Which statement is the most appropriate conclusion for David to write?

A (20*20) cm² loop has a resistance of 0.10 Ω. A magnetic field perpendicular to the loop is B = 4t - 2t², where B is in tesla a

Ilya [14]

Answer with Explanation:

We are given that

Area of loop= $(20\times 20) cm^2=400\times 10^{-4} m^2$

$1 cm^2=10^{-4} m^2$

Resistance, R= $0.1\Omega$

B= $4t-2t^2$

We know that magnetic flux

$\phi=BA$

Emf , $E=\mid \frac{d\phi}{dt}\mid =\mid\frac{d(BA}{dt}\mid =\mid A\frac{dB}{dt}=400\times 10^{-4}\times \frac{4t-2t^2}{dt}\mid =\mid400\times 10^{-4}\times(4-4t)\mid$

Current, $I=\frac{E}{R}$

Current, $I=\frac{\mid 400\times 10^{-4}(4-4t)\mid }{0.1}=1.6\mid (1-t)\mid$

Substitute t=0 s

Then, I= $1.6\mid (1-0)\mid$ =1.6 A

Substitute t=1 s

Then, I= $1.6\mid (1-1)\mid$ =0

Substitute

t=2 s

Current, I= $1.6\mid(1-2)\mid$ =1.6 A

8 0

3 years ago

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Not everything in science is clear-cut, with obvious problems, experiments, theories, and solutions. Sometimes scientists gather

sergij07 [2.7K]

A. science sometimes requires creativity

4 0

3 years ago

Alcohol of mass 33.2g and density 0.79kg/m³ or 790kg/m³ is mixed with water of 9g. What is the density of the resulting mixture?

KATRIN_1 [288]

Answer:

0.83 g/cm³

Explanation:

The volume of the alcohol is ...

(33.2 g)/(0.79 g/cm³) ≈ 42.0253 cm³

The density of water is about 1 g/cm³, so the volume of 9 g of it is ...

(9 g)/(1 g/cm³) = 9 cm³

The total volume is ...

42.0253 cm³ +9 cm³ = 51.0253 cm³

The total mass is ...

33.2 g + 9 g = 42.4 g

So, the resulting density is ...

(42.4 g)/(51.0253 cm³) ≈ 0.83 g/cm³

The resulting mixture has a density of about 0.83 g/cm³.

_____

<em>Additional comment</em>

Alcohol dissolves in water, so the total volume will be slightly less than 51.0253 cm³. The attached curve shows the result of mixing ethanol and water.

The weight of a mole of ethanol is about 46 g, of water, about 18.02 g. Then the mole fraction of alcohol is ...

(33.2/46)/(33.2/46 +9/18.02) ≈ 0.59

The volume of the mix is then estimated to be (-1.05 cc/mol)(1.221 mol), or about 1.28 cm³ less than the volume indicated above. That brings the density up to about 0.85 g/cm³.

We're not completely sure of the relevance of this calculation, since many of the applicable parameters are not specified. The point is that <em>the density of the mix will probably be slightly more than the value calculated above</em>. YMMV

3 0

3 years ago

A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km. 1. If you and your spacesuit

Law Incorporation [45]

1. 0.16 N

The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

$F=G\frac{Mm}{r^2}$

where

G is the gravitational constant

$M=8.7\cdot 10^{13}kg$ is the mass of the asteroid

m = 100 kg is the mass of the man

r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid

Substituting, we find

$F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N$

2. 1.7 m/s

In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

$m\frac{v^2}{r}=G\frac{Mm}{r^2}$

where v is the minimum speed required to stay in orbit.

Re-arranging the equation and solving for v, we find:

$v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2})(8.7\cdot 10^{13} kg)}{2000 m}}=1.7 m/s$

3 0

4 years ago

Coiling wire around an iron core and applying an electric current through the wire creates a temporary magnet called an electrom

Serggg [28]

Answer:

C) Use a battery with more voltage.

Explanation:

The equation for the magnetic field around a coil is given by,

B = μ₀NI

where,

B = Magnetic flux density

μ₀ = permeability

N = number of turns per meter

I = Current in the wire

So when using a higher voltage battery, more current passes through the battery as resistance of the wire remains the same.

5 0

3 years ago

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