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Lesechka [4]
3 years ago
13

Suppose that you find the volume of all the oceans to be 1.4×109km3 in a reference book. To find the mass, you can use the densi

ty of water, also found in this reference book, but first you must convert the volume to cubic meters. What is this volume in cubic meters?
Physics
2 answers:
Licemer1 [7]3 years ago
8 0

Answer:

Explanation:

Volume of ocean = 1.4×10^9 km3

To m^3,

1.4 × 10^9 km^3 × (1000 m)^3/(1 km)^3

= 1.4 × 10^18 m^3.

Density of water = 1000 kg/m3

Mass = density × volume

= 1000 × 1.4 × 10^18

= 1.4 × 10^21 kg.

PSYCHO15rus [73]3 years ago
4 0

Answer:

V = 1.4\times 10^{18}\,m^{3},m = 1.435\times 10^{21}\,kg

Explanation:

The volume of all oceans is:

V = (1.4\times 10^{9}\,km^{3})\cdot (\frac{1\times 10^{9}\,m^{3}}{1\,km^{3}} )

V = 1.4\times 10^{18}\,m^{3}

Ocean water has a density of 1025\,\frac{kg}{m^{3}}, the mass of all oceans is:

m = (1025\,\frac{kg}{m^{3}} )\cdot (1.4\times 10^{18}\,m^{3})

m = 1.435\times 10^{21}\,kg

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Answer:

h'=0.25m/s

Explanation:

In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).

So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of 8m^{3}/s. As you  may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

V_{cone}=\frac{1}{3} \pi r^{2}h

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

\frac {r}{h}=\frac{4}{5}

When solving for r, we get:

r=\frac{4}{5}h

so we can substitute this into our volume of a cone formula:

V_{cone}=\frac{1}{3} \pi (\frac{4}{5}h)^{2}h

which simplifies to:

V_{cone}=\frac{1}{3} \pi (\frac{16}{25}h^{2})h

V_{cone}=\frac{16}{75} \pi h^{3}

So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:

\frac{dV}{dt}= \frac{16}{75} \pi (3)h^{2} \frac{dh}{dt}

Which simplifies to:

\frac{dV}{dt}= \frac{16}{25} \pi h^{2} \frac{dh}{dt}

So now I can solve the equation for dh/dt (the rate of height over time, the velocity at which height is increasing)

So we get:

\frac{dh}{dt}= \frac{(dV/dt)(25)}{16 \pi h^{2}}

Now we can substitute the provided values into our equation. So we get:

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so:

\frac{dh}{dt}=0.25m/s

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An object with a mass of 6.0 kg accelerates 8.0 m/s^2 when an unknown force is applied to it. What is the amount of the force? R
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A steel piano wire, of length 1.150 m and mass 4.80 g is stretched under a tension of 580.0 N.
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A steel piano wire, of length 1.150 m and mass of 4.80 g is stretched under a tension of 580.0 N.the speed of transverse waves on the wire would be  372.77 m/s

<h3>What is a sound wave?</h3>

It is a particular variety of mechanical waves made up of the disruption brought on by the movements of the energy. In an elastic medium like the air, a sound wave travels through compression and rarefaction.

For calculating the wave velocity of the sound waves generated from the piano can be calculated by the formula

V= √F/μ

where v is the wave velocity of the wave travel on the string

F is the tension in the string of piano

μ is the mass per unit length of the string

As given in question a steel piano wire, of length 1.150 m and mass of 4.80 g is stretched under a tension of 580.0 N.

The μ is the mass per unit length of the string would be

μ = 4.80/(1.150×1000)

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By substituting the respective values of the tension on the string and the density(mass per unit length) in the above formula of the wave velocity

V= √F/μ

V=√(580/0.0041739)

V =  372.77 m/s

Thus,  the speed of transverse waves on the wire comes out to be  372.77 m/s

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