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denis-greek [22]

2 years ago

12

The two ends of a coiled wire are connected to the electrodes of a lightbulb.

1 answer:

barxatty [35]2 years ago

5 0

Answer:

a

Explanation:

The bar magnet moves downward with respect to the wire loop, so that the number of magnetic field lines going through the loop decreases with time. This causes an emf to be induced in the loop, creating an electric current.

in other words, the magnets motion creates a current in the loop

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An accelerating voltage of 2.47 x 10^3 V is applied to an electron gun, producing a beam of electrons originally traveling horiz

Dmitry [639]

Answer:

6.3445×10⁻¹⁶ m

Explanation:

E = Accelerating voltage = 2.47×10³ V

m = Mass of electron

Distance electron travels = 33.5 cm = 0.335 cm

$E=\frac{mv^2}{2}\\\Rightarrow v=\sqrt{\frac{2E}{m}}\\\Rightarrow v=\sqrt{\frac{2\times 2470\times 1.6\times 10^{-19}}{9.11\times 10^{-31}}}\\\Rightarrow v=29455356.08671\ m/s$

Deflection by Earth's Gravity

$\Delta =\frac {gt^2}{2}$

Now, Time = Distance/Velocity

$\Delta =\frac {g\frac{s^2}{v^2}}{2}\\\Rightarrow \Delta =\frac{9.81\frac{0.335^2}{29455356.08671^2}}{2}\\\Rightarrow \Delta =6.3445\times 10^{-16}\ m$

∴ Magnitude of the deflection on the screen caused by the Earth's gravitational field is 6.3445×10⁻¹⁶ m

3 0

3 years ago

(a) What is the intensity in W/m2 of a laser beam used to burn away cancerous tissue that, when 90.0% absorbed, puts 500 J of en

inysia [295]

Answer:

4.42 x 10⁷ W/m²

Explanation:

A = energy absorbed = 500 J

η = efficiency = 0.90

E = Total energy

Total energy is given as

E = A/η

E = 500/0.90

E = 555.55 J

t = time = 4.00 s

Power of the beam is given as

P = E /t

P = 555.55/4.00

P = 138.88 Watt

d = diameter of the circular spot = 2.00 mm = 2 x 10⁻³ m

Area of the circular spot is given as

A = (0.25) πd²

A = (0.25) (3.14) (2 x 10⁻³)²

A = 3.14 x 10⁻⁶ m²

Intensity of the beam is given as

I = P /A

I = 138.88 / (3.14 x 10⁻⁶)

I = 4.42 x 10⁷ W/m²

6 0

3 years ago

Read 2 more answers

A particle of mass m moves under an attractive central force F(r) = -Kr4 with angular momentum L. For what energy will the motio

docker41 [41]

Answer:

Angular velocity is same as frequency of oscillation in this case.

ω = $\sqrt{\frac{7K}{m} }$ x $[\frac{L^{2}}{mK}]^{3/14}$

Explanation:

- write the equation F(r) = -K $r^{4}$ with angular momentum <em>L</em>

- Get the necessary centripetal acceleration with radius r₀ and make r₀ the subject.

- Write the energy of the orbit in relative to r = 0, and solve for "E".

- Find the second derivative of effective potential to calculate the frequency of small radial oscillations. This is the effective spring constant.

- Solve for effective potential

- ω = $\sqrt{\frac{7K}{m} }$ x $[\frac{L^{2}}{mK}]^{3/14}$

3 0

3 years ago

69PTS! Which choices are also major levels of classification? Choose all answers that are correct. A. Genus B. Species C. Group

Pachacha [2.7K]

A. Genus and B. Species are also major levels of classification.

5 0

3 years ago

Read 2 more answers

What is the greatest distance an image can be located behind a convex spherical mirror?

Afina-wow [57]

Answer:

Maximum distance of image from mirror is equal to focal length of the mirror

Explanation:

As we know by the equation of mirror we have

$\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}$

here we know for convex mirror

object position is always negative as it will be placed behind the mirror always

while the focal length of the convex mirror is always taken positive

So here we have

$\frac{1}{d_i} + \frac{1}{-d_o} = \frac{1}{f}$

$\frac{1}{d_i} = \frac{1}{d_o} + \frac{1}{f}$

so here maximum value of image distance is equal to focal length of the mirror

6 0

3 years ago