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3 years ago

The substance that a wave moves through is called a __________.

2 answers:

6 0

- - A medium is the substance in which a wave moves through. This is because the medium in a wave makes it possible to generate and transfer energy from one location to another, which is why the waves are able to go out, than in.

klio [65]3 years ago

4 0

The substance that a wave moves through is called a _medium_

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In midair in the international space station a 1 kg chunk of putty moving at 1 m/s collides with and sticks to a 5 kg chunk of p

Delicious77 [7]

<span>We know that the momentum keeps constant in a inelastic collisions, so the product of mass and speed do not change:
m1 * v1 + m2 * v2 = m * v
1 * 1 + 5 * 0 = (1 + 5) * v
1 = 6 * v
v = 1/6 m/s
So the final speed of the 6 kg chunk will travel at 0.167 m/s</span>

6 0

3 years ago

4. Which of the following would be a good reference point to describe the motion of a dog?

saul85 [17]

ANOTHER RUNNING DOG

Explanation:

In the given question it is to find a suitable reference point to describe the motion of dog. Here I could suggest that it is better to compare the dog with another running dog to create the relative speed difference to get a reliable motion variation.

Because the motion of dog is in the linear with respect to the another dog and to the acceleration produced by the dog in the required interval is easy to calculate with respect to another dog which is already in motion.

Hence, I suggest that Motion of dog can be analysed better by analyse the motion variation of dog with another dog running.

4 0

3 years ago

Will mark as brainlist for only correct answer

Crank

Answer:

low, low

Explanation:

Longer wavelengths will have lower frequencies, and shorter wavelengths will have higher frequencies.

Large amplitude waves contain more energy. The other is frequency, which is the number of waves that pass by each second. If more waves( or more wiggly lines) pass by, more energy is transferred each second

4 0

3 years ago

A stone with a mass m is dropped from an airplane that has a horizontal velocity v at a height h above a lake. If air resistance

seropon [69]

Answer: Option B. R = (1/2)gt^2

Explanation:

S = R (horizontal distance)

V^2 = 2gS

V^2 = 2gR

R = V^2 / 2g

But V = gt

R = (gt)^2 / 2g

R = (g^2 x t^2) / 2g

R = gt^2 / 2

But t^2 = 2h/g

R = ( g x 2h/g) / 2

R = h

But h = (1/2)gt^2

R = h = (1/2)gt^2

4 0

3 years ago

Water moves through a constricted pipe in steady, ideal flow. At the

Irina-Kira [14]

A) Speed in the lower section: 0.638 m/s

B) Speed in the higher section: 2.55 m/s

C) Volume flow rate: $1.8\cdot 10^{-3} m^3/s$

Explanation:

To solve the problem, we can use Bernoulli's equation, which states that

$p_1 + \rho g h_1 + \frac{1}{2}\rho v_1^2 = p_2 + \rho g h_2 + \frac{1}{2}\rho v_2^2$

where

$p_1=1.75\cdot 10^4 Pa$ is the pressure in the lower section of the tube

$h_1 = 0$ is the heigth of the lower section

$\rho=1000 kg/m^3$ is the density of water

$g=9.8 m/s^2$ is the acceleration of gravity

$v_1$ is the speed of the water in the lower pipe

$p_2$ is the pressure in the higher section

$h_2 = 0.250 m$ is the height in the higher pipe

$v_2$ is hte speed in the higher section

We can re-write the equation as

$v_1^2-v_2^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}$ (1)

Also we can use the continuity equation, which state that the volume flow rate is constant:

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-section of the lower pipe, with

$r_1 = 3.00 cm =0.03 m$ is the radius of the lower pipe (half the diameter)

$A_2 = \pi r_2^2$ is the cross-section of the higher pipe, with

$r_2 = 1.50 cm = 0.015 m$ (radius of the higher pipe)

So we get

$r_1^2 v_1 = r_2^2 v_2$

And so

$v_2 = \frac{r_1^2}{r_2^2}v_1$ (2)

Substituting into (1), we find the speed in the lower section:

$v_1^2-(\frac{r_1^2}{r_2^2})^2v_1^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}\\v_1=\sqrt{\frac{2(p_2-p_1+\rho g h_2)}{\rho(1-\frac{r_1^4}{r_2^4})}}=0.638 m/s$