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3 years ago

Will mark as brainliest if correct!!!

Physics

1 answer:

Mrrafil [7]3 years ago

4 0

D. transmits

Explanation:

A medium for light is something that can easily transmit light.

Light is an electromagnetic radiation that are transverse waves which propagates by the vibration of electric and magnetic field.

A medium allows something to be transmitted without it being obstructed.
Although the velocity and other properties might be attenuated, the particular property of the wave is maintained.
A medium is not a source, reflector or absorber.
They simply transmit a wave.

learn more:

Transmission of light in the eye brainly.com/question/8032392

#learnwithBrainly

You might be interested in

Gravitational force is determined by which of the following

hram777 [196]

Gravitational force is determined by mass

Answer: Option B

Explanation:

According to Universal law of gravity, the gravitational force is directly proportional to the product of the masses of the objects and inversely proportional to the square of the distance between the objects.

$F=G \times \frac{m_{1} \times m_{2}}{r^{2}}$

Where,

G – gravitational constant = $\text { 6. } 67 \times 10^{-11} \mathrm{Nm}^{2} / \mathrm{kg}^{2}$

$\boldsymbol{m}_{1}, \boldsymbol{m}_{2}$ = masses of two objects

r – distance between the objects

So, as per this law, the gravitational force is found by mass.

8 0

4 years ago

A bicyclist starts at rest and speeds up to 30 m/s while accelerating at 4 m/s^2. Determine the distance traveled.

raketka [301]

Answer:

Distance, d = 112.5 meters

Explanation:

Initially, the bicyclist is at rest, u = 0

Final speed of the bicyclist, v = 30 m/s

Acceleration of the bicycle, $a=4\ m/s^2$

Let s is the distance travelled by the bicyclist. The third equation of motion is given as :

$v^2-u^2=2as$

$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{(30)^2}{2\times 4}$

s = 112.5 meters

So, the distance travelled by the bicyclist is 112.5 meters. Hence, this is the required solution.

6 0

3 years ago

To test the performance of its tires, a car

Rom4ik [11]

The coefficient of static friction is 0.222

Explanation:

In order for the car to remain in circular motion, the frictional force must be able to provide the necessary centripetal force. Therefore, the car will start skidding when the two forces are equal:

$\mu mg=m\frac{v^2}{r}$

where the term on the left is the frictional force, while the term on the right is the centripetal force, and where

$\mu$ is the coefficient of static friction

m is the mass of the car

g is the acceleration of gravity

v is the speed of the car

r is the radius of the track

In this problem, we have:

r = 564 m

v = 35 m/s

$g=9.8 m/s^2$

And re-arranging the equation for $\mu$ , we can find the coefficient of static friction:

$\mu = \frac{v^2}{gr}=\frac{35^2}{(9.8)(564)}=0.222$

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

5 0

3 years ago

Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant

irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

$a_g = \frac{GM}{R^2}$

Here

$M = \text{Mass inside the Orbit of the star}$

$R = \text{Orbital radius}$

$G = \text{Universal Gravitational Constant}$

Mass inside the orbit in terms of Volume and Density is

$M =V \rho$

Where,

V = Volume

$\rho =$ Density

Now considering the volume of the star as a Sphere we have

$V = \frac{4}{3} \pi R^3$

Replacing at the previous equation we have,

$M = (\frac{4}{3}\pi R^3)\rho$

Now replacing the mass at the gravitational acceleration formula we have that

$a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho$

$a_g = \frac{4}{3} G\pi R\rho$

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration. So centripetal acceleration of the star is

$a_c = \frac{4}{3} G\pi R\rho$

At the same time the general expression for the centripetal acceleration is

$a_c = \frac{\Theta^2}{R}$

Where $\Theta$ is the orbital velocity

Using this expression in the left hand side of the equation we have that

$\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2$

$\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}$

$\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R$

Considering the constant values we have that

$\Theta = \text{Constant} \times R$

$\Theta \propto R$

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

6 0

3 years ago

Can you explain that gravity pulls us to the Earth & can you calculate weight from masses on both on Earth and other planets

schepotkina [342]

I don't actually understand what your question is, but I'll dance around the subject
for a while, and hope that you get something out of it.

-- The effect of gravity is: There's a pair of forces, in both directions, between
every two masses.

-- The strength of the force depends on the product of the masses, so it doesn't matter whether there's a big one and a small one, or whether they're nearly equal.
It's the product that counts. Bigger product ==> stronger force, in direct proportion.

-- The strength of the forces also depends on the distance between the objects' centers. More distance => weaker force. Actually, (more distance)² ==> weaker force.

-- The forces are equal in both directions. Your weight on Earth is exactly equal to
the Earth's weight on you. You can prove that. Turn your bathroom scale face down
and stand on it. Now it's measuring the force that attracts the Earth toward you.
If you put a little mirror down under the numbers, you'll see that it's the same as
the force that attracts you toward the Earth when the scale is right-side-up.

-- When you (or a ball) are up on the roof and step off, the force of gravity that pulls
you (or the ball) toward the Earth causes you (or the ball) to accelerate (fall) toward the Earth.
Also, the force that attracts the Earth toward you (or the ball) causes the Earth to accelerate (fall) toward you (or the ball).
The forces are equal. But since the Earth has more mass than you have, you accelerate toward the Earth faster than the Earth accelerates toward you.

-- This works exactly the same for every pair of masses in the universe. Gravity
is everywhere. You can't turn it off, and you can't shield anything from it.

-- Sometimes you'll hear about some mysterious way to "defy gravity". It's not possible to 'defy' gravity, but since we know that it's there, we can work with it.
If we want to move something in the opposite direction from where gravity is pulling it, all we need to do is provide a force in that direction that's stronger than the force of gravity.
I know that sounds complicated, so here are a few examples of how we do it:
-- use arm-muscle force to pick a book UP off the table
-- use leg-muscle force to move your whole body UP the stairs
-- use buoyant force to LIFT a helium balloon or a hot-air balloon
-- use the force of air resistance to LIFT an airplane.

-- The weight of 1 kilogram of mass on or near the Earth is 9.8 newtons. (That's
about 2.205 pounds). The same kilogram of mass has different weights on other planets. Wherever it is, we only know one of the masses ... the kilogram. In order
to figure out what it weighs there, we need to know the mass of the planet, and
the distance between the kilogram and the center of the planet.

I hope I told you something that you were actually looking for.

7 0

3 years ago