When does The velocity of a wave change

you have two solid substances that look the same. What measurements would you take and which tests would you preform to determin

musickatia [10]

To see how much they weigh

7 0

3 years ago

1. Potential energy and a roller coaster. A 1000-kg roller coaster car moves from point A to B then C.

zalisa [80]

Answer:(a) With our choice for the zero level for potential energy of the car-Earth system when the car is at point B ,

U

B

=0

When the car is at point A, the potential energy of the car-Earth system is given by

U

A

=mgy

where y is the vertical height above zero level. With 135ft=41.1m, this height is found as:

y=(41.1m)sin40.0

0

=26.4m

Thus,

U

A

=(1000kg)(9.80m/s

2

)(26.4m)=2.59∗10

5

J

The change in potential energy of the car-Earth system as the car moves from A to B is

U

B

−U

A

=0−2.59∗10

5

J=−2.59∗10

5

J

(b) With our choice of the zero configuration for the potential energy of the car-Earth system when the car is at point A, we have U

A

=0. The potential energy of the system when the car is at point B is given by U

B

=mgy, where y is the vertical distance of point B below point A. In part (a), we found the magnitude of this distance to be 26.5m. Because this distance is now below the zero reference level, it is a negative number.

Thus,

8 0

2 years ago

2. A car with mass 2,500 kg is travelling at 15 m/s. What is the car's kinetic energy?

BabaBlast [244]

Answer:

c, 280,000J

Explanation:

You should use the equation KE (Kinetic Energy) = $\frac{1}{2} mv^2$

m = 2500

v = 15

KE = $\frac{1}{2} * 2500 * (15)^2\\$

KE = 281250 Joules.

This is consistent with the answer B, as all of the parameters given to you are rounded to 2 significant figures so naturally, the answer should also be rounded to 2 sig figs.

KE = 280,000 J

7 0

3 years ago

What is the frequency of a clock waveform whose period is 750 microseconds?

stich3 [128]

The frequency f of a wave is defined as the inverse of the period T:
$f= \frac{1}{T}$
The clock in the problem has a waveform with period $T=750 ms=0.750 s$ . Therefore, its frequency is
$f= \frac{1}{0.750 s}=1.33 Hz$

6 0

4 years ago

You are designing a new piece of fishing equipment that will allow you to catch a fish from the surface. You know that water ref

OleMash [197]

Answer:

Law of refraction

Explanation:

An experiment to analyze the refraction of light in water can easily be performed with a laser pointer and protractor.

We throw the fishing rod line into the water, place the protractor at the point where the line touches the water and use the direction of the line for the direction of the laser pointer (on), the laser is visible by the reflection on the particles in the air.

The vertical line is called Normal and all angles must be measured with respect to this reference in optics.

Having these angles and the refractive index of water we can use the law of refraction

n₁ sin θ₁ = n₂ sin θ₂

θ₂ = $sin^{-1} ( \frac{n_1}{n_2} \ sin \ \theta_1 )$

we can repeat several times to analyze several different input points (different angles) and to decrease the errors in the measurements.

the refractive index of air is n1 = 1 and n2= 1.33 (water)

4 0

3 years ago