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Korvikt [17]
3 years ago
7

What happens to the wavelength of a wave if the frequency quadruples, but the wave is in the same medium?

Physics
1 answer:
Otrada [13]3 years ago
7 0

Answer:

I think C? I'm not sure totally though...

Explanation:

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A perpendicular force is applied to a certain area and produces a pressure P. If the
Furkat [3]

Answer:

p/2

Explanation:

4 0
2 years ago
A ball of mass 0.160 kg is dropped from a height of 2.25 m. When it hits the ground it compresses 0.087 m.
Studentka2010 [4]

A) 6.64 m/s downward

B) 0.026 s

C) -40.9 N

Explanation:

A)

We can solve this problem by using the law of conservation of energy.

In fact, since the total mechanical energy of the ball must be conserved, this means that the initial gravitational potential energy of the ball before the fall is entirely converted into kinetic energy just before it reaches the floor.

So we can write:

PE=KE\\mgh = \frac{1}{2}mv^2

where

m = 0.160 kg is the mass of the ball

g=9.8 m/s^2 is the acceleration due to gravity

h = 2.25 m is the initial height of the ball

v is the final velocity of the ball before hitting the ground

Solving for v, we find:

v=\sqrt{2gh}=\sqrt{2(9.8)(2.25)}=6.64 m/s

And the direction of the velocity is downward.

B)

The motion of the ballduring the collision is a uniformly accelerated motion (= with constant acceleration), so the time of impact can be found by using a suvat equation:

s=(\frac{u+v}{2})t

where:

v is the final velocity

u is the initial velocity

s is the displacement of the ball during the impact

t is the time

Here we have:

u = 6.64 m/s is the velocity of the ball before the impact

v = 0 m/s is the final velocity after the impact (assuming it comes to a stop)

s = 0.087 m is the displacement, as the ball compresses by 0.087 m

Therefore, the time of the impact is:

t=\frac{2s}{u+v}=\frac{2(0.087)}{0+6.64}=0.026 s

C)

The force exerted by the floor on the ball can be found using the equation:

F=\frac{\Delta p}{t}

where

\Delta p is the change in momentum of the ball

t is the time of the impact

The change in momentum can be written as

\Delta p = m(v-u)

So the equation can be rewritten as

F=\frac{m(v-u)}{t}

Here we have:

m = 0.160 kg is the mass of the ball

v = 0 is the final velocity

u = 6.64 m/s is the initial velocity

t = 0.026 s is the time of impact

Substituting, we find the force:

F=\frac{(0.160)(0-6.64)}{0.026}=-40.9 N

And the sign indicates that the direction of the force is opposite to the direction of motion of the ball.

4 0
3 years ago
Josie walks 15 meters north down the freshman hallway. She then walks 25 meters east. Lastly, she walks 15 meters south. What is
hichkok12 [17]

Answer:

the answer is yours a calculated

3 0
2 years ago
If there are 3 resistors of .5 ohms, 1 ohms, and 1 ohms connected in parallel then what is the equivalent resistance
Ede4ka [16]

Answer:0.45ohms

Explanation:

Let R be there equivalent resistance

1/R=1/r+1/r+1/r

1/R=1/5+1/1+1/1

1/R=1/5+2

1/R=(1+10)/5

1/R=11/5

Cross multiplying we get

11R=5

Divide both sides by 11

11R ➗ 11=5 ➗ 11

R=0.45ohms

3 0
3 years ago
The acceleration of a particle is defined by the relation a 5 28 m/s2. Knowing that x 5 20 m when t 5 4 s and that x 5 4 m when
mamaluj [8]

Explanation:

It is given that,

The acceleration of a particle, a=-8\ m/s^2 (negative as the particle is decelerating)

Initial distance, x₁ = 20 m

Initial time, t₁ = 4 s

New distance x₂ = 4 m

Velocity, v = 10 m/s

(A) Calculating initial distance using second equation of motion as :

x_1=ut_1+\dfrac{1}{2}at^2

20=4u+\dfrac{1}{2}(-8)\times 4^2

u = 21 m/s

When velocity of the particle is zero, time taken is t (say). Using first equation of motion as :

v=u+at

0=21+(-8)t

t = 2.62 seconds

So, the velocity of the particle is zero at t = 2.62 seconds.

(B) Velocity at t = 11 s

v=21+(-8)\times 11

v = 13 m/s

Total distance covered at t = 11 s. The overall path travelled by the particle during its entire journey is called total distance covered.

d=ut+\dfrac{1}{2}at^2+|ut+\dfrac{1}{2}at^2|

d=21\times 2.62+\dfrac{1}{2}\times (-8)(2.62)^2+|21\times 8.38+\dfrac{1}{2}\times (-8)(8.38)^2|

d = 132.48 m

So, the distance travelled by the particle at t = 11 seconds is 132.48 meters.    

5 0
3 years ago
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