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Rudiy27
4 years ago
6

A bus is designed to draw its power from a rotating flywheel that is brought up to 3000 rpm by an electric motor. The flywheel i

s a solid cylinder of mass 2000 kg and diameter 1 m. If the bus requires an average power of 20 kilowatts, how long will the flywheel rotate?
620 s

830 s

980 s

1,200 s

2,500 s
Physics
1 answer:
Arada [10]4 years ago
3 0

To solve this problem we will apply the concept of rotational kinetic energy. Once this energy is found we will proceed to find the time from the definition of the power, which indicates the change of energy over time. Let's start with the kinetic energy of the rotating flywheel is

E_r = \frac{1}{2} I\omega^2

Here

I = moment of inertia

\omega = Angular velocity

Here we have that,

\omega = 3000\frac{rev}{min}(\frac{2\pi rad}{1rev})(\frac{1min}{60s})

\omega = 314.159rad/s

Replacing the value of the moment of inertia for this object we have,

E_r = \frac{1}{2} (\frac{MR^2}{2})\omega^2

E_r = \frac{1}{2} (\frac{2000(0.5)^2}{2})(314.159)^2

E_r = 1.233698*10^7J

The expression for average power is

P = \frac{E_r}{\Delta t}

\Delta t = \frac{E_r}{P}

\Delta t = \frac{1.233698*10^7}{20*10^3}

\Delta t = 616.8s \approx 620s

Therefore the correct answer is 620s.

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At a certain instant, a ball is thrown downward with a velocity of 8.0 m/s from a height of 40 m. At the same instant, another b
maks197457 [2]

Answer:

(a) The two balls collide 2\; \rm s after launch.

(b) The height of the collision is 4\; \rm m.

(Assuming that air resistance is negligible.)

Explanation:

Let vector quantities (displacements, velocities, acceleration, etc.) that point upward be positive. Conversely, let vector quantities that point downward be negative.

The gravitational acceleration of the earth points dowards (towards the ground.) Therefore, the sign of g should be negative. The question states that the magnitude of g\! is 10\; \rm m \cdot s^{-2}. Hence, the signed value of \! g should be \left(-10\; \rm m \cdot s^{-2}\right).

Similarly, the initial velocity of the ball thrown downwards should also be negative: \left(-8.0\; \rm m \cdot s^{-1}\right).

On the other hand, the initial velocity of the ball thrown upwards should be positive: \left(12\; \rm m \cdot s^{-1}\right).

Let v_0 and h_0 denote the initial velocity and height of one such ball. The following SUVAT equation gives the height of that ball at time t:

\displaystyle h(t) = \frac{1}{2}\, g \cdot {t}^{2} + v_0 \cdot t + h_0.

For both balls, g = \left(-10\; \rm m \cdot s^{-2}\right).

For the ball thrown downwards:

  • Initial velocity: v_0 = \left(-8.0\; \rm m \cdot s^{-1}\right).
  • Initial height: h_0 = 40\; \rm m.

\displaystyle  h(t) = -5\, t^{2} + (-8.0)\, t + 40 (where h is in meters and t is in seconds.)

Similarly, for the ball thrown upwards:

  • Initial velocity: v_0 = \left(12\; \rm m \cdot s^{-1}\right).
  • Initial height: h_0 = 0\; \rm m.

\displaystyle  h(t) = -5\, t^{2} + 12\, t (where h is in meters and t is in seconds.)

Equate the two expressions and solve for t:

-5\, t^{2} + (-8.0)\, t + 40 = -5\, t^{2} + 12\, t.

t = 2.

Therefore, the collision takes place 2\, \rm s after launch.

Substitute t = 2 into either of the two original expressions to find the height of collision:

h = -5\times 2^{2} + 12 \times 2 = 4\; \rm m.

In other words, the two balls collide when their height was 4\; \rm m.

3 0
3 years ago
An electron is accelerqated in the uniform field betwen two parallel charged oplates. The separation of the plates is 1.20 cm
marysya [2.9K]

Answer:

The speed of electron is 8.7\times10^{6}\ m/s

Explanation:

Given that,

Separation of the plate = 1.20 cm

Suppose the field is E=1.80\times10^{4}\ N/C.

If the electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate.

What the speed does it leave the hole?

We need to calculate the acceleration

Using formula of electric force

F = qE

ma=qE

a=\dfrac{qE}{m}

We need to calculate the speed of electron

Using equation of motion

v^2=u^2+2as

v^2=2as

Put the value of acceleration in the formula

v^2=2\times\dfrac{qE}{m}\times s

Put the value into the formula

v^2=2\times\dfrac{1.6\times10^{-19}\times1.80\times10^{4}\times1.20\times10^{-2}}{9.11\times10^{-31}}

v=\sqrt{2\times\dfrac{1.6\times10^{-19}\times1.80\times10^{4}\times1.20\times10^{-2}}{9.11\times10^{-31}}}

v=8.7\times10^{6}\ m/s

Hence, The speed of electron is 8.7\times10^{6}\ m/s

7 0
3 years ago
A person lifts a 6.50kg box off the ground a vertical distance of 1.25m and then carries it horizontally across the room by 4.75
victus00 [196]
What type of reaction is being shown in this energy diagram
6 0
3 years ago
Someone help please.....
Westkost [7]

Answer:

0.0928km/min (4dp)

Explanation:

To find the jogger's speed in km per minute, we just need to divide the number of km jogged by the time in minutes it took to jog that distance. This will give us the distance they jogged every minute which is their speed.

4km in 32 minutes:

4/32 = 0.125km/min

2km in 22 minutes:

2/22 = 0.091 (3dp)km/min

1km in 16 minutes:

0.0625km/min

Now to find the average speed of these 3 speeds, we just add them all together and divide by how many values there are (3 values).

Average (mean)  = \frac{0.125+0.091+0.0625}{3}

Average = 0.2785/3

Average speed of jogger = 0.0928 (4dp) km/min

Hope this helped!

8 0
3 years ago
Acceleration is defined as
zalisa [80]
Acceleration is the rate of change of velocity per unit of time.
7 0
3 years ago
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