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MAVERICK [17]
3 years ago
15

A 100-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope a

bout the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.400 rev/s in 2.00 s
Physics
1 answer:
olchik [2.2K]3 years ago
3 0

Answer:

The force required is equal to 30\pi N.

Explanation:

We know that

     Torque = force × perpendicular distance

                  = F × R = I×\alpha

               I = M×\frac{R^{2} }{2}

  From above equation \alpha  = \frac{FR}{I}              .............. 1

      We know that

              w=w_{o} +\alpha t                      ...........................2

               Given that t= 2 sec

                                  w_{o} = 0

                                  w=0.4 rev/s = 0.8\pi rad/s

From equation 1 and 2 , we get

                  F =  \frac{wI}{RT}  =  \frac{w\times m \times r}{2t}

    Upon substituting the above values we will be getting

          F = 30\pi N

                   

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PilotLPTM [1.2K]

Answer: E =  7,490.6 N/C

Explanation:

If we have a field E, and a particle with a charge q, the force that the particle experiences is:

F = E*q

In this case, we know that the force is:

F = 1.2*10^(-15) N

And we know that the particle is a proton, where the charge of a proton is:

q = 1.602*10^(-19) C

Then we can replace these two values in the equation to get:

1.2*10^(-15) N = E*1.602*10^(-19) C

We just need to isolate E.

(1.2*10^(-15) N)/(1.602*10^(-19) C) = E

7,490.6 N/C = E

That is the strength of the electric field.

8 0
3 years ago
A jet on an aircraft carrier can be launched from rest to 40 m/s in 2 seconds. What is the acceleration of the aircraft? Show st
Alex_Xolod [135]

Answer:

a=20\ m/s^2

Explanation:

Given that,

Initial speed, u = 0

Final speed, v = 40 m/s

Time, t = 2 s

We need to find the acceleration of the aircraft. We know that, acceleration is equal to the rate of change of velocity. So,

a=\dfrac{v-u}{t}\\\\a=\dfrac{40-0}{2}\\\\=20\ m/s^2

So, the acceleration of the aircraft is 20\ m/s^2.

5 0
2 years ago
What phase difference between two otherwise identical harmonic waves, moving in the same direction along a stretched string, wil
lora16 [44]

Answer:

\theta=145

Explanation:

The amplitude of he combined wave is:

B=2Acos(\theta/2)\\

A, is the amplitude from the identical harmonic waves

B, is the amplitude of the resultant wave

θ, is the phase, between the waves

The amplitude of the combined wave must be 0.6A:

0.6A=2Acos(\theta/2)\\ cos(\theta/2)=0.3\\\theta/2=72.5\\\theta=145

5 0
3 years ago
Halleys comet has period of 75.3 years. Using Kepler’s third law, find it’s semimajor axis expressed in astronomical units?
natta225 [31]

Answer: 17.83 AU

Explanation:

According to Kepler’s Third Law of Planetary motion <em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.  </em>

T^{2}\propto a^{3}  (1)

Talking in general, this law states a relation between the <u>orbital period</u> T of a body (moon, planet, satellite, comet) orbiting a greater body in space with the <u>size</u> a of its orbit.

However, if T is measured in <u>years</u>, and a is measured in <u>astronomical units</u> (equivalent to the distance between the Sun and the Earth: 1AU=1.5(10)^{8}km), equation (1) becomes:

T^{2}=a^{3}  (2)

This means that now both sides of the equation are equal.

Knowing T=75.3years and isolating a from (2):

a=\sqrt[3]{T^{2}}=T^{2/3}  (3)

a=(75.3years)^{2/3}  (4)

Finally:

a=17.83AU  (5)

4 0
3 years ago
A test pilot flies with an acceleration of of 5 g . what is the acceleration in meter per second
JulijaS [17]
I mean if he flies 5g that means that's his average speed too
5 0
3 years ago
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