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2 years ago

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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]2 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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Answer:

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A 0.2-kg steel ball is dropped straight down onto a hard, horizontal floor and bounces straight up. The ball's speed just before

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Answer:

Explanation:

Given

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2 years ago

Read 2 more answers

A wooden disk of mass m and radius r has a string of negligible mass is wrapped around it. If the disk is allowed to fall and th

Tju [1.3M]

Answer:

$a = \frac{2}{3}g$

$T = \frac{mg}{3}$

Explanation:

As the disc is unrolling from the thread then at any moment of the time

We have force equation as

$mg - T = ma$

also by torque equation we can say

$TR = I\alpha$

$TR = \frac{1}{2}mR^2(\frac{a}{R})$

$T = \frac{1}{2}ma$

Now we have

$mg - \frac{1}{2}ma = ma$

$mg = \frac{3}{2}ma$

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A 150 g particle at x = 0 is moving at 8.00 m/s in the +x-direction. As it moves, it experiences a force given by Fx=(0.850N)sin

krok68 [10]

Answer:

9.98 m/s

Explanation:

The force acting on the particle is defined by the equation:

$F=(0.850) sin (\frac{x}{2.00})$ [N]

where x is the position in metres.

The acceleration can be found by using Newton's second law:

$a=\frac{F}{m}$

where

m = 150 g = 0.150 kg is the mass of the particle. Substituting into the equation,

$a=\frac{0.850}{0.150}sin (\frac{x}{2.00})=5.67 sin(\frac{x}{2.00})$ [m/s^2]

When x = 3.14 m, the acceleration is:

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Now we can find the final speed of the particle by using the suvat equation:

$v^2-u^2=2ax$

where

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2 years ago

In preparation for a demonstration, your professor brings a 1.50−L bottle of sulfur dioxide into the lecture hall before class t

mina [271]

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2 years ago