How would you classify an EM wave with a frequency of 10^7 Hz?

In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1

asambeis [7]

Explanation:

It is given that,

Frequency of monochromatic light, $f=5\times 10^{14}\ Hz$

Separation between slits, $d=2.2\times 10^{-5}\ m$

(a) The condition for maxima is given by :

$d\ sin\theta=n\lambda$

For third maxima,

$\theta=sin^{-1}(\dfrac{n\lambda}{d})$

$\theta=sin^{-1}(\dfrac{nc}{fd})$

$\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})$

$\theta=4.69^{\circ}$

(b) For second dark fringe, n = 2

$d\ sin\theta=(n+1/2)\lambda$

$\theta=sin^{-1}(\dfrac{5\lambda}{2d})$

$\theta=sin^{-1}(\dfrac{5c}{2df})$

$\theta=sin^{-1}(\dfrac{5\times 3\times 10^8}{2\times 2.2\times 10^{-5}\times 5\times 10^{14}})$

$\theta=3.90^{\circ}$

Hence, this is the required solution.

8 0

3 years ago

A 1 kg object sits on the earth’s surface. What is the force of gravity between the object and the earth? (mass of the earth = 5

snow_lady [41]

Answer:

9.81N

Explanation:

the force of attraction is given by

F=GmM/R²

where m is mass of the body

M is mass of the earth

R is radius of the earth

G is the universal gravitational constant(6.67x10-¹¹)

hence we substitute the values in the formula.

you can ask questions

4 0

2 years ago

In a crossing situation, which vessel is required to maintain its course and speed?

makvit [3.9K]

Both in the domestic and international guidelines tell that when two power-driven vessels are crossing so as to contain risk of collision, the vessel which has the other on her starboard side (the give-way vessel) must keep out of the way.

If you are the give-way vessel, it is your responsibility to avoid a collision. Normally, this means you must change speed or direction to cross behind the other vessel which is the stand-on vessel.

At evening, when you perceive a red light crossing right-to-left in front of you, you need to change your course. But if you perceive a green light crossing from left-to-right, you are the stand-on vessel, and should maintain course and speed.

The leading situations of collision risk are meeting head-on, overtaking, and crossing. When one of two vessels is to keep out of the way (give-way vessel), the other, the stand-on vessel, must uphold course and speed.

3 0

3 years ago

A 2.00-gram air inflated balloon is given an excess negative charge, q1 = -3.25 × 10-8 C, by rubbing it with a blanket. It is fo

Llana [10]

Answer:

The rod`s charge must be positive, because the gravity force is pointing downwards and the electrostatic force must be pointing upwards (in order to balance the gravity force)

The charge is q_2 = 1.667 times 10^(-7) C

Explanation:

F_e = F_g

where F_g = m g and F_e= (1/4 pi e_0)*(q_1*q_2)/d^2,

please see the file attached for more details.

Download pdf

3 0

3 years ago

Does anyone know this

NARA [144]

1) The net force is 16 N to the right

2) The net force is 98 N to the left

3) The net force is 0.5 N downward

4) The net force is 170 N to the right

5) The net force is 175 N to the right

Explanation:

1)

To find the net force, we have to analyze all the forces acting on the box.

We have:

Force to the right: $F_a = 20 N$ , the applied force
Force to the left: $F_f = 4 N$ , the force of friction
Force to the bottom: $F_g = 400 N$ , the weight of the box (the weight is always downward vertically)
Force to the top: $F_N = 400 N$ . This is the normal force, which is the reaction force exerted by the table on the box: it points upward and counterbalances the weight of the box, preventing it from falling down)

Therefore, the horizontal net force is

$F_x = F_a - F_f = 20 - 4 = 16 N$ (to the right)

While the vertical force is

$F_y = F_N - F_g = 400 - 400 = 0$

So the net force is 16 N to the right.

2)

In this case, we have the following forces:

$F_g = 4 N$ downward, the weight of the ball
$F_a = 100 N$ to the left, the force that kicks the ball
$F_f = 2 N$ to the right, the force of friction
$F_N = 4 N$ upward, the normal reaction exerted by the field on the ball

Therefore, the horizontal net force is

$F_x =F_a - F_f = 100 -2 = 98 N$ (to the left)

While the vertical force is

$F_y = F_g - F_N = 4 - 4 = 0$ (downward)

And so, the net force is 98 N to the left.

3)

The force acting on the squirrel in this problem are:

$F_g = 8 N$ downward, the weight of the squirrel
$F_f = 7.5 N$ upward, the air resistance, acting upward

Both forces act vertically and there are no other forces acting in other directions, therefore the net force on the squirrel is simply equal to the net force on the vertical direction, which is:

$F_y = F_g - F_f = 8 - 7.5 = 0.5 N$

And since the weight is larger than the air resistance, the direction of the net force is downward.

4)

The forces acting on Monkey are:

$F_1=95 N$ is the force applied to the right by Bunny
$F_2 = 75 N$ is the force applied by Deer from the left (so, also on the right)
$F_g = 50 N$ is the weight of Monkey, downward
$F_N = 50 N$ is the normal reaction exerted by the surface, upward

So, the net force in the horizontal direction is

$F_x = F_1 + F_2 = 95+75=170 N$ (to the right)

While the net force in the vertical direction is

$F_y = F_N - F_g = 50 - 50 = 0$

And therefore the net force is 170 N to the right

5)

The forces acting on Deer are:

$F_a = 100 N + 100 N = 200 N$ to the right, the combined force applied by Bunny and Monkey
$F_f = 25 N$ to the left, the force of friction
$F_g = 150 N$ downward, the weight of the deer
$F_N = 150 N$ upward, the normal reaction from the surface that balances the weight