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3 years ago

6

Can you answer this question about work and force?

1 answer:

velikii [3]3 years ago

8 0

112.5 Joule....its for A

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A resistor and a capacitor are connected in series to an ideal battery of constant terminal voltage. When the system reaches its

PilotLPTM [1.2K]

Answer:

when the system reaches steady state the voltage across the resistor amounts to Zero and the voltage across the capacitor equals to the terminal voltage of the battery

Explanation:

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4 years ago

A _____ resource could run out thousands of years from now.

tresset_1 [31]

The answer is B. Nonrenewable

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3 years ago

Blood should not be stored in which way?

scZoUnD [109]

It should not be stored in a place where it is open to the air, as it can obtain diseases, bacteria and viruses just hanging out waiting for a host.

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4 years ago

Read 2 more answers

A tank, shaped like a cone has height 12 meter and base radius 1 meter. It is placed so that the circular part is upward. It is

Temka [501]

Answer:

376966.991 Joules

Explanation:

Given that :

the height = 12 m

Let assume the tank have a thickness = dh

The radius of the tank by using the concept of similar triangle is :

$\dfrac{1}{r} = \dfrac{12}{h}$

$r = \dfrac{h}{12}$

The area of the tank = $\mathbf{\pi r^2}$

The area of the tank = $\mathbf{\pi( \dfrac{h}{12})^2}$

The area of the tank = $\mathbf{ \dfrac{\pi}{144}h^2}$

The volume of the tank is = area × thickness

= $\mathbf{ \dfrac{\pi}{144}h^2 \ dh}$

Weight of the element = $\rho_ g * volume$

where;

$\rho_g$ = density of water ; which is given as 10000 N/m³

So;

Weight of the element = $\mathbf{ 10000 *\dfrac{\pi}{144}h^2 \ dh}$

Weight of the element = $\mathbf{69.44 \ \pi \ h^2 \ dh}$

However; the work required to pump this water = weight × height rise

where the height rise = 12 - h

the work required to pump this water = $\mathbf{69.44 \ \pi \ h^2 \ dh}$ (12 - h)

the work required to pump this water = $\mathbf{69.44 \pi (12h^2-h^3)dh}$

We can determine the total workdone by integrating the work required to pump this water

SO;

Workdone = $\mathbf{\int\limits^{12}_0 {69.44 \pi(12h^2-h^3)} dh}$

= $\mathbf{ 69.44 \pi \int\limits^{12}_0 {(12h^2-h^3)} dh}$

= $\mathbf{ 69.44 \pi[ \frac{12h^3}{3}- \frac{h^4}{4}]^{12}}_0} }$

= $\mathbf{69.44 \pi [ \frac{12^4}{3}-\frac{12^4}{4}]}$

= $\mathbf{69.44 \pi*12^4 [ \frac{4-3}{12}]}$

= $\mathbf{69.44 \pi*12^4 *\frac{1}{12}}$

= 376966.991 Joules

6 0

3 years ago

Suppose that a rectangular toroid has 2,000 windings and a self-inductance of 0.060 H. If the height of the rectangular toroid i

andrey2020 [161]

Answer:

0.01154 A

Explanation:

We have given the energy in the magnetic field $U=4\times 10^{-6}J$

Value of inductance L =0.060 H

Energy stored in magnetic field is given by $U=\frac{1}{2}Li^2$

$i=\sqrt{\frac{2U}{L}}$

$i=\sqrt{\frac{2\times 4\times 10^{-6}}{0.06}}=0.01154\ A$

So the current flowing through rectangular toroid will be 0.01154 A

3 0

3 years ago