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likoan [24]
3 years ago
14

What happens when a wave encounters a boundary and then goes back to the original medium?

Physics
1 answer:
lawyer [7]3 years ago
7 0
It causes a movement of convection in the water resulting in a pulling current.

Hope this helpss
You might be interested in
If the acceleration of the projective is: a = c s m/s 2 Where c is a constant that depends on the initial gas pressure behind th
Sedbober [7]

Answer:

c = 4,444.44

Explanation:

You have the following expression for the acceleration of the projectile:

a=cs   (1)

s: distance to the ground of the projectile

To find the value of the constant c you use the following formula:

v^2=v_o^2+2a \Delta s   (2)

vo: initial  velocity = 0 m/s

v: final speed = 200 m/s

Δs: distance traveled by the projectile = 3m - 1.5m = 1.5m

You replace the expression (1) into the expression (2):

v^2=2(cs)\Delta s

You do the constant c in the last equation, then you replace the values of v, s and Δs:

c=\frac{v^2}{2s\Delta s}=\frac{(200m/s)^2}{2(3m/s^2)(1.5m)}=4444.44

6 0
3 years ago
I NEED HELP!!!!!!!!!!!
nadezda [96]

1. 168.1 Hz

To find the apparent frequency heard by the driver in the car, we can use the formula for the Doppler effect:

f'=(\frac{v\pm v_o}{v\pm v_s})f

where

f is the original sound of the horn

v is the speed of sound

v_o is the velocity of the observer (the driver and the car), which is positive if the observer is moving towards the source and negative if it is moving away

v_s is the velocity of the sound source (the train), which is positive if the source is moving away from the observer and negative otherwise

In this problem we have, according to the sign convention used:

v = 343 m/s\\f = 164 Hz\\v_o = -15 m/s\\v_s = -23 m/s

Substituting, we find:

f'=(\frac{343-15}{343-23})(164)=168.1 Hz

2.  2.96\cdot 10^8 m/s

The speed of light can be calculated as

v=\frac{d}{t}

where

d is the distance travelled

t is the time taken

In this problem:

d=2\cdot 3.85\cdot 10^8 =7.7\cdot 10^8 m is the total distance travelled by the laser beam (twice the distance between the Earth and the Moon)

t = 2.60 s is the time taken

Substituting in the formula,

v=\frac{7.7\cdot 10^8 m}{2.60 s}=2.96\cdot 10^8 m/s

6 0
3 years ago
A projectile is fired with an initial velocity of 450 feet per second at an angle of 70° with the horizontal.
pav-90 [236]
<h2>After 26.28 seconds projectile returns 26.28 seconds.</h2>

Explanation:

Initial velocity = 450 ft/s = 137.16 m/s

Angle, θ = 70°

Consider the vertical motion of projectile,

When the projectile return to the ground we have

           Displacement, s = 0 m

           Acceleration, a = -9.81 m/s²

            Initial velocity, u = 137.16 x sin70 = 128.89 m/s

Substituting in s = ut + 0.5 at²

                s = ut + 0.5 at²

                0 = 128.89 x t + 0.5 x (-9.81) x t²

                t² - 26.28 t = 0

                t ( t- 26.28) = 0

               t = 0 s or t = 26.28 s

After 26.28 seconds projectile returns 26.28 seconds.

5 0
3 years ago
A ball is tossed vertically upward. When it reaches its highest point (before falling back downward) Group of answer choices the
nignag [31]

Answer:

the velocity is zero, the acceleration is directed downward, and the force of gravity acting on the ball is directed downward

Explanation:

Is this exercise in kinematics

          v = v₀ - g t

where g is the acceleration of the ball, which is created by the attraction of the ball to the Earth.

At the highest point

velocity must be zero.

The acceleration depends on the Earth therefore it is constant at this point and with a downward direction.

The force of the earth on the ball is towards the center of the Earth, that is, down

all other alternatives are wrong

7 0
3 years ago
When point charges q = +8.4 uC and q2 = +5.6 uC are brought near each other, each experiences a repulsive force of magnitude 0.6
Stella [2.4K]

Answer:

it is separated by 80 cm distance

Explanation:

As per Coulombs law we know that force between two point charges is given by

F = \frac{kq_1q_2}{r^2}

here we know that

q_1 = +8.4\mu C

q_2 = +5.6 \mu C

force between two charges is given as

F = 0.66 N

now we have

F = \frac{kq_1q_2}{r^2}

0.66 = \frac{(9\times 10^9)(8.4 \mu C)(5.6 \mu C)}{r^2}

r = 0.8 m

so it is separated by 80 cm distance

3 0
3 years ago
Read 2 more answers
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