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Mariulka [41]
2 years ago
12

A record player turntable initially rotating at 3313 rev/min is braked to a stop at a constant rotational acceleration. The turn

table has a rotational inertia of 0.031kg?m2 . When it is switched off, it slows down to 75% of its initial rotational speed in 4.0 s.A- How long does it take to come to rest?B- How much work has to be done on the turntable to bring it to rest?
Physics
1 answer:
KatRina [158]2 years ago
7 0

Answer:

15.99861 seconds

1865.58658 Joules

Explanation:

\alpha = Angular acceleration

\theta = Angle of rotation

I = Moment of inertia of disc = 0.031 kgm²

t = Time taken

Initial angular speed

\omega_i=3313\frac{2\pi}{60}\\\Rightarrow \omega_i=346.93\ rad/s

Final angular velocity

\omega_f=0.75\times \omega_i\\\Rightarrow \omega_f=0.75\times 346.93\\\Rightarrow 260.19\ rad/s

From the equation of rotational motion

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{260.19-346.93}{4}\\\Rightarrow \alpha=-21.685\ rad/s^2

\omega_f=\omega_i+\alpha t\\\Rightarrow t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-346.93}{-21.685}\\\Rightarrow t=15.99861\ s

Time taken by the turntable to slow down is 15.99861 seconds

Rotational energy is given by

K=\frac{1}{2}I\omega_i^2\\\Rightarrow K=\frac{1}{2}\times 0.031\times 346.93^2\\\Rightarrow K=1865.58658\ J

The work has to be done on the turntable to bring it to rest is 1865.58658 Joules

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James gently releases a ball at the top of a slope, but does not push the ball. The ball rolls down the slope. Which force cause
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D. Gravitational force.

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2 years ago
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In a shot-put competition, a shot moving at 15m/s has 450J of mechanical kinetic energy. What is the mass of the shot? Please he
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Answer:

Mass of shot (m) = 4 kg

Explanation:

Given:

Velocity (v) = 15 m/s

Mechanical kinetic energy (K.E) = 450 J

Find:

Mass of shot (m) = ?

Computation:

Mechanical kinetic energy (K.E) = 1/2mv²

Mechanical kinetic energy (K.E) = [1/2](m)(15)²

450 = [1/2](m)(15)²

900 = 225 m

Mass of shot (m) = 4 kg

5 0
2 years ago
The position of a particle moving along the x-axis varies with time according to x(t) = 5.0t^2 − 4.0t^3 m. Find (a) the velocity
KengaRu [80]
<h2>Answer:</h2>

(a) v(t) = [10.0t - 12.0t²] m/s  and a(t) = [10.0 - 24.0t ] m/s² respectively

(b) -28.0m/s and -38.0m/s² respectively

(c) 0.83s

(d) 0.83s

(e) x(t)  = 1.1573 m           [where t = 0.83s]

<h2>Explanation:</h2>

The position equation is given by;

x(t) = 5.0t² - 4.0t³ m           --------------------(i)

(a) Since velocity is the time rate of change of position, the velocity, v(t), of the particle as a function of time is calculated by finding the derivative of equation (i) as follows;

v(t) = dx(t) / dt = \frac{dx}{dt} = \frac{d}{dt} [ 5.0t² - 4.0t³ ]

v(t) = 10.0t - 12.0t²     --------------------------------(ii)

Therefore, the velocity as a function of time is v(t) = 10.0t - 12.0t² m/s

Also, since acceleration is the time rate of change of velocity, the acceleration, a(t), of the particle as a function of time is calculated by finding the derivative of equation (ii) as follows;

a(t) = dx(t) / dt = \frac{dv}{dt} =  \frac{d}{dt} [ 10.0t - 12.0t² ]

a(t) = 10.0 - 24.0t             --------------------------------(iii)

Therefore, the acceleration as a function of time is a(t) = 10.0 - 24.0t m/s²

(b) To calculate the velocity at time t = 2.0s, substitute the value of t = 2.0 into equation (ii) as follows;

=> v(t) =  10.0t - 12.0t²

=> v(2.0) = 10.0(2) - 12.0(2)²

=> v(2.0) = 20.0 - 48.0

=> v(2.0) = -28.0m/s

Also, to calculate the acceleration at time t = 2.0s, substitute the value of t = 2.0 into equation (iii) as follows;

=> a(t) = 10.0 - 24.0t

=> a(2.0) = 10.0 - 24.0(2)

=> a(2.0) = 10.0 - 48.0

=> a(2.0) = -38.0 m/s²

Therefore, the velocity and acceleration at t = 2.0s are respectively -28.0m/s and -38.0m/s²

(c) The time at which the position is maximum is the time at which there is no change in position or the change in position is zero. i.e dx / dt = 0. It also means the time at which the velocity is zero. (since velocity is dx / dt)

Therefore, substitute v = 0 into equation (ii) and solve for t as follows;

=> v(t) = 10.0t - 12.0t²

=> 0 = 10.0t - 12.0t²

=> 0 = ( 10.0 - 12.0t ) t

=> t = 0            or             10.0 - 12.0t = 0

=> t = 0            or             10.0 = 12.0t

=> t = 0            or             t = 10.0 / 12.0

=> t = 0            or             t = 0.83s

At t=0 or t = 0.83s, the position of the particle will be maximum.

To get the more correct answer, substitute t = 0 and t = 0.83 into equation (i) as follows;

<em>Substitute t = 0 into equation (i)</em>

x(t) = 5.0(0)² - 4.0(0)³ = 0

At t = 0; x = 0

<em>Substitute t = 0.83s into equation (i)</em>

x(t) = 5.0(0.83)² - 4.0(0.83)³

x(t) = 5.0(0.6889) - 4.0(0.5718)

x(t) = 3.4445 - 2.2872

x(t)  = 1.1573 m

At t = 0.83; x = 1.1573 m

Therefore, since the value of x at t = 0.83s is 1.1573m is greater than the value of x at t = 0 which is 0m, then the time at which the position is at maximum is 0.83s

(d) The velocity will be zero when the position is maximum. That means that, it will take the same time calculated in (c) above for the velocity to be zero. i.e t = 0.83s

(e) The maximum position function is found when t = 0.83s as shown in (c) above;

Substitute t = 0.83s into equation (i)

x(t) = 5.0(0.83)² - 4.0(0.83)³

x(t) = 5.0(0.6889) - 4.0(0.5718)

x(t) = 3.4445 - 2.2872

x(t)  = 1.1573 m            [where t = 0.83s]

8 0
2 years ago
A cubical box, 5.00 cm on each side, is immersed in a fluid. The gauge pressure at the top surface of the box is 594 Pa and the
Zolol [24]

Answer:

The density of the fluid is 1100 kg/m³.

Explanation:

Given that,

Height = 5.00 cm

Pressure at top =594 Pa

Pressure at bottom = 1133 Pa

We need to calculate the change in pressure

Using formula of change in pressure

\Delta P=P_{b}-P_{t}

Where, P_{b} = Pressure at bottom

P_{t} = Pressure at top

put the value into the formula

\Delta P=1133-594

\Delta P=539\ Pa

Using formula of pressure for density

\Delta P = \rho g h

\rho =\dfrac{\Delta P}{gh}

Where, \rho = density

P = pressure

h = height

Put the value in to the formula

\rho =\dfrac{539}{5.00\times10^{-2}\times9.8}

\rho =1100\ kg/m^3

Hence, The density of the fluid is 1100 kg/m³.

4 0
2 years ago
Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to t
yarga [219]

Answer

The rate at which the magnetic field is changing is  [\frac{dB}{dt} ] =  0.000467 T/s

Explanation

From the question we are told that

   The electric field strength is E =  3.5mV/m =  3.5 *10^{-3} \ V/m

    The radius is  r =  1.5 \ m

The rate of change of the  magnetic  field  is mathematically represented as

        \frac{d \phi }{dt}  =  \int\limits^{} {E \cdot dl}

Where dl is change of a unit length

     \frac{d \phi}{dt}  =  A *  \frac{dB}{dt}

Where A is the area which is mathematically represented as

     A = \pi r^2

    So

    E \int\limits^{} {  dl} =  ( \pi r^2) (\frac{dB}{dt} )  

  E L  =  ( \pi r^2) (\frac{dB}{dt} )  

where L is the circumference of the circle which is mathematically represented as

     L = 2 \pi r

So

     E (2 \pi r ) =  (\pi r^2 ) [\frac{dB}{dt} ]

      E  =   \frac{r}{2}  [\frac{dB}{dt} ]

       [\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }

substituting values

      [\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }

      [\frac{dB}{dt} ] =  0.000467 T/s    

8 0
2 years ago
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