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marishachu [46]
3 years ago
10

An air compressor takes normal air and pushes more and more of it in a rigid steel tank. The number of gas particles in the tank

is increased. What happens to the temperature as more and more air is pushed into a tank which does not change size?
A. It increases


B. It stays the same


C. It decreases
Physics
1 answer:
erik [133]3 years ago
3 0

Answer:

C. It decreases

Explanation:

The pressure law states that for a constant volume of gas in a sealed container the temperature of the gas is directly proportional to its pressure. This can be easily understood by visualizing the particles of gas in the container moving with a greater energy when the temperature is increased.

A common example is cooking gas when refilled, there is a perceptible change in the temperature of the cylinder.

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A 2kg bowling ball is 2.5 meter off the ground on a post when it falls. Just before it reaches the ground, it is traveling 7m/s.
shutvik [7]
<span>The mechanical energy is conserved.

I hope this helps, good luck! :)</span>
7 0
3 years ago
Answer the following question​
Ray Of Light [21]

Answer:

A) OA, AB, BC

B) 25m/s^2

C) see explanation

D) 25

E) Rest

Explanation:

From the Velocity time graph shown:

The positive slope = OA ; This is positive because, it is the point of uniform acceleration on the graph.

Constant slope = AB, the slope here is constant because, AB on the graph is the point of constant velocity.

-ve slope = BC

B) Acceleration of body in path OA.

Acceleration = change in Velocity / time

Acceleration = (150 - 0) / 6

Acceleration = 150/6 = 25m/s^2

C) Path AB is Parallel to the because it marks the period of constant velocity (that is Velocity does not increase or decrease during the time interval).

D) Length of BC

BC corresponds to the distance moved, that velocity / time

Velocity = 150 ; time = 6

Therefore Distance (BC) = 150/6 = 25

E.) Velocity =0 ; Hence body is at rest

5 0
3 years ago
A 35.9 g mass is attached to a horizontal spring with a spring constant of 18.4 N/m and released from rest with an amplitude of
lidiya [134]

Answer:

7.74m/s

Explanation:

Mass = 35.9g = 0.0359kg

A = 39.5cm = 0.395m

K = 18.4N/m

At equilibrium position, there's total conservation of energy.

Total energy = kinetic energy + potential energy

Total Energy = K.E + P.E

½KA² = ½mv² + ½kx²

½KA² = ½(mv² + kx²)

KA² = mv² + kx²

Collect like terms

KA² - Kx² = mv²

K(A² - x²) = mv²

V² = k/m (A² - x²)

V = √(K/m (A² - x²) )

note x = ½A

V = √(k/m (A² - (½A)²)

V = √(k/m (A² - A²/4))

Resolve the fraction between A.

V = √(¾. K/m. A² )

V = √(¾ * (18.4/0.0359)*(0.395)²)

V = √(0.75 * 512.53 * 0.156)

V = √(59.966)

V = 7.74m/s

8 0
3 years ago
Read 2 more answers
A common misconception is that an object always moves when a force acts on it. Why is this statement incorrect? Explain the conc
dsp73

Answer:

The statement is incorrect because, a force acting on an object does not necessarily have to produce motion.

People have the misconception that when a force acts on an object it always produces motion

Explanation:

The statement is incorrect because, a force acting on an object does not necessarily have to produce motion. It could be in static equilibrium where the net force is zero and produces not motion. The body could also be in dynamic equilibrium when  no net force acts on it moving at a constant velocity. But here we are concerned with static equilibrium since the body does not move at all.

People have the misconception that when a force acts on an object it always produces motion and, we have seen from the above tat its not always true.

3 0
3 years ago
A positively charged particle is in the center of a parallel-plate capacitor that has charge ±Q on its plates. SUppose the dista
slamgirl [31]

Answer:

Stay the same

Explanation:

First of all, let's find how the capacitance of the capacitor changes.

Initially, it is given by

C=\frac{\epsilon_0 A}{d}

where

\epsilon_0 is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

From the formula, we see that the capacitance is inversely proportional to the separation between the plates. In this problem, the distance between the plates is doubled, so the capacitance will be halved:

C' = \frac{1}{2}C

The potential difference across the capacitor is given by

V= \frac{Q}{C}

where

Q is the charge on the plates

C is the capacitance

We see that the voltage is inversely proportional to the capacitance. We said that the capacitance has halved: therefore, the potential difference across the two plates will double:

V' = 2 V

Now we can analyze the electric field between the plates of the capacitor, which is given by

E=\frac{V}{d}

we said that:

- The voltage has doubled: V' = 2 V

- The distance between the plates has doubled: d' = 2 d

therefore, the new electric field will be

E'=\frac{2V}{2d}=\frac{V}{d}=E

So, the electric field is unchanged. And since the force on the particle at the center is directly proportional to the electric field:

F = qE

Then the force on the particle will stay the same.

4 0
3 years ago
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