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Arlecino [84]
4 years ago
8

Define relative density​

Physics
1 answer:
weeeeeb [17]4 years ago
5 0

Hello There!

Relative density is another word for specific gravity. Relative density is comparing the density of one thing to the density of another. Most of the time, you look at the density of water so we look at the density of an object relative to the density of water.

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If a 54 kg sprinter can accelerate from a standing start to a speed of 10 m/s in 3 s, what average power is generated?
lora16 [44]

Answer:

Power, P = 600 watts

Explanation:

It is given that,

Mass of sprinter, m = 54 kg

Speed, v = 10 m/s

Time taken, t = 3 s

We need to find the average power generated. The work done divided by time taken is called power generated by the sprinter i.e.

P=\dfrac{W}{t}

Work done is equal to the change in kinetic energy of the sprinter.

P=\dfrac{\dfrac{1}{2}mv^2}{t}

P=\dfrac{\dfrac{1}{2}\times 54\ kg\times (10\ m/s)^2}{3\ s}

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3 0
3 years ago
What is the difference in electrical potential energy between two places in an electric field?
galben [10]
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8 0
3 years ago
A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein
RSB [31]

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:

\frac{dh}{dt}= \frac{(8m^{3}/s)(25)}{16 \pi (4m)^{2}}

so:

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

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Novosadov [1.4K]

Answer:

76,000 meters

Explanation:

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