Which type of variable is most related to the data gathered from an experiment?

An emf of 28.0 mV is induced in a 501 turn coil when the current is changing at a rate of 12.0 A/s. What is the magnetic flux th

dmitriy555 [2]

Answer:

Φ = 5.589×10⁻⁵ Wb

Explanation:

The inductance of a coil is given as

L = e/(di/dt) ..................... Equation 1

Where L = inductance of the coil, e = induced e.m.f, di/dt = rate of change of current in the coil.

Also,

The inductance of each turn of the coil when a magnetic field is step up in the coil is

L = NΦ/i ................. Equation 2

Where N = number of turns, Φ = magnetic field, i = current.

equating equation 1 and equation 2

e/(di/dt) = NΦ/i

making Φ the subject of the equation,

Φ = (e×i)/N.(di/dt) .................. Equation 3

Given: e = 28.0 mV = 0.028 V, N = 501 turns, di/dt = 12.0 A/s, i = 4.00 a

Substitute into equation 3,

Φ = (0.028×4)/(12×501)

Φ = 0.112/2004

Φ = 5.589×10⁻⁵ Weber

Φ = 5.589×10⁻⁵ Wb

6 0

3 years ago

If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surfa

shepuryov [24]

Answer:

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

Explanation:

Surface gravity is given by the following formula:

$g=G\frac{m}{r^{2}}$

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

$g_{E}=G\frac{m}{r_{E}^{2}}$

$g_{P}=G\frac{m}{r_{P}^{2}}$

The problem tells us the radius of the planet is twice that of the radius on earth, so:

$r_{P}=2r_{E}$

If we substituted that into the gravity of the planet equation we would end up with the following formula:

$g_{P}=G\frac{m}{(2r_{E})^{2}}$

Which yields:

$g_{P}=G\frac{m}{4r_{E}^{2}}$

So we can now compare the two gravities:

$\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}$

When simplifying the ratio we end up with:

$\frac{g_{P}}{g_{E}}=\frac{1}{4}$

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.

3 0

3 years ago

A CD has to rotate under the readout-laser with a constant linear velocity of 1.25 m/s. If the laser is at a position 3.7 cm fro

Savatey [412]

Answer:N=322.53 rpm

Explanation:

Given

Linear velocity (v)=1.25 m/s

Position from center is 3.7 cm

we know

$v=\omega \times r$

$1.25\times 100=\omega \times 3.7$

$\omega =\frac{125}{3.7}=33.78$

and $\frac{2\pi N}{60}=\omega$

$N=\frac{\omega \times 60}{2\pi }$

$N=\frac{33.78\times 60}{2\pi }$

N=322.53 rpm

8 0

3 years ago

1. James drives 400 km in 5 hours to his grandmothers. What are the units for speed going to be?

Airida [17]

Answer:

See the answer below

Explanation:

1. Speed is calculated as the ratio of distance and time. Hence, Jame's speed can be calculated as:

400/5 km/hr = 80 km/hr

The unit for the speed would be km/hr. This can also be converted to m/s:

80 km = 80,000 m

1 hr = 3,600 s

80 km/hr = 80,000/3600 m/s = 22.22 m/s

2. Since James drove 400 km in 5 hours, the distance he drove is 400 km.

3. The time it took for James to get there is 5 hours.

6 0

3 years ago

Electricity costs 6 cents per kilowatt hour. In one month one home uses one megawatt hour of electricity. How much will the elec

Luda [366]

Answer:

Cost of 1000 kilowatt hour = 6000 cents

Explanation:

Given that

Electricity cost is 6 cents per kilowatt hour.

And we have to found out the cost for one megawatt hour

We know that

1 kilowatt = 1000 watt

1 megawatt = = 1000000 watt

1 megawatt = 1000 kilowatt

1 megawatt hour = 1000 kilowatt hour

Given that cost of 1 kilowatt hour = 6 cents

So the cost of 1000 kilowatt hour = 6 x 1000 cents

Cost of 1000 kilowatt hour = 6000 cents

3 0

3 years ago

Which type of variable is most related to the data gathered from an experiment?​

Which type of variable is most related to the data gathered from an experiment?