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mr Goodwill [35]
4 years ago
6

Your 64-cm-diameter car tire is rotating at 3.5 rev/s when suddenly you press down hard on the accelerator. After traveling 200

m, the tire’s rotation has increased to 6.0 rev/s. What was the tire’s angular acceleration? Give your answer in rad/s2.
Physics
1 answer:
andreyandreev [35.5K]4 years ago
4 0

Answer: angular acceleration = 0.748rad/s²

Explanation: according to the question, our answer needs to be in rad/s², thus all units in rev/s will be converted to rad/s

Assuming the motion of the object is of a constant angular acceleration, then newton's laws of motion is applicable.

The formulae below is used

v² = u² + 2αθ

v = final angular speed =6rev/s = 6*2π = 12π rad/s

u =initial angular speed =3.5rev/s = 3.5 *2π = 7π rad/s

Note 1 rev = 2π rad.

α = angular acceleration.

θ = angular displacement.

Diameter = 64cm = 0.64m, radius = 64/2 = 32cm = 0.32m

The angular displacement can be gotten using the formulae below

S = rθ, where s= linear distance covered = 200m, r = radius = 0.32m

θ = S/r = 200/0.32=625 rad.

By substituting the parameter we have that

(12π)² = (7π)² + 2α(625)

1421.22 = 486.31 + 1250α

1421.22 - 486.31 = 1250α

934.91 = 1250α

α = 934.91/1250

α= 0.748 rad/s²

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Answer:

The correct answers are "chemical energy into electrical energy" and then "the electrical energy into light energy". Explanation: In the battery-powered flashlight, the battery supplies the chemical energy which makes the electrons to flow in the circuit and constitutes the current

Explanation:

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3 years ago
If we know the total energy in a system is 30 J, and we know the PE is 20 J. What is the KE?
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3 years ago
Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth’s center. Satellite A is to o
Vera_Pavlovna [14]

Answer:

Explanation:

Orbital radius of satellite A , Ra = 6370 + 6370 = 12740 km

Orbital radius of satellite B , Rb = 6370 + 19110 = 25480 km

Orbital potential energy of a satellite = - GMm / r where G is gravitational constant , M is mass of the earth and m is mass of the satellite

Orbital potential energy of a satellite A = - GMm / Ra

Orbital potential energy of a satellite B = - GMm / Rb

PE of satellite B /PE of satellite A

=  Ra / Rb

= 12740 / 25480

= 1 / 2

b ) Kinetic energy of a satellite is half the potential energy with positive value , so ratio of their kinetic energy will also be same

KE of satellite B /KE of satellite A

= 1 / 2

c ) Total energy will be as follows

Total energy = - PE + KE

- P E + PE/2

= - PE /2

Total energy of satellite B / Total energy of A

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Satellite B will have greater total energy because its negative value is less.

5 0
3 years ago
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Jobisdone [24]

The amplitude of wave-c is 1 meter.

The speed of all of the waves is (12meters/2sec)= 6 m/s.

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4 0
3 years ago
what equastion do you use to solve Riders in a carnival ride stand with their backs against the wall of a circular room of diame
Hitman42 [59]

Answer:

μsmín = 0.1

Explanation:

  • There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
  • This friction force has a maximum value, that can be written as follows:

       F_{frmax} = \mu_{s} *F_{n} (1)

       where  μs is the coefficient of static friction, and Fn is the normal force,

       perpendicular to the wall and aiming to the center of rotation.

  • This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
  • This force has the following general expression:

       F_{c} =  m* \omega^{2} * r (2)

       where ω is the angular velocity of the riders, and r the distance to the

      center of rotation (the  radius of the circle), and m the mass of the

      riders.

      Since Fc is actually Fn, we can replace the right side of (2) in (1), as

      follows:

     F_{frmax} = m* \mu_{s} * \omega^{2} * r (3)

  • When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:

       m* g = m* \mu_{smin} * \omega^{2} * r (4)

  • (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
  • Cancelling the masses on both sides of (4), we get:

       g = \mu_{smin} * \omega^{2} * r (5)

  • Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

      60 rev/min * \frac{2*\pi rad}{1 rev} *\frac{1min}{60 sec} =6.28 rad/sec (6)

  • Replacing by the givens in (5), we can solve for μsmín, as follows:

       \mu_{smin} = \frac{g}{\omega^{2} *r}  = \frac{9.8m/s2}{(6.28rad/sec)^{2} *2.5 m} =0.1 (7)

5 0
3 years ago
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