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3 years ago

10

An electron enters a region of uniform perpendicular en and bn fields. it is observed that thevelocity nv of the electron is una

ffected. a possible explanation is:

1 answer:

konstantin123 [22]3 years ago

5 0

<span>v is perpendicular to both E and B and has a magnitude E/B</span>

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From class, weight is a measurement of the force of

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Answer:

the answer is well known as gravity

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3 years ago

You submerge 100 grams of steel in 200 grams of water. If the steel has an initial temperature of 90°C and the water has an init

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Answer:

Q=cmΔT

Q1(steel)=Q2(water)

c1•m1• (t1-t) = c2•m2• (t-t2)

628•0.1•(80-t) = 4180•0.2• (t-10)

5024-62.8t = 836t -8360

5024+8360=(836+62.8)t

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3 years ago

The company where you work has obtained and stored five lasers in a supply room. You have been asked to determine the intensity

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Answer:

a) I = 5.79 10⁵ W/m² , b) I = 2.58 10² W / m², c) I = 8.03 10³ W / m² , d) I = 5.3 10⁶ W / m², e) I = 9 10¹ W / m² , f) D> A> C> B> E

Explanation:

The intensity is defined as the power per unit area

I = P / A

The area of a circle is

A = π r²

Laser A

Power P = 2.2 W

Diameter d = 2.9 mm = 2.9 10⁻³ m

Let's calculate

Area

A = π d² / 4

A = π (2.2 10⁻³)²/4

A = 3.80 10⁻⁶ m²

Let's calculate the intensity

I = 2.2 / 3.80 10⁻⁶

I = 0.579 10⁶ W / m²

I = 5.79 10⁵ W/m²

Laser B

The electric field is E = 440 V / m

Intensity average is

I = E B / 2 μ₀

The relationship of the fields with the speed of light

E / B = c

The intensity

I = EE / 2 μ₀ c

I = 440² / (2 4π 10⁻⁷ 3 10⁸)

I = 1.936 105/750

I = 2.58 10² W / m²

Laser C

The magnetic field amplitude B = 8.2 10⁻⁶ T

I = c / 2μ₀ B²

I = 3 10⁸/2 4π 10⁻⁷ (8.2 10⁻⁶)²

I = 8.03 10³ W / m²

Part D

Diameter d = 1.8 mm = 1.8 10⁻³ m

The radius is r = d / 2 = 0.9 10⁻³ m

The force is F = 9.0 10⁻⁸ N

The radiation pressure is on a reflective surface is

P = 2S / c

I = S =P c / 2

The definition of pressure is

P = F / A

I = F c / 2 A

I = 9.0 10⁻⁸ 3 10⁸ / (2π (0.9 10⁻³)²)

I = 5.3 10⁶ W / m²

Part E

Average energy density

u = 3.0 10⁻⁷ J / m³

I = S = c u

I = 3 10⁸ 3.0 10⁻⁷

I = 9 10¹ W / m²

Part F

Sort in descending order

The order is

D> A> C> B> E

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3 years ago

A 0.210-kg metal rod carrying a current of 11.0 A glides on two horizontal rails 0.490 m apart. If the coefficient of kinetic fr

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Answer:

The magnetic field is $B = 0.0764 \ T$

Explanation:

From the question we are told that

The mass of the metal is $m = 0.210 \ kg$

The current is $I = 11.0 \ A$

The distance between the rail(length of the rod ) is $d = 0.490 \ m$

The coefficient of kinetic friction is $\mu_k = 0.200$

Generally the magnetic force is mathematically represented as

$F_b = B * I * d$

Given that the rod is moving at a constant velocity, it

=> $F_b = F_k$

Where $F_k$ is the kinetic frictional force which is mathematically represented as

$F_k = \mu_k * m * g$

So

$B * I * d = \mu_k * m * g$

=> $B = \frac{\mu_k * m * g}{I * d }$

substituting values

=> $B = \frac{0.200 * 0.210 * 9.8 }{ 11 * 0.490 }$

=> $B = 0.0764 \ T$

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What should you do if you need to pull over to get some rest?

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