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sasho [114]
2 years ago
10

The pressure on a balloon holding 356 mL of an ideal gas is increased from 267 torr to 1.00 atm. What is the new volume of the b

alloon (in mL) at constant temperature
Physics
1 answer:
Natalija [7]2 years ago
4 0

Answer:

124.52 mL

Explanation:

from Boyle's Law,

PV = P'V' ................... Equation 1

Where P = Initial pressure of the gas, V = Initial volume of the gas, P' = Final pressure of the gas, V' = Final volume of the gas.

make V' the subject of the equation.

V' = PV/P'............. Equation 2

Given: P = 267 torr = (267×0.00131) = 0.34977 atm, V = 356 mL, P' = 1 atm

Substitute into equation 2

V' = (0.34977×356)/1

V' = 124.52 mL.

Hence the new volume of the balloon = 124.52 mL

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Montano1993 [528]

The motion of planets is separate to the motion of stars. Like everything in the sky, they rise in the east, and set in the west, because of the earth's rotation. But night by night, their position at a given time changes because of their orbit around the sun.

7 0
2 years ago
A 1-kg rock is suspended by a massless string from one end of a
maxonik [38]

Answer:

The weight of measuring stick is 9.8 N

Explanation:

given information:

the mass of the rock, m_{r} = 1 kg

measuring stick, x =1 m

d = 0.25 m

to find the weight of measuring stick, we can use the following equation:

τ = Fd

τ = 0

F_{r} d - F_{s}d = 0

F_{r} = the force of the rock

F_{s} = the force of measuring stick

F_{s} =F_{r}

    = m g

    = 1 kg x 9.8 m/s

    = 9.8 N

thus, the weight of measuring stick is 9.8 N

6 0
3 years ago
A car travelling at 15m/s comes to a rest in a distance of 14m when the brakes are applied. Calculate the deceleration of the ca
vlada-n [284]

Answer:

force of the breaks is 6650 N, direction opposite to direction of movement

Explanation:

6 0
3 years ago
Read 2 more answers
Please someone help, I’m very confused and it’s due soon, thanks
Anit [1.1K]

Answer:

  1. 1 s
  2. 19.6 m
  3. 2 s
  4. 0.8 m/s^2
  5. 28 m/s
  6. 79 m/s
  7. 0.37 s
  8. 26 m/s
  9. 242 m/s
  10. 19,930 m

Explanation:

In physics, many of the relationships between speed, distance, and acceleration are tied up in the equations for potential and kinetic energy. For an object of mass M* at height h in a gravity field with acceleration g, the potential energy is

  PE = Mgh

At velocity v, the kinetic energy of the object is ...

  KE = 1/2Mv^2

When an object is dropped or launched from rest, the height and velocity are related by the fact that kinetic energy gets translated to potential energy, or vice versa. This gives rise to ...

  PE = KE

  Mgh = (1/2)Mv^2

The mass (M) can be factored out of this, so we have ...

  2gh = v^2

This can be solved for height:

  h = v^2/(2g) . . . . [eq1]

or for velocity:

  v = √(2gh) . . . . [eq2]

__

When acceleration is constant, as assumed here, the velocity changes linearly (to/from 0). So, over the time of travel, the average velocity is half the final velocity. That is,

  t = 2h/v

Depending on whether you start with h or with v, this resolves to two more equations:

  t = 2(v^2/(2g))/v = v/g . . . . [eq3]

  t = 2h/(√(2gh)) = √(4h^2/(2gh)) = √(2h/g) . . . . [eq4]

The last of these can be rearranged to give distance as a function of time:

  h = gt^2/2 . . . . [eq5]

or acceleration as a function of time and distance:

  g = 2h/t^2 . . . . [eq6]

__

These 6 equations can be used to solve the problems posed. Just "plug and chug." For problems in Earth's gravity, we use g=9.8 m/s^2. (You may want to keep these equations handy. Be aware of the assumptions they make.)

_____

* M is used for mass in these equations so as not to get confused with m, which is used for meters.

_____

1) Use [eq4]: t = √(2·6 m/(9.8 m/s^2)) ≈ 1.107 s ≈ 1 s

__

2) Use [eq5]: h = (9.8 m/s^2)(2 s)^2/2 = 19.6 m

__

3) Use [eq4]: t = √(25 m/(4.9 m/s^2)) ≈ 2.259 s ≈ 2 s

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4) Use [eq6]: g = 2(10 m)/(5 s)^2 = 0.8 m/s^2

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5) Use [eq2]: v = √(2·9.8 m/s^2·40 m) = 28 m/s

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6) Use [eq2]: v = √(2·9.8 m/s^2·321 m) ≈ 79.32 m/s ≈ 79 m/s

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7) Using equation [eq3], we will find the time until Tina reaches her maximum height. Her actual off-the-ground total time is double this value. Using [eq3]: t = v/g = (1.8 m/s)/(9.8 m/s^2) = 9/49 s. Tina is in the air for double this time:

  2(9/49 s) ≈ 0.37 s

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8) Use [eq2]: v = √(2·9.8 m/s^2·33.5 m) ≈ 25.624 m/s ≈ 26 m/s

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9) Use [eq2]: v = √(2·9.8·3000) m/s ≈ 242.49 m/s ≈ 242 m/s

(Note: the terminal velocity in air is a lot lower than this for an object like a house.)

__

10) Use [eq1]: h = (625 m/s)^2/(2·9.8 m/s^2) ≈ 19,930 m

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<em>Additional comment</em>

Since all these questions make use of the same equation development, I have elected to answer them. Your questions are more likely to be answered if you restrict your posts to 3 or fewer questions each.

5 0
3 years ago
special metal hangers or stirrips called joist hangers are used when joist must be _____ the bottom of the grider or beam
djyliett [7]

Answer:

Special metal hangers or stirrups called joist hangers are used when joist must be flush with the bottom of the girder or beam.

Explanation:

  • A joist hanger also known as a beam hanger is a mechanical device which is used to fasten joists and rafters.
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Thus, special metal hangers or stirrups called joist hangers are used when joist must be flush with the bottom of the girder or beam.

Learn more about construction here:

brainly.com/question/14428327

#SPJ4

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