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Advocard [28]
3 years ago
14

A beam of red light is made to pass through two slits that are 3.55 E-3 meters apart. On a screen 2.25 meters away from the slit

s, an interference pattern appears with bands of light separated by 1.25 E-4 meters. What is the wavelength of the light?
1.97E-7m
0.197m
9.98E.7m
5.07E6m
Physics
1 answer:
JulsSmile [24]3 years ago
7 0
I am assuming you know the relation obtained between slit width, distance of screen from slits, distance of interference pattern obtained on the screen from the center and the wavelength of monochromatic light used in Young's Double Slit experiment.
λ =\frac{y*d}{D} =  \frac{3.55*10^{-3}*1.25*10^{-4}  }{2.25} = 1.97*10^{-7} m
λ ~ 1.97 ×10⁻⁷m
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A baseball with a mass of 0.15 kilograms collides with a bat at a speed of 40 meters/second. The duration of the collision is 8.
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Answer: 1687.5 N

Explanation:

From the second law of motion given by Newton, Force is the rate change of momentum.

F = \frac{dp}{dt}=\frac {m dv}{dt} = \frac{m (v_f-v_i)}{dt}

Mass of the baseball, m = 0.15 kg

Initial velocity, v_i=-40 m/s (negative because direction of initial velocity is opposite to the final velocity)

Final velocity, v_f=50 m/s

The duration of collision, dt= 8.0 \times 10^{-3} s

Force, F = \frac{0.15 kg (50-(-40) m/s)}{8.0 \times 10^{-3} s}=1687.5 N

Hence, the value of force is 1687.5 N.

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In which direction does a bag at rest move when a force of 20 newtons is applied from the right?
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sammy [17]

Answer:

the final angular velocity of the platform with its load is 1.0356 rad/s

Explanation:

Given that;

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Initial angular velocity of platform ω₀ = 1.63 rad/s

mass of banana m_{b} = 8.97 kg

at distance r = 4/5  { radius of platform }

mass of monkey m_{m} = 22.1 kg

at edge = R

R = 1.73 m

now since there is No external Torque

Angular momentum will be conserved, so;

mR²/2 × ω₀ = [ mR²/2 + m_{b} (\frac{4}{5} R)² + m_{m}R² ]w

m/2 × ω₀ = [ m/2 + m_{b} (\frac{4}{5} )² + m_{m} ]w

we substitute

w = 97.1/2 × 1.63 / ( 97.1/2 + 8.97(16/25) + 22.1

w = 48.55 × [ 1.63 / ( 48.55 + 5.7408 + 22.1 )

w = 48.55 × [ 1.63 / ( 76.3908 ) ]

w = 48.55 × 0.02133

w = 1.0356 rad/s

Therefore; the final angular velocity of the platform with its load is 1.0356 rad/s

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