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zheka24 [161]
3 years ago
5

A large centrifuge is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and at

mospheric reentries. At what angular velocity is the centripetal acceleration 10 g if the rider is 15.0 m from the center of rotation
Physics
1 answer:
melomori [17]3 years ago
6 0

Answer:

ω = 2.55 rad/sec

Explanation:

Assuming no other external forces acting in the horizontal plane, the only force keeping the  rider in a circular path of a radius equal to his distance to the center of rotation, is the centripetal force.

According to Newton's 2nd law, in the horizontal direction, we have:

F = Fc = m*a = m*ω²*r

We know that a = ac = 10*g = 98.0 m/s², and that r = 15.0 m.

Replacing these values in (1), and solving for ω, we get:

ω = √98.0m/s²/ 15.0 m = 2.55 rad/sec

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Monochromatic light of wavelength 687 nm is incident on a narrow slit. On a screen 1.65 m away, the distance between the second
Sophie [7]

Answer:

a ) 1.267 radian

b ) 1.084 10⁻³ mm

Explanation:

Distance of screen D = 1.65 m

Width of slit d = ?

Wave length of light   λ  = 687 nm.

Distance of second minimum fro centre y = 2.09 cm

Angle of diffraction = y / D

=  2.09 /1.65  

= 1.267. radian

Angle of diffraction of second minimum

= 2 λ / d

so 2 λ / d = 1.267

d = 2 λ / 1.267 = (2 x 687 ) /1.267 nm

=1084.45 nm = 1.084 x 10⁻³ mm.

3 0
3 years ago
Coherent light of wavelength 525 nm passes through two thin slits that are 4.15×10^(−2) mm apart and then falls on a screen 7
IRINA_888 [86]

A) 4.7 cm

The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

sin \theta=\frac{n \lambda}{d}

where

n is the order of the maximum

\lambda is the wavelength

a is the distance between the slits

In this problem,

n = 5

\lambda=525 nm =5.25\cdot 10^{-7} m

a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m

So we find

\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}

And given the distance of the screen from the slits,

D=75.0 cm = 0.75 m

The distance of the 5th  bright fringe from the central bright fringe will be given by

y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm

B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}

And the distance of the 8th dark fringe from the central bright fringe will be given by

y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm

5 0
3 years ago
(b) Which has more mass: 50 cm of gold or 50 cm of aluminum? Explain.
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A cohesive force between the liquids molecules is responsible for the fluids is called​
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According to Newton's law of cooling, the rate at which an object's temperature changes is directly proportional to the differen
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Answer:

dT(t)/dt = k[T5 - T(t)]

Explanation:

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Introducing the constant of proportionality

dT(t)/dt = k[T5 - T(t)]

which is the desired differential equation

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