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3 years ago

13

A ball is tossed upwards into the air with an initial upwards velocity of 3.0m/s. What is the ball's velocity when it returns to

the height from which it was tossed?

1 answer:

krek1111 [17]3 years ago

7 0

Same as initial velocity.
Acceleration due to gravity is 9.8m/s^2

Which is constant on the whole way of travel of the ball.

As no change in acceleration there is no change in velocity at end and it remains at 3m/s

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Open the Faraday Law simulation and discover what you can about induction. Make a list of ways to cause induction.

nika2105 [10]

Answer:

Current can be induced in a lot of ways. I'll try and list as much of these methods as I can here.

1. Moving a bar magnet relative to a wire coil

2. Moving a wire coil relative to a magnet.

3. Move a wire coil that has electricity flowing though it relative to a wire coil without electricity flowing through it.

4. Moving a current carrying circuit relative to a non-current carrying circuit

5. Rapidly opening and closing the switch of a current carrying circuit beside a non current carrying circuit.

6. Moving a bar magnet through the middle of a wire coil.

7. Moving a wire coil through a magnet

8. Moving a circuit relative to a magnetic field.

The main idea is to cause a change in the magnetic field, or to change the available area of the wire loop, or to change the angle between the field and the loop.

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4 years ago

A racecar driver is driving her car down the drag strip at 120 m/s. What is the shortest distance in which she can brake and sto

DaniilM [7]

Answer:

1034.78 m

Explanation:

The shortest distance is the displacement of the car from initial position to final position.

Displacement of body is given using Newton's equations of motion.

Given:

Initial velocity, $u = 120$ m/s

Final velocity, $v = 0$ m/s( As the car stops in its final position)

Coefficient of static friction, $\mu=0.71$

Acceleration due to gravity, $g=9.8\textrm{ } m/s^{2}$

Now, when brakes are applied, only friction acts on the body in a direction opposite to that of its motion.

The acceleration of the car when friction acts the stopping force is given as:

$a=- \mu g$ .

The acceleration is negative as it reduces the velocity of the motion and acts in the direction opposite to that of the motion.

Plug in 0.71 for $\mu$ and 9.8 m/s² for $g$ . Solve for $a$ .

So, $a=-0.71\times 9.8=-6.958\textrm{ }m/s^{2}$ .

Now, displacement of the car is given using the following equation of motion:

$v^{2}=u^{2} +2aS$

Here, $S$ is the displacement of the racing car.

Plug in 120 m/s for $u$ , 0 m/s for $v$ , -6.958 m/s² for $a$ . Solve for $S$ . This gives,

$0^{2} =(120)^{2}+2(-6.958)S\\13.916S=14400\\S=\frac{14400}{13.916}=1034.78\textrm{ m}$

Therefore, the shortest distance in which she can brake and stop is 1034.78 m.

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prisoha [69]

Explanation:

This is a good example of Doppler effect

Given

Source velocity Vs=20m/s

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The frequency of the horn Fh=?

But mathematically Fh=(V-Vs/V)*Fa

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Explanation:

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