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kaheart [24]
3 years ago
9

In a game of baseball, a player hits a high fly ball to the outfield. (a) Is there a point during the flight of the ball where i

ts velocity is parallel to its acceleration? (b) Is there a point where the ball’s velocity is perpendicular to its acceleration? Explain in each case.
Physics
1 answer:
Tatiana [17]3 years ago
4 0

Answer:

a) No

b) No

Explanation:

When a bat is hit in a game of baseball such that it flies out of the field that means it is going with some angle to the horizontal.

a)

Then is such a case the velocity of the ball is never parallel to the acceleration because there acts a net acceleration which is resultant of the acceleration due to the applied force and the acceleration due to gravity but a component of the velocity when the ball descends the height acts parallel to the gravity.

b)

At no point during the motion of the ball its velocity is perpendicular to its acceleration because it has an initial angle of projection form the horizontal.

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When a fixed amount of ideal gas goes through an isobaric expansion A) its internal (thermal) energy does not change.B) the gas
Bingel [31]
<h2>Answer: its temperature must increase.</h2>

Explanation:

In an isobaric process the pressure remains constant, which means the initial pressure and the final pressure will be the same.

In addition, during this thermodynamic process, the volume of the ideal gas expands or contracts in such a way that the variation of pressure \Delta P is neutralized.

Now, according to the First law of Thermodynamics that establishes the conservation of energy:

\Delta U=\Delta Q-\Delta W   (1)

Where:

\Delta U is the internal energy

\Delta Q is the heat transferred

\Delta W is the work

Now, for an isobaric process:

\Delta W=P\Delta V    (2)

Where:

P is the pressure (<u>always positive</u>)

\Delta V is the volume variation of the gas

<u />

<u>Here we have two possible results:</u>

-If the gas expands (positive \Delta V), the work is positive.

-If the gas compresses (negative \Delta V), the work is negative.

In this case we are talking about the first result (work is positive).

Then, according to the above, equation (1) can be written as follows:

\Delta U=\Delta Q - P\Delta V   (3)

Clearing \Delta Q:

\Delta Q=\Delta U+P \Delta V    (4)

Then, for an ideal gas in an isobaric process, part of the heat (Q) added to the system will be used to do work (positive in this case) and the other part <u>will increase the internal energy</u>, hence <u>the temperature will increase as well.</u>

7 0
3 years ago
A small bolt with a mass of 33.0 g sits on top of a piston. The piston is undergoing simple harmonic motion in the vertical dire
densk [106]

Answer:

0.027m

Explanation:

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ω² * A = g where ω is angular velocity, A amplitude, g acceleration due to gravity

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A= g/4π²f² depending on the value of g used either 10m/s² or 9.8m/s²,

i used 10m/s² in this answer

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Kinetic energy = mv²/2, potential energy = mgh

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v² = 2gh

v = √2gh

Where g = 9.8 m/s², h = 2.80m

v = √2×9.8×2.8 = 7.41 m/s

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