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Ivenika [448]
3 years ago
12

At 10°C, the gas in a cylinder has a volume of 0.250 L. The gas is allowed to expand to 0.285 L. What must the final temperature

be for the pressure to remain constant? (Hint °C + 273 = K.)
Physics
2 answers:
xxTIMURxx [149]3 years ago
7 0

Use the formula: <span><span><span><span>P1</span><span>V1</span></span><span>T1</span></span>=<span><span><span>P2</span><span>V2</span></span><span>T2</span></span></span>

We know that we want <span><span>P1</span>=<span>P2</span></span> so the pressure will remain constant, so we can say that: <span><span><span><span>P1</span><span>V1</span></span><span>T1</span></span>=<span><span><span>P1</span><span>V2</span></span><span>T2</span></span></span>

Plug in the values we know: <span><span><span><span>P1</span>⋅0.250L</span><span>283K</span></span>=<span><span><span>P1</span>⋅0.285L</span><span>T2</span></span></span>
(Remember that temperature must be done in Kelvin.)

Cross multiply: <span><span>T2</span>⋅<span>P1</span>⋅0.250L=283K⋅<span>P1</span>⋅0.285L</span>

Divide both sides by <span>P1</span>.

<span><span>T2</span>⋅0.250L=283K⋅0.285L</span>

<span><span>T2</span>=<span><span>283K⋅0.285L</span><span>0.250L</span></span></span>

<span><span>T2</span>=323K</span>

(This is also <span>50˚C</span>.)


Annette [7]3 years ago
3 0

323k      is the answer for sure

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time = 10s
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3 years ago
wo lacrosse players collide in midair. Jeremy has a mass of 120 kg and is moving at a speed of 3 m/s. Hans has a mass of 140 kg
Julli [10]

2.71 m/s fast Hans is moving after the collision.

<u>Explanation</u>:

Given that,

Mass of Jeremy is 120 kg (M_J)

Speed of Jeremy is 3 m/s (V_J)

Speed of Jeremy after collision is (V_{JA}) -2.5 m/s

Mass of Hans is 140 kg (M_H)

Speed of Hans is -2 m/s (V_H)

Speed of Hans after collision is (V_{HA})

Linear momentum is defined as “mass time’s speed of the vehicle”. Linear momentum before the collision of Jeremy and Hans is  

= =\mathrm{M}_{1} \times \mathrm{V}_{\mathrm{J}}+\mathrm{M}_{\mathrm{H}} \times \mathrm{V}_{\mathrm{H}}

Substitute the given values,

= 120 × 3 + 140 × (-2)

= 360 + (-280)

= 80 kg m/s

Linear momentum after the collision of Jeremy and Hans is  

= =\mathrm{M}_{\mathrm{J}} \times \mathrm{V}_{\mathrm{JA}}+\mathrm{M}_{\mathrm{H}} \times \mathrm{V}_{\mathrm{HA}}

= 120 × (-2.5) + 140 × V_{HA}

= -300 + 140 × V_{HA}

We know that conservation of liner momentum,

Linear momentum before the collision = Linear momentum after the collision

80 = -300 + 140 × V_{HA}

80 + 300 = 140 × V_{HA}

380 = 140 × V_{HA}

380/140= V_{HA}

V_{HA} = 2.71 m/s

2.71 m/s fast Hans is moving after the collision.

4 0
3 years ago
Which term below describes a measurement of how hard an object pushes against a surface?
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I hope the answer is D. Pressure

8 0
3 years ago
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An object is thrown upward from the edge of a tall building with a velocity of 10 m/s. Where will the object be 3 s after it is
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We use a fundamental kinematic equation as follows:

V = Vo + g*t. 
<span>Tr = (V-Vo)/g = (0-10)/-10 = 1 s. = </span><span>time to reach max. height </span>

<span>Tf = Tr = 1 s. = Fall time or time to fall back to edge of bldg. </span>

<span>3-Tr-Tf = 3-1-1 = 1 s. Below edge of bldg. </span>

<span>d = Vo*t + 0.5g*t^2. </span>
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3 years ago
Two wheel gears are connected by a chain. The larger gear has a radius of 8 centimeters and the smaller gear has a radius of 3 c
dezoksy [38]

Answer:

B.The linear velocity of the gears is the same. The linear velocity is 432π centimeters per minute.

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As we know that the small gear completes 24 revolutions in 20 seconds

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\omega = 2.4\pi rad/s

Now we know that the tangential speed of the chain is given as

v = r \omega

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v = (3 cm)(2.4\pi)

v = 7.2 \pi cm/s

v = 432\pi cm/min

Since both gears are connected by same chain so both have same linear speed and hence correct answer will be

B.The linear velocity of the gears is the same. The linear velocity is 432π centimeters per minute.

8 0
4 years ago
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