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3 years ago

What is the the steadiest firing position?

Physics

1 answer:

Paha777 [63]3 years ago

3 0

The closer you are to the ground the more accurate you'll be. That's why most snipers are in the "prone" position.

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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

2 years ago

Read 2 more answers

An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea

melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, $v=1.37\times10^7\ \rm m/s$

Explanation:

Given:

Linear charge density of the ring= $0.1\ \rm \mu C/m$
Radius of the ring R=0.2 m
Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

$Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C$

Potential due the ring at a distance x from the centre of the rings is given by

$V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\$

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

$\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V$

Let $\Delta U$ be the change in potential Energy given by

$\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J$

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

$\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s$

So the electron will be moving with $v=1.37\times10^7\ \rm m/s$

5 0

3 years ago

Do you think it is possible to control the magnetic properties of a magnet? Can a magnet be turned on and off?

Sunny_sXe [5.5K]

Answer:

Yes it is possible to control to some extent.

Explanation:

In general there are two types of magnets : permanent and temporary (electromagnets).

Electromagnets can be controlled since it basically depends on electricity. By switching on and off the electric supply the magnets also can be switched on and off respectively. We can also control the intensity of magnetic power.

On the other hand permanent magnet cannot be switched on and off but the magnetic properties can be altered event to an extent when it loses all its magnetic properties. It can be caused by high temperature, physical impact and also exposure to other magnetic fields. For every element there is a point of temperature called curie temperature above which the permanent magnet loses its magnetic properties. This can be brought back again by induced magnetism. The only issue is that induced magnetism work in most cases but not in all.

6 0

3 years ago

The relationship between sharks and remoras is characterized TRUE OR FALSE

Damm [24]

False, sharks and remoras have a symbiotic relationship. The remora removes parasites from the sharks scales, and the shark provides protection for the remora

8 0

3 years ago

Please choose the answer that describes the scientific notation 5098000

bonufazy [111]

So there is a decimal after the last zero and it looks like this 5098000. You have to move the decimal point six back to get in between the five and the zero which looks like this 5.098000

Scientific notation is the way that scientists easily handle very large numbers or very small numbers. For example, instead of writing 0.0000000056, we write 5.6 x 10^9.

Being that we moved the decimal six places back the answer is 5.098 x 10^6

3 0

3 years ago