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3 years ago

If you added sand to your coffee, what kind of mixture would your coffee now be?

Physics

1 answer:

Pepsi [2]3 years ago

7 0

Answer:

A heterogenous mixture.

Explanation:

If you added sand to your coffee, the kind of mixture your coffee would now

be is a heterogenous mixture.

A heterogenous mixture can be defined as any mixture which has a different or non-uniform composition and properties throughout any given sample of the mixture. This ultimately implies that, the constituents of a heterogenous mixture always remain separate in the sample and as such comprises of two or more phases.

Hence, a mixture of sand and coffee is simply a heterogenous mixture and can easily be separated into their various components.

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Virty [35]

The answer to this is Longitudinal/compression waves, such as sound waves.

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3 years ago

A ramp with a mechanical advantage of 13 lifts objects to a height of 11 meters.

Aliun [14]

It is 143 meters I think

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3 years ago

Billy is pulling his sister Joslyn in the sled. Based on your knowledge of F=ma, describe what would

lana [24]

Answer:

the motion of the sled decrease

Explanation:

the mass increase

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3 years ago

What is conflict resolution?

andriy [413]

Answer:

B) Settling a dispute or disagreement

Explanation:

A conflict is a disagreement or argument between opposing parties.

A resolution is a solution to a problem.

Knowing both these definitions, the answer choice that makes the most sense would be B) Settling a dispute or disagreement.

Hope this helps!!

3 0

2 years ago

The displacement from equilibrium of an oscillating weight suspended by a spring and subject to the damping effect of friction i

Lisa [10]

Note that the correct times are t = 0, t = 1/4, t = 1/2. You can tell from the spaces between the two digits. i.e 1&4 and 1&2

Answer:

y(0) = 2.00 cm

y(1/4) = 1.56 cm

y(1/2) = 1.21 cm

Explanation:

This is a very simple exercise, the displacement of the oscillating weight from equilibrium has already been modeled by the equation:

$y(t) = 2e^{-t} cos 4t$

Where y = displacement ( in cm)

and t = time (in seconds)

The task is to find the displacement when t = 0, 1/4 and 1/2

When t = 0 s

$y(0) = 2e^{0} cos 4(0)\\y(0) = 2* 1*1\\y(0) = 2.00 cm$

When t = 1/4 s

$y(1/4) = 2e^{-1/4} cos 4(1/4)\\y(14) = 2e^{-1/4} cos (1)\\y(1/4) = 1.56 cm$

When t = 1/2

$y(1/2) = 2e^{-1/2} cos 4(1/2)\\y(14) = 2e^{-1/2} cos (2)\\y(1/2) = 1.21 cm$

8 0

3 years ago