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ankoles [38]
3 years ago
13

What happens when white light shines onto a red filter?

Physics
1 answer:
Gnesinka [82]3 years ago
6 0
The dye in between the two pieces of glass in the filter absorbs all the other colors in the white light except the red, and the red is the only light left to come out the other side. That's why the filter usually looks kind of red. It's also a big part of the reason why we call it a "red filter".
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Definition for compression and rarefraction
s344n2d4d5 [400]
Compression and rarefaction are two phenomenon occurs in longitudunal wave!

when there is denser particle gathering in that wave , there we called it compression and the rarer part of particles is rarefaction !
6 0
3 years ago
Read 2 more answers
A car slows down from 80 km/h to 60 km/h in 2 seconds.
Andreyy89

Answer:

please mark brainlist

Explanation:

Initial speed   u=80 km/h=80×185=22.22 m/s

Final speed    v=60 km/h=60×185=16.67 m/s

Using   v=u+at

Or   16.67=22.22+α×5

⟹ a=−1.1 m/s2

6 0
3 years ago
He throws a second ball (B2) upward with the same initial velocity at the instant that the first ball is at the ceiling. c. How
Gwar [14]

Answer:

hello your question has some missing parts

A juggler performs in a room whose ceiling is 3 m above the level of his hands. He throws a ball vertically upward so that it just reaches the ceiling.

answer : c) 0.39 sec

               d)  2.25 m

               e) 1.92 m/sec

Explanation:

The initial velocity of the first ball = 7.67 m/sec ( calculated )

Time required for first ball to reach ceiling = 0.78 secs ( calculated )

Determine how long after the second ball is thrown do the two balls pass each other

Distance travelled by first ball downwards when it meets second ball can be expressed as : d = 1/2 gt^2 =  9.8t^2 / 2

hence d = 4.9t^2  ----- ( 1 )

Initial speed of second ball = first ball initial speed = 7.67 m/sec

3 - d = 7.67t - 4.9t  ---- ( 2 )

equating equation 1 and 2

3 = 7.67t   therefore t = 0.39 sec

Determine how far the balls are above the Juggler's hands ( when the balls pass each other )

form equation 1 ;

d = 4.9 t^2 = 4.9 *(0.39)^2 = 0.75 m

therefore the height the balls are above the Juggler's hands is

3 - d = 3 - 0.75 = 2.25 m

determine their velocities when the pass each other

velocity = displacement / time

velocity = d / t = 0.75 / 0.39 sec  = 1.92 m/sec

7 0
3 years ago
Forces can act on an object in the same direction or in opposite. how does each situation affect the motion of the object?
Anon25 [30]



Hi pupil here's your answer ::




➡➡➡➡➡➡➡➡➡➡➡➡➡



Action and Reaction do not act on the same body !! If they acted on the same body, the resultant force will be zero and their could be never accelerated motion.

If both the forces acted on the same body, then if they are equal to opposite direction the object will remain stationary. If on of the forces is greater than other the object will move in the direction of greater force.

If both acted in the same direction there would be an accelrated motion.




⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅





Hope this helps . . . . .
5 0
3 years ago
In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the
Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, u_0= 10 m/s.

Angle of launch with the horizontal, \theta = 53 ^{\circ}

So, the vertical component of the initial velocity,

u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i).

The horizontal component of the initial velocity,

u_0\cos\theta=10 \cos 53 ^{\circ}

Let, t be the time of flight, to the horizontal distance covered

D=10 \cos (53 ^{\circ})t\cdots(ii).

Not, applying the equation of motion in the vertical direction.

s= ut +\frac 1 2 at^2

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, u =10 \sin 53 ^{\circ} (from equation (i), s=0 (as the final height is same as the launch height) and a = -9.81 m/s^2 (negative sign is due to the downward direction).

\Rightarrow 0 = 10 (\sin 53 ^{\circ})t-\frac 1 2 (9.81)t^2

\Rightarrow t= \frac {2\times 10 (\sin 53 ^{\circ})}{9.81}=1.63 seconds.

So, the total time in the air is 1.63 seconds.

From equation (i),

Total horizontal distance covered is

D=10 \cos (53 ^{\circ})\times 1.63 = 9.81 m.

Now, for the maximum height, H, applying the equation of motion as

v^2=u^2+2as

Here, v is the final velocity and v=0 (at the maximum height), and h=H.

So, 0^2=(10 \sin 53 ^{\circ})^2-2(9.81)H

\Rightarrow H = \frac {(10 \sin 53 ^{\circ})^2}{2\times 9.81}

\Rightarrow H = 3.25 m.

Hence, the maximum height covered is 3.25 m.

8 0
3 years ago
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