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Liula [17]
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.
Physics
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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A helicopter is ascending vertically with a speed of 5.10m/s. At a height of 105m above the Earth, a package is dropped from a w
valkas [14]

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

5 0
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

8 0
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

5 0
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
svet-max [94.6K]

Answer:

0.261\ \text{m}

Explanation:

\Delta f = Change in frequency = 2.1 Hz

f = Frequency of source of sound = 440 Hz

v_m= Maximum of the microphone

v = Speed of sound = 343 m/s

T = Time period = 2 s

We have the relation

\Delta f=2f\dfrac{v_m}{v}\\\Rightarrow v_m=\dfrac{\Delta fv}{2f}\\\Rightarrow v_m=\dfrac{2.1\times 343}{2\times 440}\\\Rightarrow v_m=0.8185\ \text{m/s}

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}

The amplitude of the simple harmonic motion is 0.261\ \text{m}.

4 0
2 years ago
Which of the following correctly describes the relative air pressure at the center of a hurricane, with respect to the horizonta
vampirchik [111]

Explanation:

The pressure of hurricane which is high decreases gradually as we move higher . The pressure is maximum at the surface . Hence the relative air pressure is higher at the surface and low at the top . Like wise the pressure is low at the center and keeps on increasing when we move outside . Hence the answer is pressure is high at surface and low aloft.

6 0
3 years ago
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