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3 years ago

6

A magic medallion is suspended from a string inside a compartment of Hogwarts Express which is running straight westwards on hor

izontal tracks. a) Draw a free-free body diagram for the medallion if the train is moving at constant speed. b) Draw a free-body diagram for the medallion if the train is accelerating at a constant rate. Indicate acceleration and the direction of the net force. c) Calculate the angle the string makes with the vertical if the train accelerates at a constant rate from rest to 20 m/s in 10 seconds. d) If the mass of the medallion is doubled, how will the angle change?

1 answer:

Serggg [28]3 years ago

4 0

Answer:

a = 1 m/s² and

Explanation:

The first two parts can be seen in attachment

We use Newton's second law on each axis

Y axis

Ty - W = 0

Ty = w

X axis

Tx = m a

With trigonometry we find the components of tension

Sin θ = Ty / T

Ty = T sin θ

Cos θ = Tx / T

Tx = T cos θ

We calculate the acceleration with kinematics

Vf = Vo + a t

a = (Vf -Vo) / t

a = (20 -10) / 10

a = 1 m/s²

We substitute in Newton's equations

T Sin θ = mg

T cos θ = ma

We divide the two equations

Tan θ = g / a

θ = tan⁻¹ (g / a)

θ = tan⁻¹ (9.8 / 1)

θ = 84º

We see that in the expression of the angle the mass does not appear therefore you should not change the angle

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Light of wavelength 650 nm is normally incident on the rear of a grating. The first bright fringe (other than the central one) i

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Answer:

A

$N = 1340.86 \ slits / cm$

B

$\theta = 15.7^o$

Explanation:

From the question we are told that

The wavelength is $\lambda = 650 \ nm = 650 *10^{-9} \ m$

The angle of first bright fringe is $\theta = 5^o$

The order of the fringe considered is n =1

Generally the condition for constructive interference is

$dsin (\theta ) = n * \lambda$

=> $d = \frac{1 * 650 *10^{-9 }}{ sin(5)}$

=> $d = 7.458 *10^{-6} \ m$

Converting to cm

$d = 7.458 *10^{-6} \ m = 7.458 *10^{-6} * 100 = 0.0007458 \ cm$

Generally the number of grating pre centimeter is mathematically represented as

$N = \frac{1}{d}$

=> $N = \frac{1}{0.0007458}$

=> $N = 1340.86 \ slits / cm$

Considering question B

From the question we are told that

The first wavelength is $\lambda_1 = 650 \ nm = 650 *10^{-9} \ m$

The second wavelength is $\lambda_2 = 429 \ m = 420 *10^{-9 } \ m$

The order of the fringe is $n = 2$

The grating is $N = 5000 \ slits / cm$

Generally the slit width is mathematically represented as

$d = \frac{1}{N }$

=> $d = \frac{1}{ 5000 }$

=> $d = 0.0002 \ c m = 2.0 *10^{-6} \ m$

Generally the condition for constructive interference for the first ray is mathematically represented as

$d sin(\theta_1) = n * \lambda_1$

=> $\theta_1 = sin^{-1} [\frac{ 2 * \lambda }{d}]$

=> $\theta_1 = sin^{-1} [\frac{ 2 * 650 *10^{-9} }{ 2*10^{-6}}]$

=> $\theta_1 = 40.5 ^o$

Generally the condition for constructive interference for the second ray is mathematically represented as

$d sin(\theta_2) = n * \lambda_2$

=> $\theta_2 = sin^{-1} [\frac{ 2 * \lambda_1 }{d}]$

=> $\theta_2 = sin^{-1} [\frac{ 2 * 420 *10^{-9} }{ 2*10^{-6}}]$

=> $\theta_2 = 24.8 ^o$

Generally the angular separation is mathematically represented as

$\theta = \theta_1 - \theta_1$

=> $\theta = 42.5^o - 24.8^o$

=> $\theta = 15.7^o$

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