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ziro4ka
17 days 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:
Yuliya22 [2.4K]17 days ago
3 0

To tackle this problem, we will use the concept of rotational kinetic energy. After determining this energy, we will find the time based on the power definition, which indicates how energy changes over time. Let's begin with the formula for the kinetic energy of a rotating flywheel:

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

Where

I = moment of inertia

\omega = Angular velocity

In this scenario, we have:

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

\omega = 314.159rad/s

Substituting the moment of inertia values for this object yields:

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 average power expression is given by:

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

Hence, the correct conclusion is 620 seconds.

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inna [2205]
The static frictional force exceeds the kinetic frictional force, indicating that the static frictional force is over 1200 N. Explanation: The frictional force opposes the motion of any object on a surface, caused by interactions between the surface molecules and the object. It is known that static friction is typically stronger than kinetic friction (this is the reason initiating motion requires more force than keeping it moving along a surface). Hence, option 3 correctly describes the situation.
3 0
3 days ago
A ball collides elastically with an immovable wall fixed to the earth’s surface. Which statement is false? 1. The ball's speed i
Maru [2355]

Answer:

Statements 4, 6 & 7 are incorrect.

Explanation:

In any elastic collision, the overall momentum vector sum of the system remains zero.

In this scenario, an elastic collision occurs between the ball and a stationary wall. The ball's velocity will consistently revert after the impact, leading to a change in direction of momentum.

The initial momentum of the ball is represented as:

p=m.v

where:

m = mass of the ball

v = initial velocity of the body

post-collision for the elastic interaction:

p=m.(-v)

  • Here, the momentum changes solely in direction, thus contradicting statement 7.
  • During the impact, both the ball and the wall exert forces on each other that are equal and opposite. The wall remains motionless, while the ball is influenced by the wall's reaction force, performing work on it, which contradicts statement 4.
  • Given that this collision is elastic, the ball's form and dimensions do not alter.
  • The previous points clearly indicate that not all provided statements hold true, thus violating statement 6.
4 0
1 day ago
If the position of an object is zero at one instant, what is true about the velocity of that object?
ValentinkaMS [2425]
If the position of an object is zero at a particular moment, this does not provide any indication about its velocity. It might simply be moving through that point, and you observed it exactly when it was at zero.
6 0
29 days ago
Compressed air is used to fire a 60 g ball vertically upward from a 0.70-m-tall tube. The air exerts an upward force of 3.0 N on
Yuliya22 [2420]

Answer:

2.87 m

Explanation:

Given parameters:

Mass of the ball (m) = 60 g = 0.06 kg

Height of the tube (h) = 0.70 m

Force applied on the ball by compressed air (F) = 3.0 N

Initial velocity of the ball (u) = 0 m/s (Assumed)

Final velocity of the ball at the tube's exit (v) =?

Acceleration of the ball (a) =?

The ball's weight is derived from multiplying mass and gravity. Therefore,

Weight (W) = mg=0.06\times 9.8=0.588\ N

Thus, the total force acting on the ball equals the net of upward force minus the weight.

Net force = Air force - Weight

F_{net}=F-mg\\F_{net}=3.0-0.588 = 2.412\ N

According to Newton's second law, net force equals the mass multiplied by acceleration.

F_{net}=ma\\\\a=\frac{F_{net}}{m}=\frac{2.412\ N}{0.06\ kg}=40.2\ m/s^2

Acceleration (a) is calculated as 40.2 m/s².

Using the motion equation, we find:

v^2=u^2+2ah\\\\v^2=0+2\times 40.2\times 0.7\\\\v=\sqrt{56.28}=7.5\ m/s

Let’s denote the maximum height achieved as 'H'.

Next, we apply the principle of energy conservation from the pipe's peak to the maximum height.

A decrease in kinetic energy equals an increase in potential energy.

\frac{1}{2}mv^2=mgH\\\\H=\frac{v^2}{2g}

Substituting the values, we solve for 'H', yielding:

H=\frac{56.28}{2\times 9.8}\\\\H=\frac{56.28}{19.6}=2.87\ m

Hence, the ball ascends to a height of 2.87 m above the top of the tube.

6 0
24 days ago
Jo, Daniel and Helen are pulling a metal ring. Jo pulls with a force of 100N in one direction and Daniel with a force of 140N in
inna [2205]

Answer:

She exerts a force of 40 N.

Explanation:

The fact that the ring remains stationary indicates that the forces are in equilibrium.

Let’s denote Jo's force as x.

The equation to consider is

140 = x + 100

x = 40

5 0
14 days ago
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