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IceJOKER [234]
3 years ago
13

A.Substitute in the units for each one and combine like terms.

Physics
1 answer:
storchak [24]3 years ago
8 0
IDK     ghjfnhgfjmrmhjgfhgfmmfh
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In physics class, Carrie learns that a force, F, is equal to the mass of an object, m, times its acceleration, a. She writes the
Phantasy [73]

Answer:

2.2 m/s^2

Explanation:

Acceleration = Force / Mass

= 7.92 / 3.6 = 2.2m/s^2

Hope this help you :3

5 0
3 years ago
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What force is required to push a block (mass m) up an inclined plane that makes an angle of θ with the horizon at a constant vel
Verizon [17]

Answer:

b. mg ( μ · cos θ + sin θ)

Explanation:

Hi there!

Please, see the attached figure for a graphical description of the problem.

We have the following parallel forces acting on the block (in parallel direction to the direction of movement):

F = applied force.

Fr = friction force.

wx = parallel component of the weight.

According to Newton´s second law:

∑F = m · a

Where "m" is the mass of the block and "a" its acceleration.

Then:

F - Fr - wx = m · a

Since the block is to be pushed at a constant velocity, the acceleration is zero. Then:

F - Fr - wx = 0

F = Fr + wx

The applied force has to be equal to the friction force plus the parallel component of the weight to push the block at constant velocity.

The friction force is calculated as follows:

Fr = μ · N

Where N is the normal force and μ is the coefficient of friction.

Notice that the normal force is of the same magnitude as the perpendicular component of the weight, wy.

Let´s apply Newton´s second law in the perpendicular direction to show this:

∑F = m · a

N - wy = m · a

The acceleration of the block in the perpendicular direction is zero. Then:

N - wy = 0

N = wy

And wy can be obtained by trigonometry (see figure):

wy = W · cos θ

N = wy = mg ·  cos θ

The parallel component of the weight is calculated using trigonometry (see figure):

wx = W · sin θ

wx = mg · sin θ

Then the applied force will be:

F = Fr + wx

F = μ · N + mg · sin θ      (N = wy = mg ·  cos θ)

F = μ · mg ·  cos θ  + mg · sin θ

F = mg ( μ · cos θ + sin θ)

The correct answer is the b.

8 0
3 years ago
Two small objects, with masses m and m, are originally a distance r apart, and the gravitational force on each one has magnitude
elena-s [515]
32f. That's because the force is directly proportional to the product of the masses and inversely proportional to the square of the distance. So you get 2•(1/1/4)^2=2•16=32
6 0
3 years ago
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List five rules of golf etiquette. Explain the correct stance, grip, orientation, and parts of a swing.
Verizon [17]
- keep you eyes on the ball, after you hit the ball you look up, you have to see the club hit the ball

- your grip it really matters on what your comfortable with but first you put your left hand on the club then put your right hand below your left hand, then put your right pinky on top of your index finger, but you can inter lock your pinky and index finger

- DON'T HIT AT THE BALL HIT THREW THE BALL

- parts of the swing, can be pretty tricky to explain, but I'll try my best to explain it, okay so a golf swing is like a pendulum, you swing right to left. Once you lift the club to the right you still look down at the ball. After swimming to the right your have to hit the ball with your right hand strength, in order the ball to fly.

- the way you do your grip defines if the ball will go straight or not. But also the way you swing towards the ball also defines how far the ball is going to go.
8 0
3 years ago
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"explain why the orbital radius and the speed of a satellite in circular orbit are not independent"
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In order to achieve stable circular gravitational and centrifugal force must be in balance.
\frac{mv^2}{r}=G\frac{mM}{r^2} \\ v^2=G\frac{M}{r}\\ v=\sqrt{G\frac{M}{r}
This relationship tells closer the closser your orbit is to the surface of the planet the faster you have to go. Which makes sense because the closer you are to the planet the stronger gravitational force gets.

5 0
4 years ago
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