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4 years ago

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Increase the slit width to 1050 nm. What happens to the width of the central bright fringe? (Always remember to wait a few secon

ds for the simulation to catch up with the changes you made.)
a:The width of the central bright fringe becomes wider.
b:The width of the central bright fringe becomes narrower.
c:The width of the central bright fringe does not change.

1 answer:

djyliett [7]4 years ago

3 0

Answer:

b.

Explanation:

In case of Single Slit, diffraction will occur.

Then In Single slit Diffraction, width of central fringe is

$x_c= 2D\lambda/a
$

where D= distance b/w screen and slit

a= slit width

\lambda = wavelength

Thus if Screen width increases keeping other factors same then width of central fringe becomes narrower as

$x_c\propto 1/a$

On increasing the slit width the central bright fringe width The width of the central bright fringe becomes narrower.

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A long solenoid with 1.65 103 turns per meter and radius 2.00 cm carries an oscillating current I = 6.00 sin 90πt, where I is in

Leno4ka [110]

Answer:

The electric field is $35\cos(90\pi t)\ mV/m$

Explanation:

Given that,

Radius = 2.00 cm

Number of turns per unit length $n= 1.65\times10^{3}$

Current $I = 6.00\sin 90\pi t$

We need to calculate the induced emf

$\epsilon =\mu_{0}nA\dfrac{dI}{dt}$

Where, n = number of turns per unit length

A = area of cross section

$\dfrac{dI}{dt}$ =rate of current

Formula of electric field is defined as,

$E=\dfrac{\epsilon}{2\pi r}$

Where, r = radius

Put the value of emf in equation (I)

$E=\dfrac{\mu_{0}nA\dfrac{dI}{dt}}{2\pi r}$ ....(II)

We need to calculate the rate of current

$I=6.00\sin 90\pi t$ ....(III)

On differentiating equation (III)

$\dfrac{dI}{dt}=90\pi\times6.00\cos(90\pi t)$

Now, put the value of rate of current in equation (II)

$E=\dfrac{4\pi\times10^{-7}\times1.65\times10^{3}\times\pi\times(2.00\times10^{-2})^2\times90\pi\times6.00\cos(90\pi t)}{2\pi\times 2.00\times10^{-2}}$

$E=35\cos(90\pi t)\ mV/m$

Hence, The electric field is $35\cos(90\pi t)\ mV/m$

7 0

3 years ago

Read 2 more answers

A 50g ball is released from rest 1.0 above the bottom of thetrack

ludmilkaskok [199]

Answer:

The maximum height of the ball is 2 m.

Explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

The equation is

$y=\dfrac{1}{4}x^2$

We need to calculate the velocity

Using conservation of energy

$\Delta U_{i}+\Delta K_{i}=\Delta K_{f}+\Delta U_{f}$

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

$\Delta U_{i}=\Delta K_{f}$

Put the value into the formula

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$50\times10^{-3}\times9.8\times1.0=\dfrac{1}{2}\times50\times10^{-3}\times v^2$

$v^2=\dfrac{2\times50\times10^{-3}\times9.8\times1.0}{50\times10^{-3}}$

$v=\sqrt{19.6}$

$v=4.42\ m/s$

We need to calculate the maximum height of the ball

Using again conservation of energy

$\dfrac{1}{2}mv^2=mgh$

Here, h = y highest point

Put the value into the formula

$\dfrac{1}{2}\times50\times10^{-3}\times(4.42)^2=50\times10^{-3}\times9.8\times h$

$y=\dfrac{0.5\times(4.42)^2}{9.8}$

$y=0.996\ m$

Put the value of y in the given equation

$y=\dfrac{1}{4}x^2$

$x^2=4\times0.996$

$x=\sqrt{4\times0.996}$

$x=1.99\ m\ \approx 2 m$

Hence, The maximum height of the ball is 2 m.

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3 years ago

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Natali [406]

There are three forces acting on the book.
1. Force due to gravity
2. Force exerted downward by the hamster
3. Normal Force in reaction to the downward forces

Since the book is not moving, the net force is zero. The summation of all forces must be zero. Then we could find the normal force which is unknown (denoted as x).

∑F = -(4 kg)(9.81 m/s2) - 3 N + x =0
∑F = -39.24 N - 3N + x =0
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Therefore, the normal force is 42 N.

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3 years ago

7. Two bikes travelling in the same direction move at a speed of 30 km/hr. The bikes are separated by a distance of 5 km. What w

Fantom [35]

Answer:

Explanation:

Call the bike on the right A

Call the bike on the left B

The car begins it's time when it passes A

4 minutes later, it passes B.

But B has moved in 4 minutes and that is the key to the problem.

How far has B moved.

t = 4 minutes = 4/60 hours = 1/15 of an hour.

d = ?

rate = 30 km / hr

d = r * t

d = 30 km/hr * 1/15 hours = 2 km

The distance between the bikes is 5 km.

So the car has traveled 5 - 2 = 3 km

d = 3 km

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t = 4 minutes = 1/15 hour

r = d/t = 3/(1/15)= 3 / 0.066666666 = 45 km/hr.

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3 years ago

Sally and Sam are in a spaceship that comes to within 17,000 km of the asteroid Ceres. Determine the force Sally experiences, in

MAXImum [283]

Answer:

0.01606 Newtons

Explanation:

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m₁ = Mass of the asteroid = 8.7× 10²⁰ kg

m₂ = Mass of Sally = 80 kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

From Newton's Universal law of gravity

$F=G\frac{m_1m_2}{r^2}\\\Rightarrow F=6.67\times 10^{-11}\times \frac{8.7\times 10^{20}\times 80}{17000000^2}\\\Rightarrow F=0.01606\ N$

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3 years ago