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kow [346]
2 years ago
7

A conductor is formed into a loop that encloses an area of 1.0 m2. The loop is orientedat a 30.0° angle with the xy-plane. A var

ying magnetic field is oriented parallel to thez-axis. If the maximum emf induced in the loop is 25.0 V, what is the maximum rate atwhich the magnetic field strength is changing
Physics
1 answer:
ivolga24 [154]2 years ago
3 0

Answer:

29 T/s

Explanation:

Given:

Area enclosed by loop, A = 1 m²

Maximum induced emf, e = 25.0 V

loop is oriented at an angle 30° with the x-y plane.

varying magnetic field is oriented parallel to z-axis.

we have to find \frac{dB}{dt}_{max}

Induced emf is given by:

e =-A.cos\theta .\frac{dB}{dt}\\ \Rightarrow \frac{dB}{dt}=-\frac{e}{Acos\theta} = -\frac{25}{1\times cos30^o} = -28.86T/s≈29T/s

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Parallel light rays with a wavelength of 563 nm fall on a single slit. On a screen 3.30 m away, the distance between the first d
MrMuchimi

Answer:

The width of the slit is 0.4 mm (0.00040 m).

Explanation:

From the Young's interference expression, we have;

(λ ÷ d) = (Δy ÷ D)

where λ is the wavelength of the light, D is the distance of the slit to the screen, d is the width of slit and Δy is the fringe separation.

Thus,

d = (Dλ) ÷ Δy

D = 3.30 m, Δy = 4.7 mm (0.0047 m) and λ = 563 nm (563 ×10^{-9} m)

d = (3.30 × 563 ×10^{-9} ) ÷ (0.0047)

  = 1.8579 × 10^{-6} ÷ 0.0047

  = 0.0003951 m

d = 0.00040 m

The width of the slit is 0.4 mm (0.00040 m).

3 0
3 years ago
The earth has a vertical electric field at the surface, pointing down, that averages 100 N/C. This field is maintained by variou
Kipish [7]

Answer:

The excess charge on earth's surface was calculated to be 4.56 × 10⁵ C

Explanation:

Using the formula for an electric field;

E = kQ/r²

k = 1/(4πε₀) = 8.99 × 10⁹ Nm²/C²

E = 100N/C

r = radius of the earth = 6400 km = 6400000m

Q = Er²/k = 100 × (6400000)²/(8.99 × 10⁹)

Q = 455617.4 C = 4.56 × 10⁵ C

Hope this helps!!!

6 0
3 years ago
The strength of the gravitational pull between two object's is determined by?
iogann1982 [59]
Mass and distance are the two factors
4 0
3 years ago
Read 2 more answers
Astronaut Sarah leaves Earth in a spaceship at a speed of 0.280c relative to Earth. Sarah's destination is a star-system 12.5 li
Usimov [2.4K]

Answer:

L= 12 light years

Explanation:

for length dilation we use the formula

L=L_0\sqrt{1-\frac{v^2}{c^2} }

now calculating Lo

Lo = 12.5×365×24×3600×3×10^8

= 1.183×10^17 m

now putting the values of v and Lo in the above equation we get

L=1.183\times10^{17}\sqrt{1-\frac{0.28c^2}{c^2} }

= 1.136×10^17 m

L=  = \frac{1.136\times10^{17}}{365\times24\times3600\times3\times10^8}m

so L= 12 light years

8 0
3 years ago
Please provide the steps
elena-s [515]

Answer:

wA - T =mA(a)

second option

Explanation:

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