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2 years ago

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A solenoid has 450 loops each of radius 0.0254 m. The field increases from 0 T to 3.00 T in 1.55 s. What is the EMF generated in

the coil? (Hint: What is the area of a circle) (Unit = Volts)

1 answer:

allochka39001 [22]2 years ago

8 0

Answer:

0.175 second

Explanation:

i hope it helps

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A diffraction grating with 140 slits per centimeter is used to measure the wavelengths emitted by hydrogen gas. At what angles i

Degger [83]

Answer:

0.003181 radians

0.003005 radians

Explanation:

Number of slits = 140 /cm

λ = Wavelength = 434 nm = 434×10⁻⁹ m

m = 3 Third order spectrum

Space between slits

$d=\frac{0.01}{140} =7.14\times 10^{-5}\ m$

Now,

$dsin\theta = m\lambda\\\Rightarrow \theta=sin^{-1}\left(\frac{m\lambda}{d}\right)\\\Rightarrow \theta=sin^{-1}\left(\frac{3\times 434\times 10^{-9}}{7.14\times 10^{-5}}\right)\\\Rightarrow \theta=sin^{-1}0.018228\\\Rightarrow \theta=0.01823^{\circ}=0.01823\times \frac{\pi}{180}=0.003181 radians$

0.003181 radians

When λ = 410 nm = 410×10⁻⁹ m

$dsin\theta = m\lambda\\\Rightarrow \theta=sin^{-1}\left(\frac{m\lambda}{d}\right)\\\Rightarrow \theta=sin^{-1}\left(\frac{3\times 410\times 10^{-9}}{7.14\times 10^{-5}}\right)\\\Rightarrow \theta=sin^{-1}0.01722\\\Rightarrow \theta=0.01722^{\circ}=0.01722\times \frac{\pi}{180}=0.003005 radians$

0.003005 radians

5 0

3 years ago

One part of a freely swinging magnet always points

Sladkaya [172]

One part of a freely swinging magnet always points to the Earth's magnetic pole in the Northern Hemisphere.

8 0

4 years ago

Read 2 more answers

A block slides down a frictionless plane having an inclination of 15.0°. The block starts from rest at the top, and the length o

nalin [4]

Answer: check the pic

Explanation:

8 0

3 years ago

Scientists in the laboratory create a uniform electric field e⃗ = 6. 0×105 k^v/m in a region of space where b⃗ =0⃗

zvonat [6]

Scientist in the laboratory create uniform electric field.

PART A: The electric field in rocket frame is (0,0, 6.0 × 10 ⁵).

PART B: The magnetic field in the rocket frame is 7.3 × 10⁻⁶ j∧ .

<h3>What is uniform electric field?</h3>

The electric field whose strength is not changing with respect to time is called uniform magnetic field.

The electric field is 6.0 × 10 ⁵ k^v/m and the magnetic field is zero.

E' = E + (V×B)

= 6.0 × 10 ⁵ + (1.1 × 10⁶ × 0)

E' = 6.0 × 10 ⁵ k∧

(Ex , Ey, Ez) = (0,0, 6.0 × 10 ⁵)

The new magnetic field is represented as

B' = B - 1 /c²(v × E)

=0 - 1 / (3× 10⁸)²(1.1 × 10⁶ i∧ × 6.0 × 10 ⁵k∧ )

B' = 7.3 × 10⁻⁶ j∧

PART A: The electric field in rocket frame is (0,0, 6.0 × 10 ⁵).

PART B: The magnetic field in the rocket frame is 7.3 × 10⁻⁶ j∧ .

Learn more about uniform electric field.

brainly.com/question/26446532

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7 0

2 years ago

Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d

Tasya [4]

Answer:

I(x) = 1444×k × ${\pi}$

I(y) = 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$

Explanation:

Given data

function = x^2 + y^2 ≤ 36

function = x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to $\pi /2$ for θ

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ y²p dA

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ (a sinθ)²(k × a²) adθda

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (sin²θ)dθ

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (1-cos2θ)/2 dθ

I(x) = k $({r}^{6}/6)^(5)_0$ × ${θ/2 - sin2θ/4}^{\pi /2}_0$

I(x) = k × $({6}^{6}/6)$ × ( ${\pi /4} - sin\pi /4)$

I(x) = k × $({6}^{5})$ × ${\pi /4}$

I(x) = 1444×k × ${\pi}$ .....................1

and we can say I(x) = I(y) by the symmetry rule

and here I(o) will be I(x) + I(y) i.e

I(o) = 2 × 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$ ......................2

3 0

3 years ago