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2 years ago

Why ultraviolet light have more energy than the visible light

Physics

2 answers:

3241004551 [841]2 years ago

7 0

Answer:Cause it is ultra not regular

Explanation:

Mumz [18]2 years ago

6 0

Answer:

Explanation:

Ultraviolet radiation has shorter waves than blue or violet light, and thus oscillates more rapidly and carries more energy per photon than visible light does. Does ultraviolet rays carry more energy? The distinguishing factor among the different types of electromagnetic radiation is their energy content.

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IF YOUR GOOD AT SCIENCE THEN PLEASE ANSWER THIS ASAP I WILL MARK YOU THE BRAINLIEST

Digiron [165]

Answer:

This is an open circuit

Explanation:

An open circuit I believe

It needs to be closed for the bulb to be turned on

8 0

2 years ago

current is passed through two parallel conductors in the same direction. If the conductors are placed near each other,they will

svlad2 [7]

If current is passed through two parallel conductors in the same direction and the conductors are placed near each other, they will attract each other.

<h3>What is electric current?</h3>

Electric current can be defined as the flow of electrons.

Since electrons are easily removed from atom and are very mobile, the flow of electrons constitute an electric current.

Materials which allow electric current to flow through them are known as conductors. Examples of conductors are metals, and electrolytes.

On the other hand, materials which do not allow electric current to pass through them are known as insulators. Examples of insulators are wood and rubber.

The flow of current is known as electricity.

Parallel conductors with current flowing through them in the same direction are attracted to each other as a result of a magnetic field produced by the flow of current.

In conclusion, conductors allow electric current to pass through and the flow of current through a conductor produces a magnetic field.

Learn more about parallel conductors at: brainly.com/question/17148082

#SPJ1

6 0

2 years ago

Narysuj wykres zależności v(t) jeśli w chwili początkowej t=0 V=10m/s w każdej sekundzie szybkość zmniejsza się o 1m/s . Po jaki

irina1246 [14]

1) See graph in attachment

2) 10 s

3) 50 m

Explanation:

In this problem, we have an object initially moving with a velocity of

v = 10 m/s

when the time is

t = 0 s

Then, we are told that the speed of the object is decreasing by 1 m/s every second. This means that on a velocity-time graph, the motion will be represented by a straight line, starting from v = 10 when t = 0, and decreasing by 1 m/s every second.

The result can be found in the graph in attachment.

Moreover, we can also infer that the motion of the object is accelerated (because velocity is changing), and that the acceleration is constant and it is equal to

$a=1 m/s^2$

which is equivalent to the gradient of the line in the velocity-time graph.

In this part, we want to find after what time the body will stop its motion.

To do that, we can use the following suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

In this problem:

u = 10 m/s is the initial velocity of the body

$a=-1 m/s^2$ is the acceleration

v = 0 m/s, because we want to find the time T at which the body will stop

Re-arranging the equation, we find:

$T=-\frac{u}{a}=-\frac{10}{-1}=10 s$

In order to find the total distance covered by the body during its accelerated motion, we have to use another suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

In this problem:

u = 10 m/s is the initial velocity

$a=-1 m/s^2$ is the acceleration

t = 10 s is the time it takes for the body to stop (found in part 2)

Solving for s, we find the distance covered:

$s=(10)(10)+\frac{1}{2}(-1)(10)^2=50 m$

7 0

3 years ago

Light of wavelength 500 nm is incident perpendicularly from air on a film 10-4cm thick and of refractive index 1.375. Part of th

Marysya12 [62]

Answer

given,

wavelength (λ)= 500 n m

thickness of film= 10⁻⁴ cm

refractive index = μ = 1.375

distance traveled is double which is equal to 2 x 10⁻⁴ cm

a) Number of wave

$N = \dfrac{d}{\mu\lambda}$

$N = \dfrac{2 \times 10^{-6}}{1.375\times 500 \times 10^{-9}}$

N = 2.91

N = 3

b) phase difference is equal to

Reflection from the first surface has a 180° (½λ) phase change.

There is no phase change for the 2nd surface reflection and there is no phase difference for the 2nd wave having traveled an exact whole number of waves.

net phase difference = $180^0\times \dfrac{3}{2}$

= 270°

6 0

3 years ago

Which of the following is most closely related to an activated complex?

Anon25 [30]

Among the choices above, the one that is most closely related to an activated complex is the transition state. The answer is letter D. This formation forms quickly and does not stay in a way compound is. It usually forms during the enzyme – substrate reaction.

5 0

3 years ago