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tekilochka [14]

3 years ago

15

"Compared to infrared radiation, does ultraviolet radiation have longer or shorter wavelengths? Does ultraviolet radiation have

higher or lower energy per photon?"

1 answer:

balu736 [363]3 years ago

5 0

Answer:Ultraviolet radiation has shorter wavelengths and higher energy than infrared radiation.

Explanation: Electromagnetic radiation radiations which have both electrical and magnetic properties,they can be transmitted through space or through a medium.

It includes Gamma radiation, infra-red, visible light, Ultraviolet radiation etc they occur with different wavelength, the lower the wavelength the higher the Energy dissipated per photon. According to their order of decreasing wavelength and increased energy they are classified as follows.

RADIO WAVE, MICRO WAVE, INFRA-RED, VISIBLE LIGHT, ULTRAVIOLET RAY, X-RAY, GAMMA RAYS.

You might be interested in

Professional Application. A 96 kg football player catches a 0.900 kg ball with his feet off the ground with both of them moving

Zarrin [17]

To solve this problem it is necessary to apply the equations related to the conservation of momentum.

This definition can be expressed as

$m_1u_1+m_2u_2 = (m_1+m_2)V_f$

Where

$m_{1,2}$ = Mass of each object

$u_{1,2}$ = Initial Velocity of each object

$V_f$ = Final velocity

Rearranging the equation to find the final velocity we have,

$V_f = \frac{m_1u_1+m_2u_2}{(m_1+m_2)}$

Our values are given as

$m_1 = 96Kg\\m_2 = 0.9Kg\\u_1 = 6.3m/s\\u_2 = 27.4m/s$

Replacing we have,

$V_f = \frac{(96)(6.3)+(0.9)(27.4)}{(96+0.9)}$

$V_f = 6.4959m/s$

Therefore the final velocity is 6.5m/s

3 0

3 years ago

The same ball is shot straight up a second time from the same gun, but this time the spring is compressed only half as far befor

Mekhanik [1.2K]

Answer:

The new height the ball will reach = (1/4) of the initial height it reached.

Explanation:

The energy stored in any spring material is given as (1/2)kx²

This energy is converted to potential energy, mgH, of the ball at its maximum height.

If the initial height reached is H

And the initial compression of the spring = x

So, mgH = (1/2)kx²

H = kx²/2mg

The new compression, x₁ = x/2

New energy of loaded spring = (1/2)kx₁²

And the new potential energy = mgH₁

mgH₁ = (1/2)kx₁²

But x₁ = x/2

mgH₁ = (1/2)k(x/2)² = kx²/8

H₁ = kx²/8mg = H/4 (provided all the other parameters stay constant)

6 0

3 years ago

A uniform log of length L is inclined 30° from the horizontal when supported by a frictionless rock located 0.6L from its left e

mafiozo [28]

Answer:

x = 0.974 L

Explanation:

given,

length of inclination of log = 30°

mass of log = 200 Kg

rock is located at = 0.6 L

L is the length of the log

mass of engineer = 53.5 Kg

let x be the distance from left at which log is horizontal.

For log to be horizontal system should be in equilibrium

∑ M = 0

mass of the log will be concentrated at the center

distance of rock from CM of log = 0.1 L

now,

∑ M = 0

$m_{log} g \times 0.1 L = m_{engineer} g \times (x - 0.6 L)$

$200 \times 0.1 L = 53.5 \times (x - 0.6 L)$

$0.374 L =x - 0.6 L$

x = 0.974 L

hence, distance of the engineer from the left side is equal to x = 0.974 L

7 0

3 years ago

A .5kg bird is perched on its nest so that it has 50J of potential energy. how far is it off the of the ground?

pshichka [43]

It is 10.20 m from the ground.

<u>Explanation:</u>

<u>Given:</u>

m = 0.5 kg

PE = 50 J

We know that the Potential energy is calculated by the formula:

$P. E = m \times g \times h$

where m is the is mass in kg; g is acceleration due to gravity which is 9.8 m/s and h is height in meters.

PE is the Potential Energy.

Potential Energy is the amount of energy stored when an object is stationary.

Here, if we substitute the values in the formula, we get

$P. E = m \times g \times h$

50 = 0.5 × 9.8 × h

50 = 4.9 × h

$h = \frac {50} {4.9}$

h = 10.20 m

3 0

3 years ago

A glider with mass 0.24 kg sits on a frictionless horizontal air track, connected to a spring of negligible mass with force cons

shepuryov [24]

Answer:

$v=2.556m/s$

Explanation:

From the conservation of mechanical energy

$K_{E1}+U_1=K_{E2}+U_2$

$\frac{1}{2}m*v_1^2+\frac{1}{2}*K*x_1^2=\frac{1}{2}m*v_2^2+\frac{1}{2}*K*x_2^2$

$x_2=0.08m$

$v_1=0 m/s$

Solve to velocity v2

$m*v_2^2=k*x_1^2-k*x_2^2$

$v^2=\frac{k}{m}*(x_1^2-x_2^2)$

$v^2=\frac{5.5N/m}{0.24kg}*(0.54m^2-0.080^2)$

$v=\sqrt{6.54m^2/s^2}=2.556m/s$

4 0

3 years ago