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3 years ago

5

A remote-controlled car’s wheel accelerates at 22.2 rad/s2 . If the wheel begins with an angular speed of 11.0 rad/s, what is th

e wheel’s angular speed after exactly twelve full turns? Answer in units of rad/s.

1 answer:

wariber [46]3 years ago

4 0

The angular speed remained constant after twelve turns at 11 rad/s.

Explanation:

As per the Newtons equation of motion, the final velocity can be determined by using the below equation, provided ,the displacement and acceleration with initial velocity is given.

$v^{2}= u^{2}+2as$

Since, the present case follows circular motion, the displacement will be equal to angular displacement. And the velocity is equal to angular velocity.

As the car revolves complete twelve turns, then the displacement will be zero, since the final position is equal to the initial position.

So,

$v^{2}= (11)^{2}+2*22.2*0\\\\v = 11 rad/s$

Thus, the angular speed remained constant after twelve turns at 11 rad/s.

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Op here is another problem exactly like that. Just plug in your variables instead. And remember, time is never negative.

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2 years ago

A truck using a rope to tow a 2230-kg car accelerates from rest to 13.0 m/s in a time of 15.0s. How strong must the rope be? μk

Leokris [45]

Answer:

The rope must have a force of 10084,21 N

Explanation

Acceleration calculation

The car acceleration is equal to the acceleration of the truck

ac: car acceleration $\frac{m}{s^{2} }$

at: truck acceleration $\frac{m}{s^{2} }$ )

$ac = at= \frac{vf-vi}{t-ti}$ equation(1)

Known information:

vi = Initial speed = 0, ti = initial time = 0

vf = Final speed = 13 $\frac{m}{s}$ , t = final time =5 s

We replaced the known information in the equation(1):

$ac = at = \frac{13-0}{15-0}$

$ac=ac=\frac{13}{15} \frac{m}{s}$

Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

$Wc=mc*g$

mc = Car mass = 2230kg

g = Gravity acceleration=9.8 $\frac{m}{s^{2} }$

$Wc= 2230*9.8$

$Wc=21854 N$

Normal force calculation:

Newton's first law

$sum Fy= 0$

$N-W=0$

$N=W$

$N=21854 N$

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

$Ff=0.373*21854$

$Ff=8151.54 N$

Calculation of the tension force in the rope (T):

Newton's Second law

$sum Fx= mc*ac$

$T-Ff=mc*ac$

$T=2230(\frac{13}{15}) + 8151.54$

T=10084,21 N

Answer: The rope must have a force of 10084,21 N

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3 years ago

200 Coulombs of charge passes through a point in a circuit for 0.6 minutes. what is the magnitude of the current flowing

Tasya [4]

Answer:

5.56 A

Explanation:

From the question,

Q = it.............. Equation 1

Where Q = charges, i = current, t = time.

Make i the subject of the equation

i = Q/t.............. Equation 2

Given: Q = 200 coulombs, t = 0.6 minutes = (0.6×60) seconds

Substitite these values into equation 2

i = 200/(0.6×60)

i = 5.56 A

Hence the magnitude of the current flowing through the circuit is 5.56 A

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3 years ago

A microphone is attached to a spring that is suspended from the ceiling, as the drawing indicates. Directly below on the floor i

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Answer:

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Explanation:

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$f$ = Frequency of source of sound = 440 Hz

$v_m$ = Maximum of the microphone

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We have the relation

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Amplitude is given by

$A=\dfrac{v_m T}{2\pi}\\\Rightarrow A=\dfrac{0.8185\times 2}{2\pi}\\\Rightarrow A=0.261\ \text{m}$

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Explanation:

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