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nataly862011 [7]
3 years ago
5

A gas expands against a constant external pressure of 2.00 atm until its volume has increased from 6.00 to 10.00 L. During this

process, it absorbs 500 J of heat from the surroundings.
Required:
a. Calculate the energy change of the gas, DU.
b. Calculate the work, w, done on the gas in an irreversible adiabatic (q =0) process connecting the same initial and final.
Physics
1 answer:
mars1129 [50]3 years ago
7 0

Answer:

ΔU = - 310.6 J (negative sign indicates decrease in internal energy)

W = 810.6 J

Explanation:

a.

Using first law of thermodynamics:

Q = ΔU + W

where,

Q = Heat Absorbed = 500 J

ΔU = Change in Internal Energy of Gas = ?

W = Work Done = PΔV =

P = Pressure = 2 atm = 202650 Pa

ΔV = Change in Volume = 10 L - 6 L = 4 L = 0.004 m³

Therefore,

Q = ΔU + PΔV

500 J = ΔU + (202650 Pa)(0.004 m³)

ΔU = 500 J - 810.6 J

<u>ΔU = - 310.6 J (negative sign indicates decrease in internal energy)</u>

<u></u>

b.

The work done can be simply calculated as:

W = PΔV

W = (202650 Pa)(0.004 m³)

<u>W = 810.6 J</u>

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horsena [70]

Answer:

4.36 seconds

Explanation:

According to the question;

  • Force is 550 N
  • Mass of the car is 1200 kg
  • Velocity of the car is 2.0 m/s

We are needed to find the time the car must the tow track pull the car.

  • From Newton's second law of motion;
  • Impulsive force, F = Mv÷t , where m is the mass, v is the velocity and t is the time.

Rearranging the formula;

t = mv ÷ F

Thus;

Time = (1200 kg × 2.0 m/s²) ÷ 550 N

        = 4.36 seconds

Thus, the time needed to pull the car is 4.36 seconds

5 0
3 years ago
An object experiences a net acceleration to the left. Which of the following statements about this object are true? There may be
dedylja [7]

Answer:

  1. When an object experiences acceleration to the left, the net force acting on this object will also be to the left.
  2. If the mass of the object was doubled, it would experience an acceleration of half the magnitude

Explanation:

When an object experiences acceleration to the left, the net force acting on this object will also be to the left.

From Newton's second law of motion, the acceleration of the object is given as;

a = ∑F / m

a = -F / m

The negative value of "a" indicates acceleration to the left

where;

∑F is the net force on the object

m is the mass of the object

At a constant force, F = ma ⇒ m₁a₁ = m₂a₂

If the mass of the object was doubled, m₂ = 2m₁

a₂ = (m₁a₁) / (m₂)

a₂ = (m₁a₁) / (2m₁)

a₂ = ¹/₂(a₁)

Therefore, the following can be deduced from the acceleration of this object;

  1. When an object experiences acceleration to the left, the net force acting on this object will also be to the left.
  2. If the mass of the object was doubled, it would experience an acceleration of half the magnitude

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3 years ago
When plugging in metric facts, always remember that 1 big unit = # small units. Fill in these facts: 1.________ s= ________μs 2.
balandron [24]

Answer:

1. 1 s = 1 x 10⁶ μs

2. 1 g = 0.001 kg

3. 1 km = 1000 m

4. 1 mm = 1 x 10⁻³ m

5. 1 mL = 1 x 10⁻³ L  

6. 1 g = 100 dg

7. 1 cm = 1 x 10⁻² m

8. 1 ms = 1 x 10⁻³ s

Explanation:

1.

1 x 10⁻⁶ s = 1 μs

(1 x 10⁻⁶ x 10⁶) s = 1 x 10⁶ μs

<u>1 s = 1 x 10⁶ μs</u>

2.

1000 g = 1 kg

1 g = 1/1000 kg

<u>1 g = 0.001 kg</u>

3.

<u>1 km = 1000 m</u>

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4.

<u>1 mm = 1 x 10⁻³ m</u>

<u></u>

5.

<u>1 mL = 1 x 10⁻³ L</u>

<u></u>

6.

1 x 10⁻² g = 1 dg

(1 x 10⁻² x 10²) g = 1 x 10² dg

<u>1 g = 100 dg</u>

<u></u>

7.

<u>1 cm = 1 x 10⁻² m</u>

<u></u>

8.

<u>1 ms = 1 x 10⁻³ s</u>

4 0
3 years ago
Jeffery gains super strength and pushes two different objects with the same amount of force. Object A accelerates at 40 m/s2, an
Montano1993 [528]

Answer: Object B

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5 0
4 years ago
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DiKsa [7]

Answer:

6480 km

Explanation:

The speed of the object is

v = 7500 cm/sec

We need to convert centimetres into kilometers and seconds into days. We have:

1 cm = 1\cdot 10^5 km

1 s = \frac{1}{60\cdot 60 \cdot 24}d

Using these conversion factors, we find:

v=7500 \frac{cm}{s} \cdot 1\cdot 10^{-5} \frac{km}{cm}\cdot (24)(60)(60) \frac{s}{d}=6480 km

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