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3 years ago

Please help asap no links

Physics

1 answer:

Airida [17]3 years ago

5 0

$\huge \mathcal\pink{Answer}$

So, Speed of light in Rock salt is 1/1.54 times speed of light in vacuum. Therefor Speed of light in Rock salt is less [1.94 x 10⁸m/s ] compared to speed of light in vacuum =C = 3 x 10⁸ m/s.

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You drop a 30 g pebble down a well. You hear a splash 2.7 s later. Ignoring air resistance, how deep is the well? Assume g = 9.8

lys-0071 [83]

I will assume here that the well is sufficiently short so that the time the sound takes to come from the bottom of the well to our ear is negligible.

Since the pebble moves by uniformly accelerated motion, the distance it covers is given by
$S= \frac{1}{2}gt^2$
where
$g=9.81 m/s^2$ is the gravitational acceleration
$t=2.7 s$ is the time the pebble takes to reach the bottom of the well

Therefore, the depth of the well is
$S= \frac{1}{2}(9.81 m/s^2)(2.7 s)^2 = 35.7 m \sim 36 m$
and the correct answer is B.

8 0

3 years ago

A birdcage with mass 22.5kg at rest on a living room floor is acted on by a net horizontal force of

patriot [66]

m = mass of the birdcage = 22.5 kg

F = net force acting on birdcage to move it = 140 N

a = acceleration produced due to the force applied

Using newton's second law

a = F/m

inserting the values

a = 140/22.5

a = 6.22 m/s²

t = time of travel of crate = 10.5 s

v₀ = initial velocity of the crate = 0 m/s

X = displacement of the crate

displacement of the crate is given as

X = v₀ t + (0.5) a t²

X = 0 (10.5) + (0.5) (6.22) (10.5)²

X = 342.88 m

7 0

3 years ago

Read 2 more answers

A 28.0 kg child plays on a swing having support ropes that are 2.30 m long. A friend pulls her back until the ropes are 45.0 ∘ f

Dimas [21]

Answer

A)184.9J

B)=3.63m/s

C) Zero

Explanation:

A)potential energy of the child at the initial position, measured relative the her potential energy at the bottom of the motion, is

U=Mgh

Where m=28kg

g= 9.8m/s

h= difference in height between the initial position and the bottom position

We are told that the rope is L = 2.30 m long and inclined at 45.0° from the vertical

h=L-Lcos(x)= L(1-cosx)=2.30(1-cos45)

=0.674m

Her Potential Energy will now

= 28× 9.8×0.674

=184.9J

B)we can see that at the bottom of the motion, all the initial potential energy of the child has been converted into kinetic energy:

E= 0.5mv^2

where

m = 28.0 kg is the mass of the child

v is the speed of the child at the bottom position

Solving the equation for v, we find

V=√2k/m

V=√(2×184.9/28

=3.63m/s

C)we can find work done by the tension in the rope is given using expresion below

W= Tdcosx

where W= work done

T is the tension

d = displacement of the child

x= angle between the directions of T and d

In this situation, we have that the tension in the rope, T, is always perpendicular to the displacement of the child, d. x= 90∘ and cos90∘=0 hence, the work done is zero.

6 0

3 years ago

What must be applied to move a load?

Misha Larkins [42]

An applied force must be applied to move a load. This applied force must be large enough to overcome any opposing forces in order to move the load.

3 0

4 years ago

Read 2 more answers

If it takes a ball dropped from rest 2.261 s to fall to the ground, from what height H was it released? Express your answer in m

algol [13]

Answer:

Height, H = 25.04 meters