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Nata [24]
2 years ago
9

In a demonstration, a 4.00 cm2 square coil with 10 000 turns enters a larger square region with a uniform 1.50 T magnetic field

at a speed of 100 m/s. The plane of the coil is perpendicular to the field lines. If the breakdown voltage of air is 4 000 V/cm on that day, the largest gap you can have between the two wires connected to the ends of the coil and still get a spark is:
Physics
1 answer:
Kay [80]2 years ago
6 0

Explanation:

It is given that,

Area of square coil, A=4\ cm^2=0.0004\ m^2

Side of the square, L = 0.02 m

Number of turns, N = 10000

Uniform magnetic field, B = 1.5 T

Speed, v = 100 m/s

An emf is induced in the coil which is given by :

E=NBLv

E=10000\times 1.5\times 0.02\times 100

E = 30000 V

Breakdown voltage of air, V=4000\ V/cm=400000\ V/m

Let d is the gap between the two wires connected to the ends of the coil and still get a spark. So,

Electric field, E'=\dfrac{V}{d}

\dfrac{30000}{d}=400000

d = 0.075 m

Hence, this is the required solution.

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Each metal is illuminated with 400 nm (3.10 eV) light. Rank the metals on the basis of the maximum kinetic energy of the emitted
34kurt

Answer:

K.E(K) > K.E(Cs) > 0 (others)

Explanation:

Given the Work functions of the metal as

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Cesium (Wo) =2.1eV

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Magnesium (Wo) = 3.7eV

Potassium (Wo) = 2.3eV

Using the formula:

K.E = hf - Wo........(1)

Wo = hfo..............(2)

From these the fo can be calculated for all the metals

Where K.E =Kinetic Energy

hf = energy of illumination = 3.10eV

h is Planck constant and has the value 6.6 × 10^-34JS^-1

The frequency f of the illumination is given by

f = 3.10 × 1.6 × 10^-19/6.6 × 10^-34

f = 7.51 × 10¹⁴ Hz..........(*)

Now an electron is only ejected if the threshold frequency of the metal is reached.

The work function has a threshold frequency (fo) for all the metals and this minimum frequency required to required to remove an electron from the surface of a metal.

We need to compare f with fo

If fo >= f there is emission, otherwise there is no emission

So using (2) we calculate for all fo and compare with f

K.E(Al) = 3.10 - 4.0 - 3.10 = -0.9eV, fo = 9.70 × 10¹⁴ Hz (no emission)

K.E(Pt) = 3.10 - 6.40 = -3.30eV, fo = 1.55 × 10^15 Hz, ( no emission)

K.E(Cs) = 3.10 - 2.10 = -1.0eV, fo = 5.09×10¹⁴ Hz, (emission)

K.E(Be) =3.10-5.0 = -1.90eV, fo = 12.12 ×10^15 Hz.,(no emission)

K.E(Mg) = 3.10-3.70 = -0.6eV, fo = 8.97 × 10¹⁴Hz, (no emission)

K.E(K) = 3.10 - 2.30= 0.9eV, fo = 5.58 × 10¹⁴ Hz, (emission)

So the metals whose electron gain Kinetic energy are:

Cesium

Potassium

Others have zero kinetic energy since no electron is emitted.

Hence the rank is:

K.E(K) > K.E(Cs) > 0 (others)

6 0
2 years ago
The temperature and pressure at the surface of Mars during a Martian spring day were determined to be -50 °C and 900 Pa, respect
Sidana [21]

Answer:

T = 273 + (-50) = 273 – 50 = 223 K

R = 188.82 J / kg K for CO2

Density (Martian Atmosphere) = P / RT = 900 / 188.92 x 223 = 900 / 42129.16 = 0.0213 kg / m^{3}

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Density (Earth Atmosphere) = P / RT = 101600 / 287 x 291 = 1.216 kg / m^{3}

4 0
2 years ago
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What is the efficiency of a machine if your work on the machine is 1200 j and the machines output work is 300 j?
inysia [295]

Answer:

The efificiency is 0,25 of the machine  (25%). See the explanation below

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We calculate the efficiency with this formula:

Efficiency = energy obtained/energy supplied= 300 Joule/1200 Joule

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Good work on solving part a).
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-- Amount stored in 1kg of fat = 3.8 x 10⁷ joules

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2 years ago
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Explanation:


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2) This is, the person does not feel a pressure upward, which is he/she does not feel the weight: freefalling is a situation of apparent weigthlessness.


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