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matrenka [14]
3 years ago
11

Differences between Reflection of light and Reflection of sound​

Physics
1 answer:
Leokris [45]3 years ago
7 0

Answer:

The law of reflection of sound waves states that the angle of incidence is always equal to the angle of reflection. But unlike reflection of light on a highly smooth surface, in the reflection of the sound wave, a part of the incident wave gets transmitted to the medium where it hits.

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What studies the natural world around us
Tanzania [10]

Science studies the natural world. This includes the components of the physical universe around us like atoms, plants, ecosystems, people, societies and galaxies, as well as the natural forces at work on those things.

<h3><u>PLEASE</u><u> MARK</u><u> ME</u><u> BRAINLIEST</u><u>.</u></h3>

6 0
3 years ago
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A spring loaded toy shoots straight upward with a velocity of 4.5 m/s. Determine the maximum height it reaches. Determine the ti
Angelina_Jolie [31]

Recall that

{v_y}^2-{v_{0y}}^2=2a_y(y-y_0)

At its maximum height y_{\mathrm{max}}, the toy will have 0 vertical velocity, so that

-\left(4.5\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)y_{\mathrm{max}}

\implies y_{\mathrm{max}}=1.0\,\mathrm m

For the toy to reach this maximum height, it takes time t such that

\dfrac{0+v_0}2t=y_{\mathrm{max}}\implies t=0.46\,\mathrm s

which means it takes twice this time, i.e. t=0.92\,\mathrm s, for the toy to reach its original position.

The velocity of the toy when it falls 1.0 m below its starting point is

{v_y}^2-\left(4.5\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)(0-1.0\,\mathrm m)

\implies{v_y}^2=39.85\,\dfrac{\mathrm m^2}{\mathrm s^2}

\implies v_y=-6.4\,\dfrac{\mathrm m}{\mathrm s}

where we took the negative square root because we expect the toy to be moving in the downward direction.

6 0
3 years ago
Consider the two rotating pulley-like systems above. They are identical in every way EXCEPT the mass of the pulley on the left i
____ [38]

Answer:

Consider the two rotating pulley-like systems above. They are identical in every way EXCEPT the mass of the pulley on the left is concentrated on the outside edge (ring or hoop shape), while the mass of pulley on the right is evenly distributed throughout the pulley (disc or cylinder shape). Given R =0.15m, M =0.6 kg, and m= 0.5 kg. Both pulley systems are initially at rest and the block masses are a height of 0. 8m above the ground.

Explanation:

If both masses are released at the same time, which pulley will hit the ground first

5 0
3 years ago
Which will be attracted to (or repelled by) a magnet ?
Dima020 [189]
I believe it’s A C F
4 0
3 years ago
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- A cannon of 2000 kg fires a shell of 10 kg at
fgiga [73]

1) -0.5 m/s

We can solve the first part of the problem by using the law of conservation of momentum. In fact, the total momentum of the cannon - shell system must be conserved.

Before the shot, both the cannon and the shell are at rest, so the total momentum is zero:

p=0

After the shot, the momentum is:

p=MV+mv

where

M = 2000 kg is the mass of the cannon

m = 10 kg is the mass of the shell

v = 100 m/s is the velocity of the shell (we take as positive the direction of motion of the shell)

V = ? is the velocity of the cannon

Since momentum is conserved, we can write

0=MV+mv

And solving for V, we find the velocity of the cannon:

V=-\frac{mv}{M}=-\frac{(10)(100)}{2000}=-0.5 m/s

where the negative sign indicates that the cannon moves in the direction opposite to the shell.

2) 0.5 m

The motion of the cannon is a uniformly accelerated motion, so we can solve this part by using suvat equation:

v^2-u^2=2as

where

v is the final velocity of the cannon

u = 0.5 m/s is the initial velocity of the cannon (now we take as positive the initial direction of motion of the cannon)

a=-0.25 m/s^2 is the deceleration of the cannon

s is the distance travelled by the cannon

The cannon will stop when v = 0; substituting and solving the equation for s, we find the minimum safe distance required to stop the cannon:

s=\frac{v^2-u^2}{2a}=\frac{0-0.5^2}{2(-0.25)}=0.5 m

7 0
4 years ago
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