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sweet-ann [11.9K]
4 years ago
9

Without completing the calculations, determine what the new volume will be in the problem below. Also, explain how you were able

to determine the new volume without completing the calculations. Pay special attention to how the pressure is changing. An 80.0-mL sample of carbon monoxide gas (CO) is stored at a pressure of x kPa. The pressure is doubled to 2x kPa. What is the new volume?
Physics
1 answer:
Talja [164]4 years ago
7 0
Boyles law

Pressure and volume are inversely proportional as the new variable changes from the known.

Double the pressure equals 1/2 of original volume, assuming temperature remains the same.

So 40.0 mL is the new volume as it is compressed.
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If a snowball melts so that its surface area decreases at a rate of 10 cm2/min, find the rate at which the diameter decreases wh
liubo4ka [24]

Answer: 4/9π

Explanation:

Surfacearea of snowball(SA) =4πr^2

dSA/dT = - 10cm^2/min

d = 9, hence, r = 9/2 = 4.5

Taking the derivative of surface area with respect to time:

dSA/dT = 8πrdr/dt

dSA/dT = - 10, r = 4.5

-10 = 8π(4.5)dr/dt

-10 = 36πdr/dt

Make dr/dt subject of the formula

dr/dt = - 10/36π

d = 2r

Differentiating d with respect to time, 't'

dd/dt = 2dr/dt

Therefore, 2dr/dt = 2*(dr/dt)

2×(- 10/36π) = - 20/36π

= -4/9π

Therefore, rate by which diameter decreases when diameter is 9cm is

4/9 π

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3 years ago
True or False: Heavy elements such as Carbon and Nitrogen, that are necessary for life were created in the Big
Orlov [11]

Answer:

i believe the answer is true

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4 years ago
A horizontal spring with spring constant 210 Nm is compressed by 20 cm and then used to launch a 250 g box across the floor. The
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Answer:

ugmd = 1/2 kx²

d = (1/2 kx²) / (ugm)

= (1/2 * 250 N/m * (0.2 m)²) / (0.23 * 9.81 m/s² * 0.3 kg)

= 7.4 m

ugmd = 1/2 mv²

v = √2ugd

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A thin, circular disk of radius 30.0 cm is oriented in the yz-plane with its center at the origin. The disk carries a total char
dezoksy [38]

Answer:

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Explanation:

given,

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E = \dfrac{Q}{2\epsilon_0 \pi R^2}(1-\dfrac{x}{\sqrt{x^2+R^2}})

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