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3 years ago

A passenger on your boat falls overboard. what should you do first?

1 answer:

5 0

The very first thing that you should do when a passenger on your boat falls overboard is to throw a PFD or also known as a Personal Flotation Device. This would include anything that can help the passenger to float. But this step would differ, only if the passenger is not wearing a lifevest.

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A ball collides with a vertical, unmovable wall. There is no friction between the wall and the ball (the only force acting on th

user100 [1]

Answer: the same direction I.e to the left.

Explanation:

The component perpendicular to the contact surface is such that will stop the relative motion and, in case of elastic collision like here, return the system to the same kinetic energy. So ball hitting immovable surface will have the same speed (magnitude of velocity) as before the collision.

There will also be parallel force caused by friction, but it has to be treated separately for two reasons:

The perpendicular force is limited to coefficient of friction times the normal force. If that is not enough to stop the ball, it will skid on the surface.The perpendicular force, and this depends on the specific geometry, does not pass through the centre of mass of the ball. Therefore it imparts a moment on the ball that causes it to start rotating. And once the ball is rotating so that the point of contact is stationary, there is no momentum to cause any friction force anymore and the friction force disappears and stops decelerating the ball.

So what happens is that the vertical component of the velocity will be reversed, while the horizontal component will be somewhat reduced with the corresponding amount of kinetic energy transferred to energy of rotation. The rotation will always eliminate the friction force before the horizontal component of velocity is zeroed, so the ball will always continue in the same direction, just a bit slower.

If you instead threw an elastic box (which could not start rotating freely) it could actually bounce back.

7 0

2 years ago

An object accelerating may be changing in what two ways

Oksana_A [137]

Velocity and direction

8 0

3 years ago

A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia

aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the initial point is 19820 J

2) The potential energy at point A is 19620 J

3) The kinetic energy at point B is 10010 J

4) The potential energy at point C is zero

5) The kinetic energy at point C is 19820 J

6) The velocity of the roller coaster at point C is 19.91 m/s

1) The total energy for the roller coaster at the initial point can be found as follows:

$E_{t} = KE_{i} + PE_{i}$

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The total energy is:

$E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J$

Hence, the total energy for the roller coaster at the initial point is 19820 J.

2) The potential energy at point A is:

$PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J$

Then, the potential energy at point A is 19620 J.

3) The kinetic energy at point B is the following:

$KE_{A} + PE_{A} = KE_{B} + PE_{B}$

$KE_{B} = KE_{A} + PE_{A} - PE_{B}$

Since

$KE_{A} + PE_{A} = KE_{i} + PE_{i}$

we have:

$KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J$

Hence, the kinetic energy at point B is 10010 J.

4) The potential energy at point C is zero because h = 0 meters.

$PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J$

5) The kinetic energy of the roller coaster at point C is:

$KE_{i} + PE_{i} = KE_{C} + PE_{C}$

$KE_{C} = KE_{i} + PE_{i} = 19820 J$

Therefore, the kinetic energy at point C is 19820 J.

6) The velocity of the roller coaster at point C is given by:

$KE_{C} = \frac{1}{2}mv_{C}^{2}$

$v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s$

Hence, the velocity of the roller coaster at point C is 19.91 m/s.