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stiv31 [10]
3 years ago
7

A light wave travels at a speed of 3.0 × 108 meters/second. If the wavelength is 7.0 × 10-7 meters, what is the frequency of the

wave?
Physics
1 answer:
BaLLatris [955]3 years ago
5 0

Answer : The frequency of the wave is, 4.285\times 10^{14}s^{-1}

Solution : Given,

Speed of light = 3.0\times 10^{8}m/s

Wavelength of light = 7.0\times 10^{-7}m

Formula used :

\nu=\frac{c}{\lambda}

\nu = frequency of light

\lambda = wavelength of light

c = speed of light

Now put all the given values in the above formula, we get the frequency of light.

\nu=\frac{3\times 10^8m/s}{7\times 10^{-7}m}=4.285\times 10^{14}s^{-1}

Therefore, the frequency of the wave is, 4.285\times 10^{14}s^{-1}

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Vlad1618 [11]

Answer:

I think its the 2nt

Explanation:

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4 0
3 years ago
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The tendency of an atom to attract electrons towards itself is called
SVEN [57.7K]

Answer:

Electronegativity

Explanation:

Element: <em>An element is a substance which can not be split into simpler units by an ordinary chemical process</em>

Electronegativity: This is the power of an atom in a molecule to attract electrons. The electronegativity of an element increase across the period in the periodic table and decreases down the group. The most electronegative elements are the reactive non metals.

Examples of electronnegative element include, Oxygen, Chlorine, Fluorine, Nitrogen, sulphur etc.

8 0
4 years ago
A proton is moving in a circular orbit of radius 20 cm under a uniform magnetic field 0.3 t perpendicular to the velocity of the
Vladimir [108]

Answer:

v = 5.75 x 10⁶ m/s

Explanation:

The radius (r) of the circular orbit taken by a charged particle is related to its speed perpendicular to a magnetic field of strength B, and is given by

r = \frac{mv}{qB}       --------------(i)

Where,

q = charge of the particle

m = mass of the particle

Making v subject of the formula in equation (i) above gives

v = \frac{qBr}{m}  -------------------(ii)

Given;

r = 20cm = 0.2m

B = 0.3T

v = unknown

q = charge of proton = 1.6 x 10⁻¹⁹ C

m = mass of the proton = 1.67 x 10⁻²⁷kg

Substitute the values of m, q, B and r into equation (ii) above to get;

v = \frac{1.6 * 10^{-19} * 0.3 * 0.2} {1.67*10^{-27} }

Solving for v gives:

v = 5.75 x 10⁶ m/s

Therefore, the velocity of the proton is 5.75 x 10⁶ m/s

4 0
3 years ago
The acceleration of a particle is defined by the relation a 5 28 m/s2. Knowing that x 5 20 m when t 5 4 s and that x 5 4 m when
mamaluj [8]

Explanation:

It is given that,

The acceleration of a particle, a=-8\ m/s^2 (negative as the particle is decelerating)

Initial distance, x₁ = 20 m

Initial time, t₁ = 4 s

New distance x₂ = 4 m

Velocity, v = 10 m/s

(A) Calculating initial distance using second equation of motion as :

x_1=ut_1+\dfrac{1}{2}at^2

20=4u+\dfrac{1}{2}(-8)\times 4^2

u = 21 m/s

When velocity of the particle is zero, time taken is t (say). Using first equation of motion as :

v=u+at

0=21+(-8)t

t = 2.62 seconds

So, the velocity of the particle is zero at t = 2.62 seconds.

(B) Velocity at t = 11 s

v=21+(-8)\times 11

v = 13 m/s

Total distance covered at t = 11 s. The overall path travelled by the particle during its entire journey is called total distance covered.

d=ut+\dfrac{1}{2}at^2+|ut+\dfrac{1}{2}at^2|

d=21\times 2.62+\dfrac{1}{2}\times (-8)(2.62)^2+|21\times 8.38+\dfrac{1}{2}\times (-8)(8.38)^2|

d = 132.48 m

So, the distance travelled by the particle at t = 11 seconds is 132.48 meters.    

5 0
3 years ago
A tire swing hanging from a branch reaches nearly to the ground. How could you estimate the height of the branch using only a st
AlexFokin [52]

Answer:

Explanation:

With the help of expression of  time period of pendulum we can calculate the height of the branch . The swinging tire can be considered equivalent to swinging bob of a pendulum . Here length of pendulum will be equal to height of branch .

Let it be h . Let the time period of swing of tire be T then from the formula of time period of pendulum

T = 2\pi\sqrt{\frac{l}{g} }  where l is length of pendulum .

here l = h so

T = 2\pi\sqrt{\frac{h}{g} }  

h = \frac{T^2g}{4\pi^2}

If we calculate the time period of swing of tire , we can calculate the height of branch .

The time period of swing of tire can be estimated with the help of a stop watch . Time period is time that the tire will take in going from one extreme point to the other end and then coming back . We can easily estimate it with the help of stop watch .

4 0
4 years ago
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