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zloy xaker [14]
3 years ago
10

How does the speed of visible light compare with the speed of gamma rays

Physics
1 answer:
DochEvi [55]3 years ago
3 0

Answer:

They are the same

Explanation:

Electromagnetic waves consist of perpendicular oscillations of electric and magnetic field, which oscillate perpendicularly to the direction of motion of the wave (transverse wave). One property of the electromagnetic waves is that they travel in a vacuum always at the same speed, called speed of light:

c=3.0\cdot 10^8 m/s

Electromagnetic waves are classified into 7 different types according to their frequency; from highest to lowest frequency, we have:

Gamma rays

X-rays

Ultraviolet

Visible light

Infrared

Microwaves

Radio waves

We see that both visible light and gamma rays are electromagnetic waves, so they both travel in a vacuum at the same speed, the speed of light.

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A Cephied is a star that belongs to a class of yellow supergiant pulsating stars; these stars vary periodically in brightness, a
mihalych1998 [28]

Luminosities

Thanks to this relationship between period and luminosity, a Cepheid provide a practical and accurate method to evaluate their absolute magnitude. Once this is known, it is possible to know the distance of the Cepheid, calculating the difference with respect to the apparent magnitude.

4 0
4 years ago
Tourists covered 255 km for a 4-hour ride by car and a 7-hour ride by train. what is the speed of the train, if it is 5 km/h gre
LekaFEV [45]
The distance covered by car is equal to (assuming it is moving by uniform motion) the product between the car's speed and the time of the car ride, 4 h:
S_c = v_c t_c
where
v_c is the car's speed
t_c = 4 h is the duration of the car ride

Similarly, the distance covered by train is equal to the product between the train's speed and the duration of the train ride, 7 h:
S_t = v_t t_t

The total distance covered is S=255 km, which is the sum of the distances covered by car and train:
S=255 km = S_c + S_t
which becomes
255 = 4 v_c + 7 v_t (1)
we also know that the train speed is 5 km/h greater than the car's speed:
v_t = 5 + v_c (2)

If we put (2) into (1), we find
255 = 4v_c + 7(5+v_c)
and if we solve it, we find
v_c = 20 km/h
v_t = 25 km/h

So, the car speed is 20 km/h and the train speed is 25 km/h.

4 0
4 years ago
According to Newton’s law of universal gravitation, which statements are true?
andreyandreev [35.5K]

Before we solve this, we should know this fact:

According to Newton's Law of Gravitation, the force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The force acts along the line joining the centres of the two objects. It can be shown by this:

F ∝ \frac{Mm}{ {d}^{2} }

Now, let us check all the options.

A. As we move to higher altitudes, the force of gravity on us decreases.

<em>This </em><em>statement </em><em>is </em><em>true.</em>

The force of gravity is inversely proportional to the square of distance from the centre of the earth. If, we go up the surface of the earth, the distance from the centre of the earth increases and hence the value of force of gravity decrease. So, force of gravity decreases with altitude.

B. As we move to higher altitudes, the force of gravity on us increases.

<em>This </em><em>statement</em><em> </em><em>is </em><em>false.</em>

We have already got the result in option A. that the force of gravity decreases with altitude. It never increases with altitude.

C. As we gain mass, the force of gravity on us decreases.

<em>This </em><em>statement</em><em> </em><em>is </em><em>false.</em>

The force of gravity is directly proportional to the product of the masses. So, if increase our mass, then the force of gravity will also increase and if we decrease our mass, then the force of gravity decreases.

D. As we gain mass, the force of gravity on us increases.

<em>This </em><em>statement</em><em> is</em><em> </em><em>true.</em>

As mentioned earlier in option C., the force of gravity is directly proportional to the product of the masses of the earth and another object. So, as we gain mass, the force of gravity on us increases.

E. As we move faster, the force of gravity on us increases.

<em>This </em><em>statement</em><em> is</em><em> </em><em>true</em><em>.</em>

Here, we have to consider a different formula. According to Newton's Second Law,

F = ma, where F is the force, m is the mass and a is the acceleration.

In other words,

F ∝ a, i.e., force is directly proportional to acceleration.

We know, acceleration is the rate of change of velocity of an body within a time period.

So, if speed is increased, then acceleration will also be greater, which results in the increase of force. So, as we move faster, the force of gravity on us increases.

<u>Answers:</u>

A: As we move to higher altitudes, the force of gravity on us decreases.

D: As we gain mass, the force of gravity on us increases.

E: As we move faster, the force of gravity on us increases.

Hope you could understand.

If you have any query, feel free to ask.

7 0
2 years ago
Need help identifying the rest of the elements!
gregori [183]
I think that number five is lithium
6 0
3 years ago
Calculate the electric field at the center of a square
pantera1 [17]

Answer:

E_y=1175510.2\ N.C^{-1}

The Magnitude of electric field is in the upward direction as shown directly towards the charge q_1.

Explanation:

Given:

  • side of a square, a=52.5\ cm
  • charge on one corner of the square, q_1=+45\times 10^{-6}\ C
  • charge on the remaining 3 corners of the square,q_2=q_3=q_4=-27\times 10^{-6}\ C

<u>Distance of the center from each corners</u>=\frac{1}{2} \times diagonals

diagonal=\sqrt{52.5^2+52.5^2}

diagonal=74.25\cm=0.7425\ m

∴Distance of center from corners, b=0.3712\ m

Now, electric field due to charges is given as:

E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}

<u>For charge q_1 we have the field lines emerging out of the charge since it is positively charged:</u>

E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}

  • E_1=2938775.5\ N.C^{-1}

<u>Force by each of the charges at the remaining corners:</u>

E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}

  • E_2=E_3=E_4=1763265.3\ N.C^{-1}

<u> Now, net electric field in the vertical direction:</u>

E_y=E_1-E_4

E_y=1175510.2\ N.C^{-1}

<u>Now, net electric field in the horizontal direction:</u>

E_y=E_2-E_3

E_y=0\ N.C^{-1}

So the Magnitude of electric field is in the upward direction as shown directly towards the charge q_1.

8 0
3 years ago
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