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3 years ago

What happens when two forces act in the same direction?

Physics

2 answers:

SSSSS [86.1K]3 years ago

7 0

The two forces are added when in hook ground the same direction

Wewaii [24]3 years ago

5 0

Using the picture provided the forces are added together, because they are putting force on an object the same direction, thus the forces are added.

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The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the

tigry1 [53]

Answer:

A) The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) The maximum speed of the plastic sphere will be 2.2 m/s

Explanation:

Hi there!

I´ve found the complete problem on the web:

A toy launcher that is used to launch small plastic spheres horizontally contains a spring with a spring constant of 50. newtons per meter. The spring is compressed a distance of 0.10 meter when the launcher is ready to launch a plastic sphere.

A) Determine the elastic potential energy stored in the spring when the launcher is ready to launch a plastic sphere.

B) The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the plastic sphere will be launched. [Neglect friction.] [Show all work, including the equation and substitution with units.]

A) The elastic potential energy (EPE) is calculated as follows:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = compressing distance

EPE = 1/2 · 50 N/m · (0.10 m)²

EPE = 0.25 J

The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) Since there is no friction, all the stored potential energy will be converted into kinetic energy when the spring is released. The equation of kinetic energy (KE) is the following:

KE = 1/2 · m · v²

Where:

m = mass of the sphere.

v = velocity

The kinetic energy of the sphere will be equal to the initial elastic potential energy:

KE = EPE = 1/2 · m · v²

0.25 J = 1/2 · 0.10 kg · v²

2 · 0.25 J / 0.10 kg = v²

v = 2.2 m/s

The maximum speed of the plastic sphere will be 2.2 m/s

6 0

4 years ago

Maria's mother was deported 6 months ago, and she is staying with a cousin. She has to share a room with 3 other children and is

svetlana [45]

The answer would be A.

8 0

3 years ago

Can someone help me with one through seven I will mark you the brainly

ki77a [65]

Answer:

1. F = M x A

2. Force

3. 2nd Law: Force

4. a, b, c (in order)

5. 3rd Law: Action and Reaction

6. b, c, a (in order)

7. 1st Law: Inertia

3 0

3 years ago

Read 2 more answers

Suppose you are playing baseball and hit a home run. What forces slow down the ba while it is in motion? If you hit a home run i

mart [117]

Answer:

Air resistance slows down the ball

In space, there would be no air resistance

Explanation:

When an object moves through the air, there is a force acting in the opposite direction to the motion of the ball: this force is called air resistance.

Air resistance is due to the friction between the molecules of air and the molecules at the surface of the object - because of this frictional force, the object is slows down in its motion and loses some energy (which is converted into thermal energy of the surrounding air).

There is also the force of gravity (downward) that acts on the ball: however, this force does not slows down the ball in its motion, instead it accelerates it towards the ground.

In space, however, there is no air and no gravity. This means that there are no forces acting on the ball: therefore, the ball will not be slowed down, and therefore will continue its motion forever, at constant velocity, according to Newton's first law:

An object at rest (or in motion) will stay at rest (or in motion at constant velocity) when the net external force acting on it is zero

4 0

3 years ago

A 1100 kg car rounds a curve of radius 68 m banked at an angle of 16 degrees. If the car is traveling at 95 km/h, will a frictio

Mariulka [41]

Answer:

Yes. Towards the center. 8210 N.

Explanation:

Let's first investigate the free-body diagram of the car. The weight of the car has two components: x-direction: towards the center of the curve and y-direction: towards the ground. Note that the ground is not perpendicular to the surface of the Earth is inclined 16 degrees.

In order to find whether the car slides off the road, we should use Newton's Second Law in the direction of x: F = ma.

The net force is equal to $F = \frac{mv^2}{R} = \frac{1100\times (26.3)^2}{68} = 1.1\times 10^4~N$

Note that 95 km/h is equal to 26.3 m/s.

This is the centripetal force and equal to the x-component of the applied force.

$F = mg\sin(16) = 1100(9.8)\sin(16) = 2.97\times10^3$

As can be seen from above, the two forces are not equal to each other. This means that a friction force is needed towards the center of the curve.

The amount of the friction force should be $8.21\times 10^3~N$

Qualitatively, on a banked curve, a car is thrown off the road if it is moving fast. However, if the road has enough friction, then the car stays on the road and move safely. Since the car intends to slide off the road, then the static friction between the tires and the road must be towards the center in order to keep the car in the road.

5 0

4 years ago