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ELEN [110]
4 years ago
15

When a gear rotates 20 revolutions, it achieves an angular velocity of 30 rad/s, starting from rest. determine its constant angu

lar acceleration in units of rad/s2?
Physics
1 answer:
Temka [501]4 years ago
4 0
Initial angular velocity  = 0  (starting from rest)
Final angular velocity = 30 rad/s
Distance traveled = (20 rev)*(2π rad/rev) = 40π rad

Let the angular acceleration be α rad/s².
Then
(30 rad/s)² = (0 rad/s)² + 2*(α rad/s²)*(40π rad)
α = 3.58 rad/s²

Answer: 3.58 rad/s² (nearest hundredth)
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Suppose a particle moves along a straight line with velocity v(t)=t2e−2tv(t)=t2e−2t meters per second after t seconds. How many
dimulka [17.4K]

Explanation:

It is given that,

Velocity of the particle moving in straight line is :

v(t)=t^2e^{-2t}\ m/s

We need to find the distance (x)  traveled by the particle during the first t seconds. It is given by :

x=\int\limits {v.dt}

x=\int\limits {t^2e^{-2t}dt}

Using by parts integration, we get the value of x as :

x=\dfrac{-(2t^2+2t+1)e^{-2t}}{4}\ meters

Hence, this is the required solution.

6 0
4 years ago
Two coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves with wavelength 6.00 meters. Consid
Korolek [52]

Answer:

a

    z= 2.5 \ m

b

   z =  (1 \ m ,  4 \ m )

Explanation:

From the question we are told that

     Their distance apart is  d =  5.00 \ m

      The  wavelength of each source wave \lambda =  6.0 \ m

Let the distance from source A  where the construct interference occurred be z

Generally the path difference for constructive interference is

              z - (d-z) =  m \lambda

Now given that we are considering just the straight line (i.e  points along the line connecting the two sources ) then the order of the maxima m =  0

  so

        z - (5-z) =  0

=>     2 z - 5 =  0

=>     z= 2.5 \ m

Generally the path difference for destructive  interference is

           |z-(d-z)| = (2m + 1)\frac{\lambda}{2}

=>         |2z - d |= (0 + 1)\frac{\lambda}{2}

=>        |2z - d| =\frac{\lambda}{2}

substituting values

          |2z - 5| =\frac{6}{2}

=>      z  =  \frac{5 \pm 3}{2}

So  

      z =  \frac{5 + 3}{2}

      z =  4\ m

and

      z =  \frac{ 5 -3 }{2}

=>   z =  1 \ m

=>    z =  (1 \ m ,  4 \ m )

7 0
3 years ago
An early method of measuring the speed of light makes use of a rotating slotted wheel. A beam of light passes through one of the
Readme [11.4K]

Answer:

4.7 x 10³ rad / s

Explanation:

During the time light goes and comes back , one slot is replaced by next slot while rotating  before the light source

Time taken by light to travel a distance of 2 x 500 m is

= (2 x 500) / 3 x 10⁸

= 3.333 x 10⁻⁶ s .

In this time period, two consecutive slots come before the source of light one after another by rotation. There are 400 slots so time taken to make one rotation

= 3.333 x 10⁻⁶ x 400

= 13.33 x 10⁻⁴ s

This is the time period so

T = 13.33 X 10⁻⁴

Angular speed

= 2π / T

= \frac{2\times3.14}{13.33\times10^{-4}}

4.7 x 10³ rad / s

5 0
3 years ago
Sound waves require a ________ to travel through
Karolina [17]

Answer:

sound Requires a medium in order to travel.

7 0
3 years ago
Read 2 more answers
An elevator is moving down with an acceleration of 3.36 m/s2.
sergeinik [125]

Answer : 413.44N

Here it is given that an elevator is moving down with an acceleration of 3.36 m/s² . And we are interested in finding out the apparent weight of a 64.2 kg man . For the diagram refer to the attachment .

  • From the elevator's frame ( non inertial frame of reference) , we would have to think of a pseudo force.
  • The direction of this force is opposite to the direction of acceleration the frame and its magnitude is equal to the product of mass of the concerned body with the acceleration of the frame .
  • When a elevator accelerates down , the weight recorded is less than the actual weight .

From the Free body diagram ,

\sf\longrightarrow Weight = mg - ma \\

\sf\longrightarrow Weight = m ( g - a ) \\

  • Mass of the man = 64.2 kg

\sf\longrightarrow Weight = 64.2( 9.8 - 3.36) N\\

\sf\longrightarrow Weight = 64.2 * 6.44 N\\

\sf\longrightarrow \underline{\boxed{\bf Weight_{apparent}= 413.44 N }} \\

5 0
2 years ago
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