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2 years ago

8

Coherent microwaves of wavelength 5.00 cm enter a tall, narrow window in a building otherwise essentially opaque to the microwav

es. if the window is 36.0 cm wide, what is the distance from the central maximum to the first-order minimum along a wall 6.50 m from the window?

1 answer:

zzz [600]2 years ago

8 0

The solution for this problem is:

For 1st minimum, let m be equal to 1.

d = slit width

D = screen distance.

Θ = arcsin (m * lambda/ (d))

= 0.13934 rad, 7.9836 deg

y = D*tan (Θ)

y = 6.50 * tan (7.9836)

= 0.91161 m is the distance from the central maximum to the first-order minimum

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Answer:

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Explanation:

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A 0.5 m diameter wagon wheel consists of a thin rim having a mass of 7 kg and six spokes, each with a mass of 1.2 kg. 1.2 kg 7 k

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Explanation:

It is given that,

Mass of the rim of wheel, m₁ = 7 kg

Mass of one spoke, m₂ = 1.2 kg

Diameter of the wagon, d = 0.5 m

Radius of the wagon, r = 0.25 m

Let I is the the moment of inertia of the wagon wheel for rotation about its axis.

We know that the moment of inertia of the ring is given by :

$I_1=m_1r^2$

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The moment of inertia of the rod about one end is given by :

$I_2=\dfrac{m_2l^2}{3}$

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$I_2=\dfrac{m_2r^2}{3}$

$I_2=\dfrac{1.2\times (0.25)^2}{3}=0.025\ kgm^2$

For 6 spokes, $I_2=0.025\times 6=0.15\ kgm^2$

So, the net moment of inertia of the wagon is :

$I=I_1+I_2$

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2 years ago

Unbalanced forces cause a change in motion.<br><br> How do you know this?

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2 years ago

A charge of 32.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.30x 104 V/

hodyreva [135]

A) The work done by the electric field is zero

B) The work done by the electric field is $9.1\cdot 10^{-4} J$

C) The work done by the electric field is $-2.4\cdot 10^{-3} J$

Explanation:

A)

The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).

The work done by a force is given by the equation

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the particle

$\theta$ is the angle between the direction of the force and the direction of the displacement

In this problem, we have:

The force is directed vertically upward (because the field is directed vertically upward)
The charge moves to the right, so its displacement is to the right

This means that force and displacement are perpendicular to each other, so

$\theta=90^{\circ}$

and $cos 90^{\circ}=0$ : therefore, the work done on the charge by the electric field is zero.

B)

In this case, the charge move upward (same direction as the electric field), so

$\theta=0^{\circ}$

and

$cos 0^{\circ}=1$

Therefore, the work done by the electric force is

$W=Fd$

and we have:

$F=qE$ is the magnitude of the electric force. Since

$E=4.30\cdot 10^4 V/m$ is the magnitude of the electric field

$q=32.0 nC = 32.0\cdot 10^{-9}C$ is the charge

The electric force is

$F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N$

The displacement of the particle is

d = 0.660 m

Therefore, the work done is

$W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J$

C)

In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is

$\theta=90^{\circ}+45^{\circ}=135^{\circ}$

Moreover, we have:

$F=1.38\cdot 10^{-3} N$ (electric force calculated in part b)

While the displacement of the charge is

d = 2.50 m

Therefore, we can now calculate the work done by the electric force:

$W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J$

And the work is negative because the electric force is opposite direction to the displacement of the charge.

Learn more about work and electric force:

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2 years ago

you are piloting a small plane and you want to reach an airport 450 km due south in 3.0 h a wind is blowing from the west 50.0 k

alex41 [277]

Answer:

You should choose airspeed 158.11 km/h at 18.4° west of south

Explanation:

The distance to the air port is 450 km due to south

You should to reach the airport in 3 hours

→ Velocity = distance ÷ time

→ Distance = 450 km , time = 3 hours

→ The velocity of your plane = 450 ÷ 3 = 150 km/h due to south

A wind is blowing from west 50 km/h

We need to know what heading and airspeed you should choose to

reach your destination

At first we must find the resultant velocity of your plane and the wind

The south and west are perpendicular, then the resultant velocity is

→ $v_{R}=\sqrt{(v_{p})^{2}+(v_{w})^{2}}$

→ $v_{p}=150$ km/h , $v_{w}=50$ km/h

→ $v_{R}=\sqrt{(150)^{2}+(50)^{2}}=158.11$ km/h

To cancel the velocity of the wind, the pilot should maintain the velocity

of the plane at 158.11 km/h

The direction of the velocity is the angle between the resultant velocity

and the vertical (south)

→ The direction of the velocity is $tan^{-1}\frac{50}{150}=18.4$ °

The direction of the velocity is 18.4° west of south

<em>You should choose airspeed 158.11 km/h at 18.4° west of south</em>

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2 years ago