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erastovalidia [21]
3 years ago
10

During a testing process, a worker in a factory mounts a bicycle wheel on a stationary stand and applies a tangential resistive

force of 115 N to the tire's rim. The mass of the wheel is 1.70 kg and, for the purpose of this problem, assume that all of this mass is concentrated on the outside radius of the wheel. The diameter of the wheel is 50.0 cm. A chain passes over a sprocket that has a diameter of 9.50 cm. In order for the wheel to have an angular acceleration of 4.90 rad/s2, what force, in Newtons, must be applied to the chain? (Enter the magnitude only.) N
Physics
1 answer:
Alex3 years ago
5 0

Answer:

616.223684211 N

Explanation:

F_r = Resistive force on the wheel = 115 N

F = Force acting on sprocket

r_2 = Radius of sprocket = 4.75 cm

r_1 = Radius of wheel = 25 cm

Moment of inertia is given by

I=mr_1^2\\\Rightarrow I=1.7\times 0.25^2\\\Rightarrow I=0.10625\ kgm^2

Torque

\tau=I\alpha\\\Rightarrow \tau=0.10625\times 4.9\\\Rightarrow \tau=0.520625\ Nm

Torque is given by

\tau=Fr_2-F_rr_1\\\Rightarrow F=\dfrac{\tau+F_rr_1}{r_2}\\\Rightarrow F=\dfrac{0.520625+115\times 0.25}{0.0475}\\\Rightarrow F=616.223684211\ N

The force on the chain is 616.223684211 N

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A particle moving along a straight line is subjected to a deceleration ???? = (−2???? 3 ) m/???? 2 , where v is in m/s. If it ha
dlinn [17]

Answer:

(a). The velocity is 0.099 m/s.

(b). The position is 19.75 m.

Explanation:

Given that,

The deceleration is

a=(-2v^3)\ m/s^2

We need to calculate the velocity at t = 25 s

The acceleration is the first derivative of velocity of the particle.

a=\dfrac{dv}{dt}

\dfrac{dv}{dt}=-2v^3

\dfrac{dv}{-v^3}=2dt

On integrating

int{\dfrac{dv}{-v^3}}=\int{2dt}

\dfrac{1}{2v^2}=2t+C

v^2=\dfrac{1}{4t+2C}....(I)

At t = 0, v = 10 m/s

10^2=\dfrac{1}{4\times0+2C}

C=\dfrac{1}{200}

Put the value of C in equation (I)

v^2=\dfrac{1}{4\times25+2\times\dfrac{1}{200}}

v=\sqrt{\dfrac{1}{4\times25+2\times\dfrac{1}{200}}}

v=0.099\ m/s

The velocity is 0.099 m/s.

(b). We need to calculate the position at t = 25 sec

The velocity is the first derivative of position of the particle.

\dfrac{ds}{dt}=v

On integrating

\int{ds}=\int(\sqrt{\dfrac{200}{800t+1}})dt

s=\dfrac{\sqrt{200}\times2\sqrt{800t+1}}{800}+C'

At t = 0, s = 15 m

15=\dfrac{200}{800}+C'

C'=15-\dfrac{200}{800}

C'=14.75

Put the value in the equation

s=\dfrac{\sqrt{200}\times2\sqrt{800\times25+1}}{800}+14.75

s=19.75\ m

The position is 19.75 m.

Hence, (a). The velocity is 0.099 m/s.

(b). The position is 19.75 m.

3 0
3 years ago
How long must a pendulum be to have a period of 4.6 5 on Neptune, where g = 11.15 m/s?​
Anon25 [30]

Answer:

5.98 m

Explanation:

T=2\pi \sqrt\frac{L}{g}

T^2=4\pi ^2L/g\\L = gT^2/(4\pi^2) = 11.15*4.6²/(4π²) = 5.98 m

5 0
2 years ago
Big Bang theory flow chart
swat32

Answer:

I don't understand

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6 0
3 years ago
The conductors that carry the current to electrical devices and ? equipment are the heart of all electrical systems. There are a
Aleksandr [31]

Answer:

Utilization, effects

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The conductors that carry the current to electrical devices and utilization equipment are the heart of all electrical systems. There are associated effects whenever current flows through a conductor.

7 0
4 years ago
Which of the following statements is true?
Maurinko [17]

Answer:

Concrete absorbs visible light and re-radiates it as infrared energy.

Explanation:

As we know that concrete will not reflect 100% light incident on it but it will absorb few energy and remaining energy is reflected back.

So here when visible light incident on the concrete then few of its energy is absorbed by is and then the remaining energy is reflected back.

Since the reflected energy must be less than the energy which incident on the concrete so here we will say that the reflected waves must be of less frequency range

so here most appropriate energy range of reflected wave is infrared waves

so here correct answer will be

Concrete absorbs visible light and re-radiates it as infrared energy.

6 0
3 years ago
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