V₂ = 5 m/s

Explanation:

Given,

Your velocity, V₁ = 2 m/

The velocity of the person, V₂ =?

The velocity of the person relative to you, V₂₁ = 3 m/s

According to the relative velocity of two

                                V₂₁ = V₂ -V₁

∴                               V₂ =  V₂₁ + V₁

On substitution

                                 V₂ = 3 + 2

                                      = 5 m/s

Hence, the velocity of the man from the frame of reference of a stationary observe is, V₂ = 5 m/s

8 0
3 years ago
On a sunny day, a student poured a cup of water on the sidewalk to make a puddle. When he returned later, the puddle was gone. T
ivanzaharov [21]

Answer:

The students conclusion is wrong.

Explanation:

The students conclusion is wrong because liquids can not soak into concrete. The water evaporated because it was a hot sunny day. The water turned into a gas called water vapor.

8 0
3 years ago
Waves are observed passing under a dock. Wave crests are 8.0 meters apart. The time for a complete wave to pass by is 4.0 second
ki77a [65]
To answer that question, we don't care what the highest and lowest
levels of the wave are, or how far apart they are.  We only need to be
able to identify the highest point on the wave, and keep track of how
often those pass by us.

You said it takes 4 seconds for a complete wave to pass by.
Through the sheer power of intellect, I'm able to take that information
and calculate that  1/4  of the wave passes by in 1 second.

There's your frequency . . .  1/4 per second, or  0.25 Hz.
6 0
2 years ago
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