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2 years ago

You are riding on a school bus and suddenly get thrown forward. What did the bus just do?

2 answers:

8 0

You are thrown the direction that the bus was originally going.

If the bus is going straight and it turns left your body will keep trying to go straight. This is the same if the bus turns to the left.

We know the answer is not A and D however because when turning you are not 'suddenly' thrown forward.

We know it cannot be that the bus was going backward and suddenly stopped because you would fly in the original direction of the bus which is backwards.

The answer is B because you are going forward, then suddenly stop, which means you are suddenly flung in the direction the bus was originally going which was forward.

Alexeev081 [22]2 years ago

3 0

B. The bus is going forward and suddenly stopped.

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if the velocity of a car is halved the fc required to keeo it in a path of constant radius is multiplied/divided by?

damaskus [11]

I believe this is it
The centripetal force is given by
F = mv^2 / r
When v' = v/2,
F' = mv'^2/r = m(v/2)^2/r = mv^2/4r = F/4.
So the centripetal force is divided by 4.

7 0

2 years ago

Read 2 more answers

Item 19 Question 1 A Ferris wheel has a radius of 75 feet. You board a car at the bottom of the Ferris wheel, which is 10 feet a

sergey [27]

Answer:

$y = 104.4 ft$

Explanation:

As we know that we board in the car of ferris wheel at the bottom position

So we will have

final height of the car at angular displacement given as

$y = y_o + R + R sin(270 - 255)$

$y = y_o + R + R sin 15$

here we know that

$y_o = 10 ft$

$R = 75 ft$

so we have

$y = 10 + 75 + 75 sin15$

$y = 104.4 ft$

4 0

2 years ago

A stone with a mass m is dropped from an airplane that has a horizontal velocity v at a height h above a lake. If air resistance

seropon [69]

Answer: Option B. R = (1/2)gt^2

Explanation:

S = R (horizontal distance)

V^2 = 2gS

V^2 = 2gR

R = V^2 / 2g

But V = gt

R = (gt)^2 / 2g

R = (g^2 x t^2) / 2g

R = gt^2 / 2

But t^2 = 2h/g

R = ( g x 2h/g) / 2

R = h

But h = (1/2)gt^2

R = h = (1/2)gt^2

4 0

2 years ago

The two masses in the Atwood's machine shown in the figure are initially at rest at the same height. After they are released, th

Inga [223]

According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.

To solve this problem we must apply the concept related to the conservation of energy theorem.

PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so

$E_i = E_f$

$0 = \frac{1}{2} (m_1+m_2)v_f^2-m_2gh+m_1gh$

$v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}$

PART B) Replacing the values given as,

$h= 1.7m\\m_1 = 3.5kg\\m_2 = 4.3kg \\g = 9.8m/s^2 \\$

$v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}$

$v_f = \sqrt{2(9.8)(1.7)(\frac{4.3-3.5}{3.5+4.3})}$

$v_f = 1.8486m/s$

Therefore the speed of the masses would be 1.8486m/s

6 0

2 years ago

Explain the difference between what Isaac Newton and Louis de Broglie would have to say about the momentum of a particle that is

pishuonlain [190]

Answer:

Explained

Explanation:

Newton would resort to the classical mechanics and say that the momentum of the particle that is moving with a constant velocity will be given by: momentum = mass x velocity

this approach will highlight the particle nature and will not be relativistic.

De-Broglie will say that the momentum of the particle is related to its associated matter wave and the relation between them is given by:

$p = \frac{h}{\lambda}$

where \lambda = wavelength of the matter wave associated to the particle, h = planck's constant

and $p = \gamma\times mv$

thus, this highlights the wave nature of the particle and is also relativistic.

6 0

2 years ago