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ludmilkaskok [199]
3 years ago
6

Light illuminates two closely spaced thin slits and produces an interference pattern on a screen behind. How will the distance b

etween the fringes of the pattern differ for red light and blue light? 1. The same spacing for both 2. Closer for red light 3. Farther apart for red light
Physics
1 answer:
umka21 [38]3 years ago
3 0

Answer:

3. Farther apart for red light

Explanation:

The distance y between fringes is

y = \dfrac{\lambda L }{d},

where \lambda is the wavelength of light, L is the distance to the screen, and d is the slit separation.

Now, the wavelength \lambda_r of the red light is greater than for the wavelength \lambda_b for blue light:

\lambda_r> \lambda_b,

which means

\dfrac{\lambda_r L }{d}>\dfrac{\lambda_b L }{d},

In other words, the distance between the fringes is greater for red light, which from the options is choice 3.

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Cart A and cart B are traveling in the same direction on a straight track. Cart A is moving at 9.25 m/s and cart B is moving at
Liula [17]

Answer:

The speed of cart B is 11.21 m/s.

Explanation:

Given that,

Speed of cart A = 9.25 m/s

Speed of cart B = 7.15 m/s

Mass of cart A = 72.0 kg

Mass of cart B = 55.0 kg

Speed of card A after collision = 6.15 m/s

We need to calculate the speed of cart B

Using conservation of momentum

m_{A}v_{A}+m_{B}v_{B}=m_{A}v_{A}+m_{B}v_{B}

Put the value into the formula

72.0\times9.25+55.0\times7.15=72.0\times6.15+55.0\times v_{B}

v_{B}=\dfrac{72.0\times9.25+55.0\times7.15-72.0\times6.15}{55.0}

v_{B}=11.21\ m/s

Hence, The speed of cart B is 11.21 m/s.

7 0
3 years ago
While skydiving, your parachute opens and you slow from 50.0 m/s to 8.0 m/s in 0.75 s . Determine the distance you fall while th
Anestetic [448]

Answer:

21.75 m

Explanation:

t = Time taken for the car to slow down = 0.75 s

u = Initial velocity = 50 m/s

v = Final velocity = 8 m/s

s = Displacement

a = Acceleration

Equation of motion

v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{8-50}{0.75}\\\Rightarrow a=-56\ m/s^2

Acceleration is -56 m/s²

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{8^2-50^2}{2\times -56}\\\Rightarrow s=21.75\ m

The distance covered in the 0.75 seconds is 21.75 m

8 0
3 years ago
What happens when a star exhausts its core hydrogen supply?
Allushta [10]

Answer:

It becomes a giant or supergiant.

Explanation:

Once all the hydrogen supply is gone, fusion of hydrogen into helium stops. The core starts to contract and liberates energy, which heats the superior layer until it becomes hot enough to start the fusion of hydrogen into helium.

6 0
3 years ago
A stone is launched from the ground, at a 70° angle, with an initial velocity of 120 m/s.
zavuch27 [327]
<span>A) x = 41t
    The classic equation for distance is velocity multiplied by time. And unfortunately, all of your available options have the form of that equation. In fact, the only difference between any of the equations is what looks to be velocity. And in order to solve the problem initially, you need to divide the velocity vector into a vertical velocity vector and a horizontal velocity vector. And the horizontal velocity vector is simply the cosine of the angle multiplied by the total velocity. So H = 120*cos(70) = 120*0.34202 = 41.04242 So the horizontal velocity is about 41 m/s. Looking at the available options, only "A" even comes close.</span>
3 0
3 years ago
Read 2 more answers
A runner has an original velocity of 6 m/s and slows to a final velocity of 0 m/s. If the runner covers a
myrzilka [38]

Answer:

4 s

Explanation:

Given:

Δx = 12 m

v₀ = 6 m/s

v = 0 m/s

Find: t

Δx = ½ (v + v₀) t

12 m = ½ (0 m/s + 6 m/s) t

t = 4 s

7 0
3 years ago
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