PHYSICS MIDDLE SCHOOL

A force of 500 N is exerted on a baseball by the bat for 0.001 s. What is the change in momentum of the baseball?

Answers

Answer 1

Answer:

The change in momentum of this baseball is equal to 0.5 Ns.

Given the following data:

Force = 500 Newton.

Time = 0.001 seconds.

To calculate the change in momentum of this baseball:

What is momentum?

Momentum is the product of the mass possessed by an object such as a baseball and its velocity.

In Physics, the change in momentum of an object is always equal to the impulse experienced by the object due to the force acting on an object.

Mathematically, change in momentum is given by this formula:

$Impulse = Change\;in\;momentum\n\nChange\;in\;momentum=m \Delta V=Force * time$

Substituting the given parameters into the formula, we have;

$Change\;in\;momentum=500 * 0.001$

Change inmomentum = 0.5 Ns.

Read more on momentum here: brainly.com/question/13822610

Answer 2

Answer: Answer: Δp = F*Δt = 500N*0.001s = 0.5Ns

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Wispy, feathery clouds made mostly of ice crystals that form at high altitudes, usually above 6 kilometers.

Answers

Cirrus clouds are the clouds mostly made of ice becasue of the high altitudes

5) Consider pushing a 50.0 kg box through a 5.00 m displacement on both a flat surface and up aramp inclined to the horizontal by 15.0°. In both cases, you apply a force of 100.N parallel to the
surface (parallel to the floor or parallel to the ramp). Calculate the work done by:
a) the gravitational force as the box is pushed across the flat ground
b) the gravitational force as the box is pushed up the ramp
c) the force you apply as the box is pushed across the flat ground
d) the force you apply as the box is pushed up the ramp

Answers

a) The work done by the gravitational force on the flat surface is zero

b) The work done by the gravitational force on the ramp is -634 J

c) The work done by the applied force on the flat surface is 500 J

d) The work done by the applied force on up along the ramp is 500 J

Explanation:

The work done by a force is given by the equation

$W=Fdcos \theta$

where

F is the magnitude of the force

d is the dispalcement of the object

$\theta$ is the angle between the direction of the force and of the displacement

In this problem, we want to calculate the work done by the gravitational force as the box is pushed across the flat ground.

We immediately notice that the gravitational force acts downward, while the displacement is horizontal: therefore, the angle between force and displacement is $90^(\circ)$ ; this means that $cos 90^(\circ)=0$ , and therefore, the work done is zero:

$W=0$

In this case, the box is pushed along the ramp. We have:

$F=mg=(50.0)(9.8)=490 N$ is the magnitude of the force of gravity, where

m = 50.0 kg is the mass of the box

$g=9.8 m/s^2$ is the acceleration of gravity

d = 5.00 m is the displacement of the box along the ramp

The ramp is inclined to the horizontal by $15.0^(\circ)$ , therefore the angle between the force of gravity and the displacement of the box (moving up along the ramp) is:

$\theta=90^(\circ)+15^(\circ)=105^(\circ)$

Therefore, the work done by gravity in this case is:

$W=(490)(5.00)(cos 105^(\circ))=-634 J$

In this case, we want to calculate the work done by the force you apply as the box is pushed across the flat ground.

Here we have:

F = 100.0 N (force applied)

d = 5.00 m (displacement of the box)

$\theta=0^(\circ)$ (the force is applied parallel to the flat surface, therefore force and displacement have same direction)

Therefore, the work done by the force you apply on the flat ground is:

$W=(100.0)(5.00)(cos 0^(\circ))=500 J$

In this last case, we want to calculate the work done by the force you apply as the box is pushed up along the ramp.

This time we have:

F = 100.0 N (force applied is the same)

d = 5.00 m (displacement of the box is also the same)

$\theta=0^(\circ)$ (the force is applied parallel to the ramp, therefore force and displacement have again same direction)

Therefore, the work done by the force you apply while pushing the box along the ramp is:

$W=(100.0)(5.00)(cos 0^(\circ))=500 J$

Learn more about work:

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Why is it important that the mouth and stomach are near the start of the digestive system? thx

Answers

Answer: To complete the process of digestion after the food is ingested by mouth.

Explanation:

The food that we eat is ingested by mouth and here the mechanical degradation of food takes place. This breaks down the food into smaller pieces.

Mouth also contains the enzyme that is responsible for the digestion of starch.

Then the food from the mouth reaches the stomach where various types of digestive juices and enzymes are secreted to digested the food.

Hence, mouth and stomach must be near in the digestive system.

Because the mouth breaks down the food and the stomach uses acid to break food down even more

Which bundles of cells in the lungs that take in oxygen

Answers

Hello,

The bundle of cells in the lungs that take in oxygen is "Red blood cells".

Have A Good Day.

Which circuits are parallel circuits?

Answers

two or more electrical devices in a circuit can be connected by series connections or by parallel connections. When all the devices are connected using parallel connections, the circuit is referred to as a parallel circuit.

They're the circuits in which each end of one component is connected to one end of the other component.

If you have a picture of some circuits to pick the parallel ones from, it's a mystery to me why you haven't posted the picture too.

A car is strapped to a rocket (combined mass = 661 kg), and its kinetic energy is 66,120 J.At this time, the rocket runs out of fuel and turns off, and the car deploys a parachute to slow down, and the parachute performs 36,733 J of work on the car.

What is the final speed of the car after this work is performed?

Answers

Answer:

9.43 m/s

Explanation:

The law of conservation of energy requires that the total energy before and after transformation must be equal. Kinetic energy of car and rocket combined equals the parachute energy and that of car.

Final energy of car will be 66120-36733=29,387 J

Kinetic energy, $KE=0.5mv^(2)$ and making v the subject of the formula then $v=\sqrt{\frac {2KE}{m}}$

Where m is the mass and v is velocity.

Substituting 29387 J for KE and 661 kg for m then

$v=\sqrt{\frac {2* 29387}{661}}=9.42957012478699\approx 9.43\ m/s$

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