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Tema [17]
1 year ago
6

Shows the position-versus-time graph of a particle in SHM. Positive direction is the direction to the right.

Physics
1 answer:
BaLLatris [955]1 year ago
3 0

A) t = 0 s, 4 s, 8 s

B) t = 2 s, 6 s

C) t = 1 s, 3 s, 5 s, 7 s

Explanation:

A)

The figure is missing: find it in attachment.

A particle is said to be in Simple Harmonic Motion (SHM) when it is acted upon a restoring force proportional to its displacement, and therefore, the acceleration of the particle is directly proportional to its displacement (but in the opposite direction):

a\propto -x

Also, the displacement of a particle in SHM can be described by a sinusoidal function, as shown in the figure for the particle in this problem.

For the particle in this problem, from the figure we can see that the displacement is described by a function in the form

x(t)=A sin(\omega t)

where

A is the amplitude

\omega=\frac{2\pi}{T} is the angular frequency, where T is the period. From the graph, we see that the particle completes 1 oscillation in 4 seconds, so

T = 4 s

So the angular frequency is

\omega = \frac{2\pi}{4}=\frac{\pi}{2}rad/s

So

x(t)=Asin(\frac{\pi}{2}t)

The velocity of a particle in SHM can be found as the derivative of the displacement, so here we find:

v(t)=x'(t)=A\omega cos(\omega t) = \frac{A\pi}{2}cos(\frac{\pi}{2}t)

So, the particle is moving to the right at maximum speed when

cos(\frac{\pi}{2}t)=+1

(because this is the maximum positive value for the cosine part). Solving for t,

\frac{\pi}{2}t=0\\t=0 s

We also have another time in which this occurs, when

\frac{\pi}{2}t=2\pi\\\rightarrow t=\frac{4\pi}{\pi}=4 s

And also when

\frac{\pi}{2}t=4\pi\\\rightarrow t=8s

B)

In this case, we want to find the time t at which the particle is moving to the left at maximum speed.

This occurs when the cosine part has the maximum negative value, so when

cos(\frac{\pi}{2}t)=-1

Which means that this occurs when:

\frac{\pi}{2}t=\pi\\\rightarrow t=2 s

Also when

\frac{\pi}{2}t=3\pi\\\rightarrow t=6 s

So, the particle has maximum speed moving to the left at 2 s and 6 s.

C)

The particle is instantaneously at rest when the speed is zero, this means when the cosine part is equal to zero:

cos(\frac{\pi}{2}t)=0

This occurs when the argument of the cosine is:

\frac{\pi}{2}t=\frac{\pi}{2}\\\rightarrow t = 1 s

Also when

\frac{\pi}{2}t=\frac{3\pi}{2}\\\rightarrow t = 3 s

And when

\frac{\pi}{2}t=\frac{5\pi}{2}\\\rightarrow t = 5 s

And finally when

\frac{\pi}{2}t=\frac{7\pi}{2}\\\rightarrow t = 7s

So, the particle is at rest at t = 1 s, 3 s, 5 s, 7 s.

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Answer:

H = 109.14 cm

Explanation:

given,                                                            

Assume ,                                                            

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Balance after third collision = 0.78 ^3 unit

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A resultant vector is 8.00 units long and makes an angle of 43.0 degrees measured ������� – ��������� with respect to the positi
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223 degree

Explanation:

We are given that

Magnitude of resultant vector= 8 units

Resultant vector makes an angle with positive -x in counter clockwise direction

\theta=43^{\circ}

We have to find the magnitude and angle of the equilibrium vector.

We know that equilibrium vector is equal in magnitude  and in opposite direction  to the given vector.

Therefore, magnitude of equilibrium vector=8 units

x-component of a  vector=v_x=vcos\theta

Where v=Magnitude of vector

Using the formula

x-component of resultant  vector=v_x=8cos43=5.85

y-component of resultant vector=v_y=vsin\theta=8sin43=5.46

x-component of equilibrium vector=v_x=-5.85

y-component of equilibrium vector=-v_y=-5.46

Because equilibrium vector lies in III quadrant

\theta=tan^{-1}(\frac{v_x}{v_y})=tan^{-1}(\frac{-5.46}{-5.85})=43^{\circ}

The angle \theta'lies in III quadrant

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4 0
1 year ago
A 150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the w
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Answer:

Frictional force, F = 45.9 N

Explanation:

It is given that,

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Acceleration, a=3\ m/s^2

The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.  

It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,

m=\dfrac{W}{g}

m=\dfrac{150}{9.8}

m=15.3\ kg

Frictional force is given by :

F=ma

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So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.

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Answer:

The terminal speed of this object is 12.6 m/s

Explanation:

It is given that,

Mass of the object, m = 80 kg

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The terminal speed of an object is attained when the gravitational force is balanced by the gravitational force.

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So, the terminal speed of this object is 12.6 m/s. Hence, this is the required solution.

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