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AURORKA [14]
3 years ago
14

What happens to the current as we increase the amount of stepping of our transformer? Does this help explain why the primary was

warm but not the secondary? How does this result demonstrate that we can’t just use a step-up transformer to create free energy?
Physics
1 answer:
matrenka [14]3 years ago
7 0

Answer:

Current will decrease.

Explanation:

When we increase the number of stepping in transformer, the voltage will increase as its is directly proportional to the number of turn of stepping. Thus as the voltage will increase, current will decrease. As per the equation of ideal transformer,   E1 / E2 = I2 / I1

E1 and E2 are the voltages in primary and secondary winding and I1 and I2 are the current.

As the number of turns will be increased more inevitable losses will be generated that dissipates heat thus warming the primary.

Though the conservation of energy is obeyed but losses occur in this scenario hence step-up transformers cannot be used to create free energy.

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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

3 0
2 years ago
Object 1 has a momentum of 10 kg m/s and Object 2 has a momentum of 25 kg m/s. Will it be easier to change the direction of move
svetoff [14.1K]

Answer:

I think its object 1

Explanation:

Because the object that has more weight has a greater momentum and the lightest object that has a less momentum will be easier to change because its lighter.

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3 years ago
What formula should I use?
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<span>Fossil Record</span>

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Work done=mgh
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