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4 years ago

6

a train travles at a speed of 30m/s. the train starts at an initial position of 1000 meters and travels for 30 seconds. what is

its final position?

1 answer:

ycow [4]4 years ago

5 0

Answer:

1900 meters

Explanation:

30m/s x 30 second = 900 meters

+ 1000 meters starting position

= 1900meters

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Coherent light of wavelength 525 nm passes through two thin slits that are 4.15Ã—10^(âˆ’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

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

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3 years ago

A thin, uniform metal bar, 2.00 m long and weighing 90.0 N, is hanging vertically from the ceiling by a frictionless pivot. Sudd

matrenka [14]

Answer:

(a) The "angular speed" is 5.88 rad/s.

Explanation:

Given values,

The length of the bar is L = 2m

The weight of the bar is w = 90 N

The metal bar is hanging vertically from the ceiling by a frictionless pivot

The mass of the ball is m = 3kg

The distance between the ceiling and the ball is d = 1.5m

$\text { The "initial speed of the ball" is } V_{i}=10 \mathrm{m} / \mathrm{s}$

$\text { The "final speed of the ball" is } V_{f}=6 \mathrm{m} / \mathrm{s}$

(a) Calculating the angular speed:

$W_{\mathrm{f}}=\left(3 \mathrm{mg} \mathrm{d} \frac{V_{i}+V_{f}}{w l^{2}}\right)$

$W_{f}=\left(3 \times 3 \times 9.8 \times 1.5 \times \frac{10+6}{90 \times 2^{2}}\right)$

$W_{f}=\frac{2116.8}{360}$

$\mathrm{W}_{\mathrm{f}}=5.88 \mathrm{rad} / \mathrm{s}$

The angular speed is 5.88 rad/s.

(b) The "angular momentum" is conserved because the torque is not exerted by "the pivot" on the system about the "axis of rotation" but the "linear momentum" is not conserved because "the pivot" exerts a "vertical" and a "horizontal force" on the system during the collision.

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4 years ago

In science and physics what is the standard unit of measure for speed?

attashe74 [19]

meters per second or m/s

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Show that liquid pressure is directly<br>proportional to height of liquid in a vessel.

Norma-Jean [14]

Answer:

P=F/A where F is the weight of the water and A is the area on which it is resting. The weight of the water is mg. The mass of the water is dv where d is the density and v is the volume. Finally, the volume of the water in a vessel is equal to the area of the base of the vessel times the height of the vessel. (v=Ah)

Plugging everything in we get:

P = dAhg/A

So

P=dhg

So we have shown that liquid pressure is directly proportional to height of liquid in a vessel.

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3 years ago

If earth had no atmosphere , would a falling object ever reach terminal velocity?

Bas_tet [7]

The right answer is no.

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3 years ago