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3 years ago

A 50 kg cart has 324 kg•m/s of momentum. How fast is the cart going?

Chemistry

1 answer:

PilotLPTM [1.2K]3 years ago

5 0

Answer: 6.48m/s

Explanation:

From the question given, we obtained the following:

M = 50 kg

Velocity =?

Momentum = 324 kg•m/s

Momentum = Mass x Velocity

Velocity = momentum /Mass

Velocity = 324 / 50

Velocity = 6.48m/s

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What is a certain indicator that a molecule is polar

snow_lady [41]

Anything asymmetrical is polar

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garri49 [273]

Answer:

D.)

Explanation:

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I hope that this helps.

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2 years ago

Question: In experiment your teacher has provided you with a sample of sandy, salty, ocean water and

liq [111]

Answer:

Explanation:

Homogeneous mixture is a mixture in which the components of the mixture are in the same proportion throughout any sample extracted from the mixture while an heterogeneous mixture is a mixture in which the components of the mixture differ in term of proportion when different samples of the mixture are extracted and compared.

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3 years ago

10.0 g Cu, C Cu = 0.385 J/g°C 10.0 g Al, C Al = 0.903 J/g°C 10.0 g ethanol, Methanol = 2.42 J/g°C 10.0 g H2O, CH2O = 4.18 J/g°C

Mazyrski [523]

Answer:

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

Explanation:

$Q=mc\Delte T$

Q = Energy gained or lost by the substance

m = mass of the substance

c = specific heat of the substance

ΔT = change in temperature

1) 10.0 g of copper

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the copper = c = 0.385 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 0.385J/g^oC}=25.97^oC$

2) 10.0 g of aluminium

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the aluminium= c = 0.903 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 0.903 J/g^oC}=11.07^oC$

3) 10.0 g of ethanol

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the ethanol= c = 2.42 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 2.42 J/g^oC}=4.13 ^oC$

4) 10.0 g of water

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the water = c = 4.18J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 4.18 J/g^oC}=2.39 ^oC$

5) 10.0 g of lead

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the lead= c = 0.128 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 0.128 J/g^oC}=78.125^oC$

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

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3 years ago

Which of these is an example of a physical change ?

Gre4nikov [31]

It's 1. Melting a substance. The rest are chemical changes

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3 years ago