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Svetlanka [38]
3 years ago
5

A student decides to give his bicycle a tune up. He flips it upside down (so there's no friction with the ground) and applies a

force of 34 N over 0.6 seconds to the pedal, which has a length of 16.5 cm. If the back wheel has a radius of 33.0 cm and moment of inertia of 1200 kg cm^2, what is the tangential velocity of the rim of the back wheel in m/s
Physics
1 answer:
aleksklad [387]3 years ago
5 0

Answer:

V=9.2565m/s

Explanation:

From the question we are told that:

Force F = 34 N  

Time t = 0.6 s

Length of pedal l_p=16.5cm \approx0.165m

Radius of wheel r = 33 cm = 0.33 m

Moment of inertia, I = 1200 kgcm2 = 0.12 kg.m2

Generally the equation for Torque on pedal \mu is mathematically given by

\mu=F*L\\\mu=34*0.165

\mu=5.61N.m

Generally the equation for  angular acceleration \alpha is mathematically given by

 \alpha=\frac{\mu}{l}

 \alpha=\frac{5.61}{0.12}

 \alpha=46.75

Therefore Angular speed is \omega

\omega=\alpha*t

\omega=(46.75)*(0.6)

\omega=28.05rad/s

Generally the equation for  Tangential velocity V is mathematically given by

V=r\omega

V=(0.33)(28.05)

V=9.2565m/s

 

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Answer:

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The following reasons are reason not to worry

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B. As an incentive for thinking hard about the problem, your instructor may choose to apply a small hint penalty, but this penalty is the same whether the hint simply gives information or asks another question (In a situation where you decide to use a hint, the instructor may have put a penalty for using the hint, so whether it asks a question or help in the solution of the question, as long as the hint is consulted, the hint penalty still applies)

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4 years ago
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Yakvenalex [24]

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3 0
2 years ago
A diffraction grating is illuminated simultaneously with red light of wavelength 670 nm and light of an unknown wavelength. The
marusya05 [52]

Answer:

dsin∅ = m× λ

so, dsin∅red = 3(670nm)

also, dsin∅? =5λ?

however ,if they overlap then dsin∅red = dsin∅?

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∴λ = 402nm

Explanation:

4 0
3 years ago
A ball is thrown straight up from ground level. It passes a 2-m-high window. The bottom of the window is 7.5 m off the ground. T
slamgirl [31]

Answer:

u=14.48m/s

Explanation:

From the question we are told that:

Height of window h=2m

Height of window off the ground h_g=7.5m

Time to fall and drop t=1.3s

 

Generally the Newton's equation motion  is mathematically given by

 s=ut+\frac{1}{2}at^2

Where

h=ut+\frac{1}{2}at^2

2=u1.3-\frac{1}{2}*9.8*1.3^2

2=u1.3-8.281

u=7.91m/s^2  

Generally the Newton's equation motion  is mathematically given by

2as=v^2-u^2

Where

-2gh_g=v^2-u^2

-2*9.8*7.5=(7.91)^2-u^2

-147=62.5681-u^2

u=\sqrt{209.5681}

u=14.48m/s

Therefore the  ball’s initial speed

u=14.48m/s

8 0
3 years ago
A biker first accelerates from 0.0 m/s to 6.0 m/s in 6 s, then continues at this speed for 5 s. What is the total distance trave
Svetradugi [14.3K]

Answer:

48m

Explanation:

Given the following data;

Initial velocity = 0m/s

Final velocity = 6m/s

Time, t = 6 secs

Time, T2 = 5 secs

Mathematically, acceleration is given by the equation;

Acceleration (a) = \frac{final \; velocity  -  initial \; velocity}{time}

Substituting into the equation;

a = \frac{6 - 0}{6}

a = \frac{6}{6}

Acceleration, a = 1m/s²

<u>To find the distance covered in the first phase;</u>

<em>Solving for distance, we would use the second equation of motion;</em>

S = ut + \frac {1}{2}at^{2}

<em>Substituting the values into the equation;</em>

S = 0(6) + \frac {1}{2}*1*(6)^{2}

S = 0 + \frac {1}{2}*1*36

S = 0.5 *36

Distance, S1 = 18m

<u>For the second phase, time T2 = 5 secs;</u>

<em>Mathematically, speed is given by the equation;</em>

Speed = \frac{distance}{time}

<em>Making distance the subject of formula, we have;</em>

Distance, S = speed * time

<em>Substituting into the above equation;</em>

Distance, S = 6 * 5

Distance, S2 = 30m

Total distance = S1 + S2 = 18m + 30m = 48m

Total distance = 48m

<em>Therefore, the total distance traveled by the biker is 48m.</em>

4 0
3 years ago
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