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matrenka [14]
2 years ago
9

Please help me! As quickly as possible

Physics
1 answer:
k0ka [10]2 years ago
7 0

Answer:

1. matter

2. kilograms

3. same

4. gravitational

5. gravity

6. space

7. weightlessness

8. Newton

9. weight

10. more

I HOPE THESE ARE CORRECT AND IT HELPS

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You have gas in a container with a movable piston. The walls of the container are thin enough so that its temperature stays the
bearhunter [10]

Answer:

New pressure of the gas increases by 26.5% with respect to initial pressure, new volume decreases 27% with respect to initial volume and new temperature decreases 8% with respect to initial volume.

Explanation:

If we assume the gas is a perfect gas we can use the perfect gas equation:

PV=nRT

  • For Isothermal process:

\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}(1)

Where subscripts 1 shows before the isothermal process and 2 after it, because isothermal means constant temperature T1=T2, and pressure increases by 10% means P2=1,1*P1, using these facts on (1) we have:

V_{2}=\frac{V_{1}}{1.1} (2)

  • For Isobaric process:

\frac{P_{2}V_{2}}{T_{2}}=\frac{P_{3}V_{3}}{T_{3}} (3)

Where subscripts 2 shows before the isobaric process and 3 after it, because isobaric means constant pressure P2=P3, and volume decreases by 20% means V3=0.8*V2, using these facts on (3) we have:

T_{3}=0.8T_{2} (4)

  • For Isochoric process:

\frac{P_{3}V_{3}}{T_{3}}=\frac{P_{4}V_{4}}{T_{4}} (5)

Where subscripts 3 shows before the isochoric process and 4 after it, because isochoric means constant volume V3=V4, and temperature increases by 15% means T4=1.15*T3, using these facts on (5) we have:

P_{4}=1.15P_{3} (6)

So now because P4=1.15*P3, P2=P3 and P2=1.1*P1:

P_{4}=1.15*1.1P_{1}=1.265P1

This is, the new pressure of the gas increases by 26.5%  with respect to initial pressure.

Similarly, we have V3=V4, V3=0.8*V2 and V1=1,1*V2:

V_{4}=\frac{0.8}{1.1}V_{1}=0.72V1

so the final volume decreases 27% with respect to initial volume.

T4=1,15*T3, T3=0.8*T2 and T1=T2:

T_{4}=1.15*0.8T_{1}=0.92T1

The new temperature decreases 8% with respect to initial volume.

3 0
3 years ago
If a ball is given a push so that it has an initial velocity of 5 m/s down a certain inclined plane, then the distance it has ro
iren [92.7K]

Answer:

v= 13 m/s

Explanation:

Velocity is defined as the derivative of displacement with respect to time

v= ds/dt

Known data

s(t) = 5t + 2t²  : distance that the ball has rolled after t seconds

vi= 5 m/s : initial velocity

t= 2 s

Problem develoment

s(t) = 5t + 2t²

v= ds/dt= 5 + 4t : velocity of the ball in function of the time

We replace t =2 s in the equation of velocity

v= 5 + 4(2)

v= 13 m/s : velocity after 2 seconds

4 0
3 years ago
Question 16 of 20 You plan to use a slingshot to launch a ball that has a mass of 0.023 kg. You want the ball to accelerate stra
liubo4ka [24]

Answer:

this is a simple application of Newton's 2nd Law: F = ma.

F = 0.023(25)

So,

F =0.575 N.

Therefore

The answer is A.

If rounded up/off.

Explanation:

HOPE IT HELPS.

PLEASE MARK AS BRAINLIEST.

4 0
2 years ago
4. Look at the graph above. What information is on the graph and what does it mean? Please help!!!
Aliun [14]

Answer:

CO2 increases in the winter

Explanation:

6 0
3 years ago
6.If a 250. gram cart moving to the right with a velocity of .31 m/s collides inelastically with a 500. gram cart traveling to t
spin [16.1K]

Answer:

The total momentum of the system before the collision is 0.0325 kg-m/s due east direction.    

Explanation:

Given that,

Mass of the cart, m = 250 g = 0.25 kg

Initial velocity of the cart, u = 0.31 m/s (due right)

Mass of another cart, m' = 500 g = 0.5 kg

Initial velocity of the another cart u' = -0.22 m/s (due left)

Let p is the total momentum of the system before the collision. It is given by :

p=mu+m'u'\\\\p=0.25\times 0.31+0.5\times (-0.22)\\\\p=-0.0325\ kg-m/s

So, the total momentum of the system before the collision is 0.0325 kg-m/s due east direction.            

5 0
3 years ago
Read 2 more answers
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