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sweet-ann [11.9K]
3 years ago
14

A car is designed to get its energy from a rotating flywheel with a radius of 1.50 m and a mass of 430 kg. Before a trip, the fl

ywheel is attached to an electric motor, which brings the flywheel's rotational speed up to 5,200 rev/min.
Required:
a. Find the kinetic energy stored in the flywheel.
b. If the flywheel is to supply energy to the car as would a 15.0-hp motor, find the length of time the car could run before the flywheel would have to be brought back up to speed.
Physics
1 answer:
fiasKO [112]3 years ago
3 0

Answer:

a

  KE  =  7.17 *10^{7} \ J

b

 t = 6411.09 \ s

Explanation:

From the question we are told that

    The radius of the flywheel is  r =  1.50 \ m

      The mass of the flywheel is m  = 430 \ kg

          The rotational speed of the flywheel is w  =  5,200 \ rev/min = 5200 *  \frac{2 \pi }{60} =544.61 \ rad/sec

      The power supplied by the motor is  P  =  15.0 hp =  15 * 746 =  11190 \ W

         

     Generally the moment of inertia of the flywheel is  mathematically represented as

       I  = \frac{1}{2} mr^2

substituting values

       I  = \frac{1}{2}  ( 430)(1.50)^2

       I  = 483.75 \  kgm^2

The kinetic energy that is been stored is  

       KE  =  \frac{1}{2} * I * w^2

substituting values

        KE  =  \frac{1}{2} * 483.75 * (544.61)^2

        KE  =  7.17 *10^{7} \ J

Generally power is mathematically represented as

          P =  \frac{KE}{t}

=>      t =  \frac{KE}{P}

substituting the value

        t = \frac{7.17 *10^{7}}{11190}

        t = 6411.09 \ s

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3 0
1 year ago
Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T
djverab [1.8K]

Answer:

v_{2} will be less than v_{1} and P_{2} will be greater than P_{1}.

Explanation:

As we know from the conservation of mass, the rate at which any amount of fluid mass (m_{1}) is entering in a system is equal to the rate at which the same amount of fluid mass (m_{2}) is leaving the system.

Rate of mass flow can be written as,

m = \rho A v

where \rho is the density of the fluid, A is the area through which the fluid is flowing and v is the velocity of the fluid.

Now, according to the problem, as the density of the fluid does not change, we can write

&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}

where A_{1} and A_{2} are the cross-sectional areas through which the fluid is passing and v_{1} and v_{2} are the velocities of the fluid through the respective cross-sectional areas.

As according to the problem, A_{2} > A_{1}, so from the above formula v_{2} < v_{1}.

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

So, as in this case v_{2} < v_{1} the pressure in the large cross-sectional area P_{2} will be greater than the pressure  P_{1} in the small cross sectional area, i.e.,

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6 0
3 years ago
Please help me with this question​
vovangra [49]

Answer:

1. 12 V

2a. R₁ = 4 Ω

2b. V₁ = 4 V

3a. A = 1.5 A

3b. R₂ = 4 Ω

4. Diagram is not complete

Explanation:

1. Determination of V

Current (I) = 2 A

Resistor (R) = 6 Ω

Voltage (V) =?

V = IR

V = 2 × 6

V = 12 V

2. We'll begin by calculating the equivalent resistance. This can be obtained as follow:

Voltage (V) = 12 V

Current (I) = 1 A

Equivalent resistance (R) =?

V = IR

12 = 1 × R

R = 12 Ω

a. Determination of R₁

Equivalent resistance (R) = 12 Ω

Resistor 2 (R₂) = 8 Ω

Resistor 1 (R₁) =?

R = R₁ + R₂ (series arrangement)

12 = R₁ + 8

Collect like terms

12 – 8 =

4 = R₁

R₁ = 4 Ω

b. Determination of V₁

Current (I) = 1 A

Resistor 1 (R₁) = 4 Ω

Voltage 1 (V₁) =?

V₁ = IR₁

V₁ = 1 × 4

V₁ = 4 V

3a. Determination of the current.

Since the connections are in series arrangement, the same current will flow through each resistor. Thus, the ammeter reading can be obtained as follow:

Resistor 1 (R₁) = 4 Ω

Voltage 1 (V₁) = 6 V

Current (I) =?

V₁ = IR₁

6 = 4 × I

Divide both side by 4

I = 6 / 4

I = 1.5 A

Thus, the ammeter (A) reading is 1.5 A

b. Determination of R₂

We'll begin by calculating the voltage cross R₂. This can be obtained as follow:

Total voltage (V) = 12 V

Voltage 1 (V₁) = 6 V

Voltage 2 (V₂) =?

V = V₁ + V₂ (series arrangement)

12 = 6 + V₂

Collect like terms

12 – 6 = V₂

6 = V₂

V₂ = 6 V

Finally, we shall determine R₂. This can be obtained as follow:

Voltage 2 (V₂) = 6 V

Current (I) = 1.5 A

Resistor 2 (R₂) =?

V₂ = IR₂

6 = 1.5 × R₂

Divide both side by 1.5

R₂ = 6 / 1.5

R₂ = 4 Ω

4. The diagram is not complete

7 0
2 years ago
An open pipe, 0.29 m long, vibrates in the second overtone with a frequency of 1,227 Hz. In this situation, the fundamental freq
Andru [333]

Answer:

f = 409 Hz

Explanation:

We have,

Length of the open organ pipe, l = 0.29 m

Frequency of vibration of second overtone, f_2 = 1227 Hz

It is required to find the fundamental frequency of the pipe. For the open organ pipe, the frequency of second overtone is given by :

f_2=\dfrac{3v}{2l}

v is speed of sound

Let f is the fundamental frequency. It is given by :

f=\dfrac{v}{2l}

The relation between f and f₂ can be written as :

f_2=3f\\\\f=\dfrac{f_2}{3}\\\\f=\dfrac{1227}{3}\\\\f=409\ Hz

So, the fundamental frequency of the pipe is 409 Hz.              

6 0
2 years ago
During which segments is kinetic energy decreasing?<br> A)1,2,3<br> B) 4,5<br> C)1,3,5<br> D)2,4
Nutka1998 [239]

Answer:

c

Explanation:

kinetic energy is energy an object has due to its movement. for instance, if someone was riding down a hill, when the motion of the bike begins to decrease so does the kinetic energy

3 0
3 years ago
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