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3 years ago

PLEASE ANSWER ASAP!!!

1 answer:

3 0

The total momentum of the system has to be conserved to satisfy the principle of conservation of momentum. Before the ball hits the bottle, the momentum of the system is 0.4 x 18 = 7.2 kg m/s

The momentum of the bottle after being hit is 0.2 x 25 = 5 kg m/s

So the momentum of the ball now is 7.2 - 5 = 2.2 kg m/s

Hence its velocity is 2.2/0.4 = 5.5 m/s

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Between the ball and the player’s head, there are forces. Which of Newton’s laws does this represent? Support your choice.

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According to Newton's third law

Explanation:

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2 years ago

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Mimi makes a table about the two types of beta decay. A 5 column table with 2 rows. The first column is labeled type of decay wi

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No change in mass number occurs in beta minus decay

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3 years ago

Suppose that you and three classmates are discussing the design of a roller coaster. One says that each hill must be lower than

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Answer:

I would say that I agree with the one that said that each hill must be lower than the previous one and use the principle of conservation of energy to explain.

Explanation:

Roller coaster are usually designed such that its total energy remains conserved at any point on the track. Now, the law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time. At certain height on the track, the total energy of the roller coaster is in form of potential energy, which gets converted to kinetic energy as soon as it starts sliding down the hill till get to the hill's endpoint where it has maximum kinetic energy. The cycle of sliding from a high point on the track to a low point on the track means there is potential energy is converted to kinetic energy and kinetic energy then converts back to potential energy and the cycle continues.

However, due to the effect of gravity and frictional force between the track and the coaster, the energy of the coaster is gradually reduces, so it becomes a bit difficult for the coaster to move to the next hill of the same height. It is for this reason that each hill must be lower than the previous one, so that the coaster can overcome the next hill's height with its reduced energy until it loses all its energy and comes to a stop.

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3 years ago

A 2.0 kg wood block is launched up a wooden ramp that is inclinedat a 35* angle. The block’s initial speed is 10m/s. (Giventhat

IgorC [24]

a) 4.0 m

We can solve this part by writing the equations of motion along the two directions: perpendicular to the slope and parallel to the slope.

Perpendicular to the slope:

$N-mgcos \theta =0$ (1)

where N is the normal reaction, m = 2.0 kg is the mass of the block, g = 9.8 m/s^2 is the acceleration of gravity, $\theta=35^{\circ}$ is the angle.

Parallel to the slope:

$-\mu_k N -mgsin \theta = ma$ (2)

where $\mu_k=0.20$ is the coefficient of friction, and a the acceleration, and where we have chosen up the slope as positive direction, so both forces are negative.

From (1) we get

$N=mg cos \theta$

And substituting into (2), we can find the acceleration:

$-\mu_k mg cos \theta -mgsin \theta = ma\\a=-\mu_k g cos\theta - g sin \theta = -(0.20)(9.8)(cos 35^{\circ})-(9.8)(sin 35^{\circ})=-7.2 m/s^2$

where the negative sign means the direction is down the slope.

Now we can find the distance travelled along the slope by using the SUVAT equation

$v^2-u^2=2ad$

where

v = 0 is the velocity when the block comes to rest

u = 10 m/s is the initial velocity

d is the distance travelled along the slope

Solving for d,

$d=\frac{v^2-u^2}{2a}=\frac{0-(10)^2}{2(-7.2)}=6.9 m$

And so, the vertical heigth gained by the block is

$h=d sin \theta = (6.9)(sin 35^{\circ})=4.0 m$

b) 7.4 m/s

The equation of motion along the direction parallel to the slope in this case is

$-\mu_k N +mgsin \theta = ma$

where this time we have taken down the slope as positive direction, so the component of the weight is positive while the frictional force is negative since the block slides downward (while friction acts upward). Solving for a, we find the new acceleration:

$a=-\mu_k g cos \theta + g sin \theta = -(0.20)(9.8)(cos 35^{\circ})+(9.8)(sin 35^{\circ})=4.0 m/s^2$

Now we can use again the SUVAT equation

$v^2-u^2=2ad$

where

v is the final velocity

u = 0 is the initial velocity

d = 6.9 m is the distance travelled along the slope

a = 4.0 m/s^2 is the acceleration

Solving for v,

$v=\sqrt{u^2+2ad}=\sqrt{0+2(4.0)(6.9)}=7.4 m/s$

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3 years ago

A lorry with a mass of 5000kg accelerates to overtake a caravan on the motorway. Its acceleration is 1.5m/s². What force must th

AlladinOne [14]

Force = mass * acceleration

Force = 5000 * 1.5

Force = 7500 N

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3 years ago