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3 years ago

A physicist wants to study the motion of a car.

Physics

1 answer:

Aleonysh [2.5K]3 years ago

3 0

Answer:

Explanation:

in a

$1 mi = 1.61km$

so,in 50 miles

$50 mi = 50 \times 1.61$

$= 80.5 \: km$

in b

$1 \: mi = 1760 \: yd$

so in 22.3 miles

$22.3 = 22.3 \times 1760$

$= 39248 \: yd$

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Practice questions, will mark brainliest!

andrew-mc [135]

Answer:

266.67Watts

Explanation:

Time = 2.5hr to seconds

3600s = 1hr

2.5hrs = 3600×2.5= 9000s

Force = 32N

Distance = 75km to m

1000m = 1km

75km = 1000×75 = 75000m

Power = workdone / time

Work = force × distance

Therefore work = 32N × 75000m

Work = 2400000Nm

Power = work ➗ time

Power = 2400000Nm ➗ 9000s

Power = 266.67Watts

Watts is the S. i unit of power

I hope this was helpful, please mark as brainliest

4 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 $\vartriangle r=m \lambda$