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Arturiano [62]
3 years ago
5

f the CD rotates clockwise at 500 rpmrpm (revolutions per minute) while the last song is playing, and then spins down to zero an

gular speed in 2.60 ss with constant angular acceleration, what is ααalpha, the magnitude of the angular acceleration of the CD, as it spins to a stop?
Physics
1 answer:
vazorg [7]3 years ago
3 0

Answer:

The magnitude of the angular acceleration is  a =  20.14 rad/s^2

Explanation:

From the question we are told that

   The angular speed of CD is  w_{CD} =  500 rpm =  \frac{500  rpm}{\frac{60 \ s }{1 \ min} } * \frac{2 \pi }{ 1 \ rev} = 52.37 rad/s

    time taken to decelerate is t_{CD} =  2.60\ s

    The final angular speed is  w_f= 0 \ rad/s

The angular acceleration is mathematically represented as

         a =  \frac{w_f - w_{CD}}{t}

substituting values

          a =  \frac{0 - 52.37}{2.60}

         a =  - 20.14 rad/s^2

The negative sign show that the CD is decelerating  but the magnitude is

       a =  20.14 rad/s^2

   

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A small rock is thrown vertically upward with a speed of 17.0m/s from the edge of the roof of a 26.0m tall building. The rock do
Pachacha [2.7K]

Answer:

A) v = 28.3 m/s

B) t =  4.64 s

Explanation:

A)

  • Assuming no other forces acting on the rock, since the accelerarion due to gravity close to the surface to the Earth can be taken as constant, we can use one of the kinematic equations in order to get first the maximum height (over the roof level) that the ball reaches:

        v_{f}^{2} - v_{o}^{2} = 2* g* \Delta h  (1)

  • Taking into account that at this point, the speed of the rock is just zero, this means vf=0 in (1), so replacing by the givens and solving for Δh, we get:

       \Delta h = \frac{-v_{o} ^{2}}{2*g} = \frac{-(17.0m/s)^{2} }{2*(-9.8m/s2)} = 14.8 m (2)

  • So, we can use now the same equation, taking into account that the initial speed is zero (when it starts falling from the maximum height) and that the total vertical displacement is the distance between the roof level and the ground (26.0 m) plus the maximum height that we have just found in (2) , 14.8m:
  • Δh = 26.0 m + 14. 8 m = 40.8 m (3)
  • Replacing now in (1), we can solve for vf, as follows:

       v_{f} =\sqrt{2*g*\Delta h} = \sqrt{2*9.8m/s2*40.8m} = 28.3 m/s (4)

B)

  • In order to find the total elapsed from when the rock is thrown until it hits the street, we can divide this time in two parts:
  • 1) Time elapsed from the the rock is thrown, until it reaches to its maximum height, when vf =0
  • 2) Time elapsed from this point until it hits the street, with vo=0.
  • For the first part, we can simply use the definition of acceleration (g in this case), making vf =0, as follows:

       v_{f} = v_{o} + a*\Delta t = v_{o} - g*\Delta t = 0 (5)

  • Replacing by the givens in (5) and solving for Δt, we get:

       \Delta t = \frac{v_{o}}{g} = \frac{17.0m/s}{9.8m/s2} = 1.74 s (6)

  • For the second part, since we know the total vertical displacement from (3), and that vo = 0 since it starts to fall, we can use the kinematic equation for displacement, as follows:

       \Delta h = \frac{1}{2} * g * t^{2}  (7)

  • Replacing by the givens and solving for t in (7), we get:

       t_{fall} =\sqrt{\frac{2*\Delta h}{g}} = \sqrt{\frac{2*40.8m}{9.8m/s2} } = 2.9 s (8)

  • So, total time is just the sum of (6) and (8):
  • t = 2.9 s + 1.74 s = 4.64 s
5 0
2 years ago
A positively charged particle initially at rest on the ground accelerates upward to 160 m/s in 2.10 s. The particle has a charge
mamaluj [8]

Answer:

The magnitude of the electric field is 8.6\times10^{2}\ N/C

Explanation:

Given that,

Time t = 2.10 s

Speed = 160 m/s

Specific charge =Ratio of charge to mass = 0.100 C/kg

We need to calculate the acceleration

Using equation of motion

a=\dfrac{v-u}{t}

Put the value into the formula

a=\dfrac{160-0}{2.10}

a=76.19\ m/s^2

We need to calculate the magnitude of the electric field

Using formula of electric field

E=\dfrac{F}{q}

E=\dfrac{ma}{q}

E=\dfrac{a+g}{\dfrac{q}{m}}

Put the value into the formula

E=\dfrac{76.19+9.8}{0.100}

E=8.6\times10^{2}\ N/C

The direction is upward.

Hence, The magnitude of the electric field is 8.6\times10^{2}\ N/C

4 0
3 years ago
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Which equation represents a neutralization reaction?
weeeeeb [17]
Is b= acid + base - water + salt.
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2 years ago
A person is using a rope to lower a 5.0-n bucket into a well with a constant speed of 2.0 m/s. What is the magnitude of the forc
OLga [1]

Answer:

5 N

Explanation:

The bucket is moving at a constant speed of 2m/s Therefore F=ma is 0 N for this to be correct the magnitude of the force exerted by the rope must be equal to the weight of the bucket which is 5 N

7 0
3 years ago
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The power in a circuit is 2 W and the voltage is 20 VDC. What's the circuit current?
Aliun [14]

Answer:

Power in the circuit is 0.1 amp

Explanation:

The power in the circuit is given by the formula

      P = V x I

Where P is Power, V is voltage supplied and I is current in circuit.

so,      I = P/v

            = 2/20

            =0.1 A

Study more about power

<u>https://brainly.in/question/1063947</u>

3 0
2 years ago
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