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oksano4ka [1.4K]
3 years ago
13

When we look into the sky every day we get to see the results of all the behaviors of waves. The blue color of the sky results f

rom the scattering of sunlight by the air molecules. The blue light has a frequency of about 7.5 X 10^14 Hertz. What is the wavelength (nm) of this radiation
Physics
1 answer:
mr_godi [17]3 years ago
4 0

Answer:

λ = 0.4 x 10⁻⁶ m = 400 nm

Explanation:

The relationship between frequency, wavelength and speed of an electromagnetic wave is given as follows:

c = f\lambda

where,

c = speed of light = 3 x 10⁸ m/s

f = frequency of the light wave = 7.5 x 10¹⁴ Hz

λ = wavelength of the light = ?

Therefore,

3\ x\ 10^8\ m/s = (7.5\ x\ 10^{14}\ Hz)\lambda\\\\\lambda = \frac{3\ x\ 10^8\ m/s}{7.5\ x\ 10^{14}\ Hz}

<u>λ = 0.4 x 10⁻⁶ m = 400 nm</u>

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A brick of mass 1.15 kg is attached to a thin (massless) cable and whirled around in a circle in a vertical plane. The circle ha
stira [4]

Explanation:

It is given that,

Mass of the brick, m = 1.15 kg

Radius of the circle, r = 1.44 m

The cable will break if the tension exceeds 43.0 N

Let v is the maximum sped can have at the bottom of the circle before the cable will break. At the bottom of the circle, the net force is equal to the centripetal force along with the weight of the brick. So,

T=\dfrac{mv^2}{r}+mg

\dfrac{T}{m}-g=\dfrac{v^2}{r}

v=\sqrt{(\dfrac{T}{m}-g)r}

v=\sqrt{(\dfrac{43}{1.15}-9.8)\times 1.44}

v = 6.30 m/s

So, the maximum speed of the brick at the bottom of the circle before the cable will break is 6.3 m/s. Hence, this is the required solution.

8 0
3 years ago
on the surface of theearth ,the weight of a boy is 400N but on a mountainpeak his weight is 360N,Calculatethe value of ''g' on t
andreyandreev [35.5K]

The value of g at sea level is 9.81 ms^-2.

The boy's mass is constant wherever he is in the universe but his weight will depend on the strength gravity where he is.

By proportion its value on the mountain peak  is (360 /400) * 9.81

= 0.9 * 9.81 = 8.83  ms^-2  to nearest hundredth,  (answer).

7 0
2 years ago
Examples of increase in pressure due to increase in applied force<br>​
Nina [5.8K]

Answer:

injecting

Explanation:

5 0
3 years ago
A block of mass m is pushed a horizontal distance D from position A to position B, along a horizontal plane with friction coeffi
Wewaii [24]

Answer:

The total work done by friction is -2 · μ · m · g · D

Explanation:

Hi there!

The work done by a force is calculated as follows:

W = F · d · cos θ

Where:

W = work.

F = force that does the work.

d = displacement.

θ = angle between the displacement and the force.

If the force is horizontal, as in this case, cos θ = 1

The friction force is calculated as follows:

Ffr = μ · m · g

Where:

μ = friction coefficient.

m = mass of the object.

g = acceleration due to gravity.

Then, in this case, the work done by friction when pushing the block from A to B will be:

W AB = -Ffr · D

W AB = - μ · m · g · D

Notice that the friction force is negative because it is opposite to the pushing force P.

When the block is pushed from B to A, the work done by friction will be:

W BA = Ffr · (-D)

W BA = -μ · m · g · D

Now, the displacement is negative and the friction force is positive (in the opposite direction to -P).

The total work done by friction will be:

W AB + W BA = - μ · m · g · D  - μ · m · g · D  = -2 μ · m · g · D

5 0
3 years ago
Giving brainiest to correct answer.
mixas84 [53]

Answer:

5.33\ m/s

Explanation:

We\ know\ that,\\Momentum=Mass*Velocity\\p=mv\\Hence,\\Lets\ first\ consider\ the\ case\ of\ the\ two\ balls\ 'Before\ Collision':\\\\Mass\ of\ the\ green\ ball=0.2\ kg\\Initial\ Velocity\ of\ the\ green\ ball=5\ m/s\\Initial\ Momentum\ of\ the\ green\ ball=5*0.2=1\ kg\ m/s\\\\Mass\ of\ the\ pink\ ball=0.3\ kg\\Initial\ Velocity\ of\ the\ pink\ ball=2\ m/s\\Initial\ Momentum\ of\ the\ pink\ ball=0.3*2=0.6\ kg\ m/s\\\\Total\ momentum\ of\ both\ the\ balls\ 'Before\ Collision'=1+0.6=1.6\ kg\ m/s

Hence,\\Lets\ now\ consider\ the\ case\ of\ the\ two\ balls\ 'After\ Collision':\\\\Mass\ of\ the\ green\ ball=0.2\ kg\\Final\ Velocity\ of\ the\ green\ ball=0\ m/s\\Final\ Momentum\ of\ the\ green\ ball=0\ kg\ m/s\\\\Mass\ of\ the\ pink\ ball=0.3\ kg\\Final\ Velocity\ of\ the\ pink\ ball=v\ m/s\\Final\ Momentum\ of\ the\ pink\ ball=0.3*v=0.3v\ kg\ m/s\\\\Total\ momentum\ of\ both\ the\ balls\ 'After\ Collision'=0+0.3v=0.3v\ kg\ m/s

As\ we\ know\ that,\\Through\ the\ law\ of\ conservation\ of\ momentum,\\In\ an\ isolated\ system:\\Total\ Momentum\ Before\ Collision=Total\ Momentum\ After\ Collision\\Hence,\\1.6=0.3v\\v=\frac{1.6}{0.3}=5.33\ m/s

5 0
3 years ago
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