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Ber [7]
3 years ago
8

Two masses are joined by a massless string. A 30-N force applied vertically to the upper mass gives the system a constant [Ans :

490 N] [Ans : 1.3 × 10−23 m] [Ans : 19 cm] [Ans : 1.96 ms−2 ] [Ans : 680 m] upward acceleration of 3.2 m/ s 2 . If the string tension is 18 N, what are the two masses?
Physics
1 answer:
Vilka [71]3 years ago
8 0

Answer:

12/13 kg, 18/13 kg

Explanation:

Let the masses are m and m'.

acceleration, a = 3.2 m/s^2

Force, F = 30 N

Tension, T = 18 N

By the diagram, using Newton's second law

F - T - m'g = m' a .... (1)

T - mg = ma ..... (2)

Substitute the values of F, T and a in equation (1) and equation (2)

30 - 18 = m' (9.8 + 3.2)

12 = 13 m'

m' = 12/ 13 kg

And, 18 = m (9.8 + 3.2)

m = 18/13 kg

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1. During the Middle Ages, armies often attacked castles using large siege engines such as the counterweight trebuchet at left.
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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
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A) 4.7 cm

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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}

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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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