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Umnica [9.8K]
2 years ago
11

If you double the wavelength, the electromagnetic radiation energy will double. The energy of the electromagnetic radiation will

halve if you halve the wavenumber. When the frequency of the light is doubled, its energy will .
Physics
1 answer:
Sedbober [7]2 years ago
7 0

Answer:

It's energy will double.

Explanation:

This is because energy, E, is related to frequency, f, by:

E = hf

Where h = Planck's constant

So, double frequency will be 2f

=> E(2f) = 2hf = 2E.

Hence, energy is doubled.

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The length of a rectangular sheet of metal decreases by 34.5 cm. Its width decreases proportionally. If the sheets original widt
devlian [24]
The original width was 94.71 cm 
<span>The area decreased 33.1% </span>

<span>The equation for the final size is </span>
<span>2X^2 = 1.2 m^2 </span>
<span>X^2 - 0.6 m^2 </span>
<span>X^2 = 10000 * .6 cm </span>
<span>X = 77.46 cm (this is the width) </span>

<span>The length is 2 * 77.46 = 154.92 cm </span>

<span>The original length was 154.92 + 34.5 = 189.42 cm </span>
<span>The original width was 189.42 / 2 = 94.71 cm </span>

<span>The original area was 94.71 * 189.92 = 17939.9 cm^2 </span>
<span>The new area is 79.46 * 154.92 = 12000.1 cm^2 </span>

<span>The difference between the original and current area is 17939.9 - 12000.1 = 5939.86 cm^2 </span>

<span>The percentage the area decreased is 5939.86 ' 17939.9 = 33.1%</span>
6 0
3 years ago
A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the
Margaret [11]

Hi there!

\boxed{\omega = 0.38 rad/sec}

We can use the conservation of angular momentum to solve.

\large\boxed{L_i = L_f}

Recall the equation for angular momentum:

L = I\omega

We can begin by writing out the scenario as a conservation of angular momentum:

I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)

I_m = moment of inertia of the merry-go-round (kgm²)

\omega_m = angular velocity of merry go round (rad/sec)

\omega_f = final angular velocity of COMBINED objects (rad/sec)

I_b = moment of inertia of boy (kgm²)

\omega_b= angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

I = MR^2

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

\omega = \frac{v}{r}

L_b = MR^2(\frac{v}{R}) = MRv

Plug in the given values:

L_b = (20)(3)(5) = 300 kgm^2/s

Now, we must solve for the boy's moment of inertia:

I = MR^2\\I = 20(3^2) = 180 kgm^2

Use the above equation for conservation of momentum:

600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}

8 0
2 years ago
What shape is the unit cell of ruby?
sleet_krkn [62]
Ruby, a variety of the mineral corundum is in the trigonal crystal system, with hexagonal scalenohedra crystals
7 0
3 years ago
Your house is 45.0 m from a powerline carrying 152 A of current. How much magnetic field does the current create at your house?
Sedaia [141]

This question involves the concepts of th magnetic field and current.

The magnetic field created by the current at the house is "6.75 x 10⁻⁷ T".

<h3>Magnetic Field</h3>

The magnetic field created by a current carrying wire can be given by the following formula:

B=\frac{\mu_o I}{2\pi r}

where,

  • B = magnetic field = ?
  • \mu_o= permeabiliy of free space =4π x 10⁻⁷
  • I = current = 152 A
  • r = distance = 45 m

B=\frac{4\pi x\ 10^{-7}(152)}{2\pi(45)}

B = 6.75 x 10⁻⁷ T

Learn more about magnetic field here:

brainly.com/question/23096032

#SPJ1

7 0
1 year ago
A ball rolls with a speed of 2.0m/s across a level table that is 1.0m above the floor. Upon reaching the edge of the table it fo
Mekhanik [1.2K]

Answer:0.58 m

Explanation:

The initial velocity of the ball is u = 2.0 m/s

The height of the table is, h = 1.0 m

The ball falls in vertical direction under acceleration due to gravity.

Time taken for ball to hit the floor:

h= ut + 0.5gt² ( from the equation of motion)

1.0 m=2.0 m/s × t+0.5 × 9.8 m/s²× t²

Solving this for t,

t = 0.29 s ( we have neglected the negative value of t)

In the same time, the ball would cover a horizontal distance of :

s = u t

⇒s = 2.0 m/s×0.29 s = 0.58 m

Thus, the landing spot is 0.58 m away from the table.

6 0
3 years ago
Read 2 more answers
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