The boy on the tower throws a ball 20 meters downrange as shown.What is his pitching speed? (Take g = 9.8 m/s²)
Show work clearly and completely (including formulas, calculations, units, etc. as done in class.)

Answers

Answer 1

Answer:

The problem states that the ball is thrown 20 meters downrange. Since the ball is thrown horizontally, we can assume that the initial vertical velocity is zero. The only force acting on the ball is gravity, which causes it to accelerate downwards at a rate of 9.8 m/s² 1.

To calculate the pitching speed, we need to find the time it takes for the ball to travel 20 meters horizontally. Unfortunately, this information is not provided in the problem statement. Therefore, we cannot calculate the pitching speed.

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Find the percentage of the total work lost to friction if 28.7 J of work is put into pushing a block up a ramp resulting in 14.2 J of stored potential energy at the top.
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Heat transfer occurs

Answers

It depends on the tempature of something or an object,
The Object has to go from hot to cold, that is a primary example of how heat transfer occurs

To produce a second overtone when strumming a guitar, you will want to _____________.

Answers

To produce the second overtone when strumming a guitar, you will want to pluck the closest string that is pretty much close to the top of the guitar. The answer would be plucking the closest string from the top of the guitar.

Light is traveling through the different media shown. In which medium does light travel fastest?

Answers

Light travelling in a vacuum is the fastest thing in the universe. The speed would be 2.99x10^8 m/s. The answer to this question is 'vacuum', where light can travel the fastest. I hope this helps you. You're welcome!

Answer:

Aluminium or Iron or any metal really

Explanation:

I did Gradpoint :)

Finding Impulse when u hav force and time

Answers

Simply multiply them. Impulse is defined as (force) x (duration).
Of course, you have to be careful with the units.

If a compasses moved close to a magnet, the compass will always point?towards the earths north pole
away from the earths south pole
along the magnets field lines
none of the above

Answers

The answer is C. Along the magnetic field lines

Carlos gets tired of pushing and instead begins to pull with force Fpull at an angle to the horizontal.The block slides along the rough horizontal surface at a constant speed. A free-body diagram for the
situation is shown below. Blake makes the following claim about the free-body diagram:
Blake: “The velocity of the block is constant, so the net force exerted on the block must be zero.
Thus, the normal force FN equals the weight Fmg, and the force of friction Ff equals the applied
force Fpull.”
What, if anything, is wrong with this statement? If something is
wrong, identify it and explain how to correct it. If this statement is
correct, explain why.

Answers

Answer:

The wrong items are;

1) The normal for FN equals the weight Fmg

2) The force of friction, Ff, equals the applied force Fpull

The corrected statements are;

1) The normal force is weight less the vertical component of the applied force Fpull

FN = Fmg - Fpull × sin(θ)

2) The force of friction equals the horizontal component of the applied force Fpull

Ff = Fpull × cos(θ)

Explanation:

The given statement was;

The velocity of the block is constant, so the net force exerted on the block must be zero. Thus, the normal force FN equals the weight Fmg, and the force of friction Ff equals the applied force Fpull

By the equilibrium of forces actin on the system, given that the applied force acts at an angle, θ, with the horizontal, we have;

The normal force is equal to the weight less the vertical component of the applied force;

That is we have, FN = Fmg - Fpull × sin(θ)

The friction force similarly, is equal to the horizontal component of the applied force;

Ff = Fpull × cos(θ)

The wrong items are therefore as follows;

1) The normal for FN equals the weight Fmg

1 i) The normal force is weight less the vertical component of the applied force Fpull

FN = Fmg - Fpull × sin(θ)

2) The force of friction, Ff, equals the applied force Fpull

2 i) The force of friction equals the horizontal component of the applied force Fpull

Ff = Fpull × cos(θ).

Final answer:

While Blake's statement about the normal force is correct, his claim about the applied force and friction force is partially accurate. In reality,the horizontal component of the applied force should equate to the friction force for the block to maintain a constant velocity.

Explanation:

Blake's claim that the normal force FN equals the weight Fmg is correct as these forces balance each other in the vertical direction. However, his claim that the force of friction Ff equals the applied force Fpull is only partially accurate. In reality, the horizontal component of Fpull (i.e., Fpull * cos(θ)) should equate to the friction force Ff, to maintain the constant velocity (the block is not accelerating). The vertical component of Fpull (i.e., Fpull * sin(θ)) reduces the effective weight of the block and thereby, the normal force.

To correct Blake's claim, the normal force FN is equal to the weight of the block minus the vertical component of the applied force, and the applied force's horizontal component equals the friction force. Hence, this is the correct solution considering both vertical and horizontal components of forces.