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Vilka [71]
3 years ago
15

URGENT! PLEASE ANSWER!!

Physics
1 answer:
Marrrta [24]3 years ago
7 0

Answer:

C

Explanation:

The answer is obvious from the given figure

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Moving a magnet inside a coil of wire will induce a voltage in the coil. How can the voltage in the coil be increased?
mr_godi [17]
As the magnet is moved inside a coil of wire, the number of lines of magnetic field passing through the coil changes. Faraday stated that : it is the change in the number of field lines passing through the the coil of wire that induces emf in the loop. Specifically, it is the rate of change in the number of magnetic field lines passing through the loop that determines the induced emf. There is a term called magnetic flux same as electric flux, this magnetic flux can be a measure of the number of field lines passing through a surface. It is given by ( Φ=ΣB. dA. Where B is magnetic field and dA is small elementary area). The induced emf is given by (ξ = dΦ/dt). This equation states that THE MAGNITUDE OF THE INDUCED CURRENT IN A CIRCUIT IS EQUAL TO THE RATE AT WHICH THE MAGNETIC FLUX THROUGH THE CIRCUIT IS CHANGING WITH TIME. So more rapid you move the coil, more will be the change in flux and hence more emf will be produced. So option D is the correct answer. I hope this long description will help you out.
5 0
3 years ago
Read 2 more answers
A student pushes a 21-kg box initially at rest, horizontally along a frictionless surface for 10.0 m and then releases the box t
Marrrta [24]

Answer:v=3.08 m/s

Explanation:

Given

mass of student m=21 kg

distance moved d=10 m

Force applied F=10 N

acceleration of system during application of force is a

a=\frac{F}{m}=\frac{10}{21}=0.476 m/s^2

using v^2-u^2=2 as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

v^2-0=2\times 0.476\times 10

v=\sqrt{9.52}

v=3.08 m/s

6 0
3 years ago
A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)
wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m           [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = 75\times 10^{-6}\ \mu C    [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}

Plug in the given values for each point and solve.

(a)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C

(b)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C

(c)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C

(d)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C

7 0
3 years ago
Consider the position vs. time graph below for a woman's movement in a hallway. What is the woman's velocity from 4 to 5 s?
Ksenya-84 [330]

Answer:

The answer is "6\  \frac{m}{s}"

Explanation:

The formula for velocity:

\to \overline{v}={\frac{\Delta x}{\Delta t}}

      =\frac{6}{1}\\\\=6\  \frac{m}{s}

7 0
3 years ago
Find the velocity. 10 points. Will give brainliest!
BARSIC [14]

Answer:

6.060606...

Explanation:

To figure out velocity, you divide the distance by the time it takes to travel that same distance, then you add your direction to it. So the distance would be 1000m and the time would be 2 minutes and 45 seconds and if you convert the minutes into fractions you would get 165 seconds than you would divide 1000m by 165 seconds and you would get 6.060606... seconds as her average velocity

4 0
3 years ago
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