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galina1969 [7]
3 years ago
6

Why is the melting of ice a physical change?

Physics
1 answer:
svetlana [45]3 years ago
7 0

It does not change the chemical composition of water.

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How does paving parking lots and roads with concrete or asphalt affect surface water and groundwater?
lidiya [134]
Parking lots with roads and concrete reduce infiltration. Infiltration is the process by which water penetrates the soil. Reducing the amount of water that enters the soil can eventually impact groundwater levels in some areas by decreasing it over time. Paved roads lead to increased surface runoff which increases the possibility of flooding in periods of heavy rainfall. This is known as urban flooding. 
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Ana walks from 4 m to 200 cm. Which of the following statements is true about
Gemiola [76]
Distance=2m

because 200cm = 2m
so 4m-2m=2m
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How is most of the electricity we use at home generated?
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Nuclear power plants, wind farms, water farms, and geothermal heating
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3 years ago
A mass of 250 N is on a piston of 2.0 m^2. What force is needed to lift this piston if the area of the second piston is 0.5 m^2?
prohojiy [21]

Answer:

<h3>62.5N</h3>

Explanation:

The pressure at one end of the piston is equal to the pressure on the second piston.

Pressure = Force/Area

F1/A1 = F2/A2

Given

F1 = 250N

A1 = 2.0m²

A2 = 0.5m²

F2 = ?

Substituting the given values in the formula;

250/2 = F2/0.5

cross multiply

250*0.5 = 2F2

125 = 2F2

F2 = 125/2

F2 = 62.5N

Hence the  force needed to lift this piston if the area of the second piston is 0.5 m^2 is 62.5N

8 0
4 years ago
You wiggle a string,that is fixed to a wall at the other end, creating a sinusoidalwave with a frequency of 2.00 Hz and an ampli
FinnZ [79.3K]

Answer:

Explanation:

A general wave function is given by:

f(x,t)=Acos(kx-\omega t)

A: amplitude of the wave = 0.075m

k: wave number

w: angular frequency

a) You use the following expressions for the calculation of k, w, T and λ:

\omega = 2\pi f=2\pi (2.00Hz)=12.56\frac{rad}{s}

k=\frac{\omega}{v}=\frac{12.56\frac{rad}{s}}{12.0\frac{m}{s}}=1.047\ m^{-1}

T=\frac{1}{f}=\frac{1}{2.00Hz}=0.5s\\\\\lambda=\frac{2\pi}{k}=\frac{2\pi}{1.047m^{-1}}=6m

b) Hence, the wave function is:

f(x,t)=0.075m\ cos((1.047m^{-1})x-(12.56\frac{rad}{s})t)

c) for x=3m you have:

f(3,t)=0.075cos(1.047*3-12.56t)

d) the speed of the medium:

\frac{df}{dt}=\omega Acos(kx-\omega t)\\\\\frac{df}{dt}=(12.56)(1.047)cos(1.047x-12.56t)

you can see the velocity of the medium for example for x = 0:

v=\frac{df}{dt}=13.15cos(12.56t)

7 0
3 years ago
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