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yan [13]
2 years ago
12

A 817 kg car has four 8.91 kg wheels. When the car is moving, what fraction of the total kinetic energy of the car is due to rot

ation of the wheels about their axles
Physics
1 answer:
KATRIN_1 [288]2 years ago
8 0

Answer:

0.0107

Explanation:

We know that

The rotational kinetic energy due to four wheel is

1/2ဃ²I x 4

So

1/4mR²(v/R)² = mv²

But kinetic energy along straight path of the car is 1/2mv²

=> 1/2( 817)v ²

Kc= 408.5v²

So The fraction of total kinetic energy that is due to rotation of the wheel about their axis

Is Kw/Kw+Kc

and Kw = 1/2* 8.91v²= 4.45v²

So 4.45v²/ 4.45v²+ 408.5v²

= 0.0107 as fraction of total kinetic energy

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60 mph east...........

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A jetboat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m ​ 5, point, 0, start fraction, start text, m, end tex
love history [14]

The question is incomplete. Here is the entire question.

A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?

Answer: Δx = - 42m

Explanation: The jetboat is moving with an acceleration during the time interval, so it is a <u>linear</u> <u>motion</u> <u>with</u> <u>constant</u> <u>acceleration</u>.

For this "type" of motion, displacement (Δx) can be determined by:

\Delta x = v_{i}.t + \frac{a}{2}.t^{2}

v_{i} is the initial velocity

a is acceleration and can be positive or negative, according to the referential.

For Referential, let's assume rightward is positive.

Calculating displacement:

\Delta x = 5(6) - \frac{4}{2}.6^{2}

\Delta x = 30 - 2.36

\Delta x = - 42

Displacement of the boat for t=6.0s interval is \Delta x = - 42m, i.e., 42 m to the left.

8 0
2 years ago
A race car makes one lap around a track of radius 50 m in 9.0 s. What is the average velocity? *
Oksi-84 [34.3K]

Given that,

Radius of track, r = 50 m

time , t = 9 s

velocity, v = ?

Distance covered by car in one lap around a track is equal to the circumference of the track.

C = 2 π r = 2 * 3.14 * 50

C = 314.159 m

Distance covered by car, s = 314.159 m

Velocity = distance/ time

V = 314.159 / 9

V = 34.9 m/s

The average velocity of car is 34.9 m/s.

7 0
1 year ago
A semi is traveling down the highway at a velocity of v = 26 m/s. The driver observes a wreck ahead, locks his brakes, and begin
Dovator [93]

Answer:

fcosθ + Fbcosθ  =Wtanθ

Explanation:

Consider the diagram shown in attachment

fx= fcosθ (fx: component of friction force in x-direction ; f: frictional force)

Fbx= Fbcosθ ( Fbx: component of braking force in x-direction ; Fb: braking force)

Wx= Wtanθ (Wx: component of weight in x-direction ; W: Weight of semi)

sum of x-direction forces = 0

fx+ Fbx=Wx

fcosθ + Fbcosθ  =Wtanθ

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2 years ago
A 10-turn conducting loop with a radius of 3.0 cm spins at 60 revolutions per second in a magnetic field of 0.50T. The maximum e
bogdanovich [222]

Answer:

Maximum emf = 5.32 V

Explanation:

Given that,

Number of turns, N = 10

Radius of loop, r = 3 cm = 0.03 m

It made 60 revolutions per second

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We need to find maximum emf generated in the loop. It is based on the concept of Faraday's law. The induced emf is given by :

\epsilon=\dfrac{d(NBA\cos\theta)}{dt}\\\\\epsilon=NBA\dfrac{d(\cos\theta)}{dt}\\\\\epsilon=NBA\omega \sin\omega t\\\\\epsilon=NB\pi r^2\omega \sin\omega t

For maximum emf, \sin\omega t=1

So,

\epsilon=NB\pi r^2\omega \\\\\epsilon=NB\pi r^2\times 2\pi f\\\\\epsilon=10\times 0.5\times \pi (0.03)^2\times 2\pi \times 60\\\\\epsilon=5.32\ V

So, the maximum emf generated in the loop is 5.32 V.

3 0
2 years ago
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