When an object is turning around, is it also at rest at the point?

Two toroidal solenoids are wound around the same form so that the magnetic field of one passes through the turns of the other. S

dedylja [7]

The concept that we need here to give a proper solution is mutual inductance.

The mutual inductance is given by the expression

$M=\frac{N\Phi}{I}$

Where,

I = current

N = Number of turns

$\Phi =$ Flux through the solenoid.

Part A) Then we have in our values that,

$I=6.6A$

$\Phi= 3.50*10^{-2}Wb$

$N=450$

Replacing in the equation,

$M = \frac{450*350*10^{-2}}{6.60}$

$M = 2.39H$

Part B) Here is required the Flux, then using the same expression we have that

$\Phi = \frac{IM'}{N}$

We conserve the same value for the Inductance but now we have a current of 2.6, then

$\Phi = \frac{2.6*2.39}{690}$

$\Phi = 9*10^{-3}Wb$

Therefore the flux in Solenoid 1 is $9*10^{-3}Wb$

8 0

3 years ago

1. What does the endocrine system have in common with the muscular system?

Lostsunrise [7]

Answer:

B Both are directly related to movement.

7 0

3 years ago

How long has eggs been around

Paraphin [41]

Answer:

About six million years

Explanation:

3 0

3 years ago

Read 2 more answers

Water is leaking out of an inverted conical tank at a rate of 10,500 cm3/min at the same time that water is being pumped into th

satela [25.4K]

The tank has a volume of $\dfrac\pi3R^2H$ , where $H=6\,\rm m$ is its height and $R=\dfrac d2=2\,\rm m$ is its radius.

At any point, the water filling the tank and the tank itself form a pair of similar triangles (see the attached picture) from which we obtain the following relationship:

$\dfrac26=\dfrac rh\implies r=\dfrac h3$

The volume of water in the tank at any given time is

$V=\dfrac\pi3r^2h$

and can be expressed as a function of the water level alone:

$V=\dfrac\pi3\left(\frac h3\right)^2h=\dfrac\pi{27}h^3$

Implicity differentiating both sides with respect to time $t$ gives

$\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9h^2\,\dfrac{\mathrm dh}{\mathrm dt}$

We're told the water level rises at a rate of $\dfrac{\mathrm dh}{\mathrm dt}=20\,\frac{\rm cm}{\rm min}$ at the time when the water level is $h=2\,\mathrm m=200\,\mathrm{cm}$ , so the net change in the volume of water $\dfrac{\mathrm dV}{\mathrm dt}$ can be computed:

$\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9(200\,\mathrm{cm})^2\left(20\,\dfrac{\rm cm}{\rm min}\right)=\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}$

The net rate of change in volume is the difference between the rate at which water is pumped into the tank and the rate at which it is leaking out:

$\dfrac{\mathrm dV}{\mathrm dt}=(\text{rate in})-(\text{rate out})$

We're told the water is leaking out at a rate of $10,500\,\frac{\mathrm{cm}^3}{\rm min}$ , so we find the rate at which it's being pumped in to be

$\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}=(\text{rate in})-10,500\,\dfrac{\mathrm{cm}^3}{\rm min}$

$\implies\text{rate in}\approx289,753\,\dfrac{\mathrm{cm}^3}{\rm min}$

4 0

3 years ago

1. Describe your egg drop apparatus and at least three (3) design features you implemented in order to try to help your egg surv

Olegator [25]

Answer:

Explanation:

There are three basic ways to increase the likelihood of safely dropping an egg:

Slow down the descent speed.

Parachutes are an obvious method for slowing the decent speed, as long as the design includes a way to keep the parachute open.

Cushion the egg so that something other than the egg itself absorbs the impact of landing.

The largest end of the egg has an area of air trapped between the egg's two membranes. This air space forms when the contents of the egg cool and contract after the egg is laid. It accounts for the crater you often see at the end of a hard-cooked egg. Upon impact the heavier spherical yolk continues moving towards the ground. The compression of the airspace acts like an air bag for the eggs' valuable contents. Building an artificial cushioning device will also help absorb the impact of landing.

Orient the egg so that it lands on the strongest part of the shell.

The arch structure at either end of the egg is stronger than its sides. Pressure is distributed down (or up) the arches so that less pressure acts on any one point. Orienting the arch downwards will increase the egg's survival.

Hope this helps you

5 0

2 years ago