A parked car begins to roll down a hill, what can you conclude from that observation?

Riders on a ferris wheel move in a circle with a speed of 4.0 m/s. As they go around, they

Vanyuwa [196]

Answer: D = 16m

Explanation: given values: a = 2 m/s2, v = 4 m/s
In this case we have to determine the diameter of the Ferris wheel.
Diameter of circle is given as: D = 2.r.
First we have to find radius of wheel. The best way to find that is using the centripetal acceleration equation: a = v2/r
Plug in values in above equation to find radius: 2 m/s2 = (4 m/s)2/r 2 m/s2 = (16 m2/s2)/r r = (16 m2/s2)/2 m/s2
r = 8.0m
Diameter of Ferris wheel is:
D = 2.r.
D = 2.8m
D = 16m

6 0

3 years ago

The viewing screen in a double-slit experiment with monochromatic light. Fringe C is the central maximum. The fringe separation

makvit [3.9K]

Answer:

Part A:

a) If the wavelength of the light is decreased the fringe spacing Δy will decrease.

Part B:

b) If the spacing between the slits is decreased the fringe spacing Δy will increase.

Part C:

a) If the distance to the screen is decreased the fringe spacing will decrease.

Part D:

The dot in the center of fringe E is $920\ x\ 10^{-9} m$ farther from the left slit than from the right slit.

Explanation:

In the double-slit experiment there is a clear contrast between the dark and bright fringes, that indicate destructive and constructive interference respectively, in the central peak and then is less so at either side.

The position of bright fringes in the screen where the pattern is formed can be calculated with

$\vartriangle y =\frac{m \lambda L}{d}$

$m=0,\pm 1,\pm 2,\pm 3,.....$

m is the order number.
$\lambda$ is the wavelength of the monochromatic light.
L is the distance between the screen and the two slits.
d is the distance between the slits.

Part A: a) In the above equation for the position of bright fringes we can see that if the wavelength of the light $\lambda$ is decreased the overall effect will be that the fringes are going to be closer. That means that the fringe spacing Δy will decrease.

Part B: b) In the above equation for the position of bright fringes we can see that if the spacing between the slits d is decreased the fringes are going to be wider apart. That means the fringe spacing Δy will increase.
Part C: a) In the above equation we can see that if the distance to the screen L is decreased the fringes are going to be closer. That means the fringe spacing Δy will decrease.

Part D: We are told that the central maximum is the fringe C that corresponds with m=0. That means that fringe E corresponds with the order number m=2 if we consider it to be the second maximum at the rigth of the central one. To calculate how much farther from the left slit than from the right slit is a dot located at the center of the fringe E in the screen we use the condition for constructive interference. That says that the path length difference Δr between rays coming from the left and right slit must be $\vartriangle r=m \lambda$

We simply replace the values in that equation :

$\vartriangle r= m \lambda =2.\ 460\ nm$

$\vartriangle r= 920\ x\ 10^{-9} m$

The dot in the center of fringe E is $920\ x\ 10^{-9}m$ farther from the left slit than from the right slit.

3 0

3 years ago

A basketball star covers 3.05 m horizontally in a jump to dunk the ball (see figure). His motion through space can be modeled pr

rosijanka [135]

anonymous 4 years agoAny time you are mixing distance and acceleration a good equation to use is ΔY=Viyt+1/2at2 I would split this into two segments - the rise and the fall. For the fall, Vi = 0 since the player is at the peak of his arc and delta-Y is from 1.95 to 0.890. For the upward part of the motion the initial velocity is unknown and the final velocity is zero, but motion is symetrical - it takes the same amount of time to go up as it does to go down. Physiscists often use the trick "I'm going to solve a different problem, that I know will give me the same answer as the one I was actually asked.) So for the first half you could also use Vi = 0 and a downward delta-Y to solve for the time. Add the two times together for the total. The alternative is to calculate the initial and final velocity so that you have more information to work with.

5 0

3 years ago

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What is the main determining factor in defining boundaries between layers of earth's atmosphere?

konstantin123 [22]

The main determining factor in defining boundaries between layers of earth's atmosphere would be temperature changes in these layers. Temperature is one essential property that varies in the atmosphere. Based from this variation, the atmosphere is divided into four major layers and further to three smaller layers - troposphere, tropopause, the stratosphere, stratopause, the mesosphere, mesopause, and the thermosphere.The troposphere is the layer that is nearest to the surface of the Earth. It is the part where humans, plants and animals survive. Also, it is the warmest layer of the atmosphere. And as we go higher the atmosphere, the temperature would drop making it much cooler.

5 0

3 years ago

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Please answer the question

AfilCa [17]

The answer is Cㅤㅤㅤㅤㅤ

4 0

3 years ago