Question 7 of 25

An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic

malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

$\omega_f = \omega_0 + \alpha t$

Where,

$\omega_f =$ Final Angular Velocity

$\omega_0 =$ Initial Angular velocity

$\alpha =$ Angular acceleration

t = time

The relation between the tangential acceleration is given as,

$a = \alpha r$

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

$\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s$

$\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s$

$t = 11s$

Replacing the previous equation with our values we have,

$\omega_f = \omega_0 + \alpha t$

$9.8436 = 6.5973 + \alpha (11)$

$\alpha = \frac{9.8436- 6.5973}{11}$

$\alpha = 0.295rad/s^2$

The tangential velocity then would be,

$a = \alpha r$

$a = (0.295)(0.2)$

$a = 0.059m/s^2$

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

$\omega_f^2=\omega_0^2+2\alpha\theta$

Replacing with our values and re-arrange to find $\theta,$

$\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}$

$\theta = \frac{9.8436^2-6.5973^2}{2*0.295}$

$\theta = 90.461rad$

That is equal in revolution to

$\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev$

The linear displacement of the system is,

$x = \theta*(2\pi*r)$

$x = 14.397*(2\pi*\frac{0.25}{2})$

$x = 11.3m$

5 0

2 years ago

Could someone explain to me how to got the answer B, thank you very much

Mnenie [13.5K]

Answer:

since -6 lasted for 5 seconds, multiplying both would result in -30

3 lasted for 10 seconds, so multiplying both would give +30

average = ( 30 + (-30) ) / 2

30 -30 is already equal to zero, so the answer should be 0

4 0

2 years ago

The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increa

Nostrana [21]

Answer: The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increased by 2.0 kg, the period is found to be 3.0 s. Then the mass m is 0.625kg.

Explanation: To find the answer, we need to know more about the simple harmonic motion.

<h3>What is simple harmonic motion?</h3>

A particle is said to execute SHM, if it moves to and fro about the mean position under the action of restoring force.
We have the equation of time period of a SHM as,

$T=2\pi \sqrt{\frac{m}{k} }$

Where, m is the mass of the body and k is the spring constant.

<h3>How to solve the problem?</h3>

Given that,

$T_1=2s\\m_1=m\\m_2=m+2kg\\T_2=3s$

We have to find the value of m,

$T_1=2\pi \sqrt{\frac{m}{k} } \\T_2=2\pi \sqrt{\frac{m+2}{k} } \\\frac{T_1}{T_2} =\sqrt{\frac{m}{m+2} }\\\frac{2}{3} =\sqrt{\frac{m}{m+2} }\\\\$

$m=\frac{5}{8} =0.625kg$

Thus, we can conclude that, the mass m will be 0.625kg.

Learn more about simple harmonic motion here:

brainly.com/question/28045110

#SPJ4

3 0

1 year ago

A horizontal beam of light of intensity 25 W/m2 is sent through two polarizing sheets. The polarizing direction of the first mak

Zina [86]

Answer:

option (B)

Explanation:

Intensity of unpolarised light, I = 25 W/m^2

When it passes from first polarisr, the intensity of light becomes

$I'=\frac{I_{0}}{2}=\frac{25}{2}=12.5 W/m^{2}$

Let the intensity of light as it passes from second polariser is I''.

According to the law of Malus

$I'' = I' Cos^{2}\theta$

Where, θ be the angle between the axis first polariser and the second polariser.

$I'' = 12.5\times Cos^{2}15$

I'' = 11.66 W/m^2

I'' = 11.7 W/m^2

7 0

2 years ago

A box of weight 74 N is sliding on a horizontal surface at a constant velocity due to an external force F-> of magnitude 4.8

n200080 [17]

Answer

4.8 N

If the box is moving with a constant velocity, then we can say that the system is in equilibrium. This is because if the external force (F->) was greater than other forces the box would be accelerating. This tells us that this force (F->) is just enough to overcome friction and so it must be equal to 4.8 N.

The normal force has no effect to the horizontal velocities or forces. It is equal to -Weight. That is -74 N. The negative sign shows that the force is in opposite direction.

6 0

2 years ago