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Paraphin [41]
3 years ago
12

Two samples of water can have different densities. Which of these statements would best explain this fact?

Physics
1 answer:
emmainna [20.7K]3 years ago
8 0
C, the two samples are in different states!
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The narrator of a story
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The answer is d! I’m happy to help :)
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A train travels 120 km in 2 hours and 30 minutes. What is its average speed?
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13.33 or 13 1/3m/s (meters per second)

Explanation:

In physics, we use the basic units of meters and seconds. So first convert (km) into meters (m) and also hours and minutes into seconds (s). We end up with 120000m and 9000s. Then divide the 120000m by the 9000s and you end up with 13.33 or 13 1/3 m/s.

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If you shout at a cliff wall that is 440 m away and the air temperature is at 25 °C, how long will it take before you hear your
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The speed of sound at T=25°C is Vs=346 m/s. So the sound has to reach the cliff and return back to you so the path it needs to travel is s=2*440 m = 880 m.
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Electromagnetic induction means that moving a magnet through a loop of wire creates an electric current.
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Suppose that a balloon is being filled with air at a rate of 10 cm3/s. (Assume that theballoon is a perfect sphere.) At what rat
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Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

i.e \frac{dv}{dt} =10 cm^3/s

Consider r be the radius of the balloon.

The volume of of a sphere is

v=\frac{4}{3} \pi r^3

Differentiate with respect to t

\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}

\Rightarrow 10 =4\pi r^2\frac{dr}{dt}

\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}

The surface of area of the balloon is(S) = 4\pi r^2

S=4\pi r^2

Differentiate with respect to t

\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}

\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}

Putting the value of \frac{dr}{dt}

\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}

\Rightarrow \frac{dS}{dt} =\frac{20}{ r}

Given that r = 5 cm

[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}  =4 cm³/s

Therefore the surface area of the balloon is increased at 4 cm³/s.

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