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3 years ago

8. What is the frequency of green light waves that have a wavelength of 5.2 x 10-7 m.? The speed of light is 3.0 x 108 m/s

Physics

1 answer:

o-na [289]3 years ago

4 0

Answer:

$f=5.76\times 10^{14}\ Hz$

Explanation:

We need to find the frequency of green light having wavelength o $5.2\times 10^{-7}\ m$ . It can be calculated as follows :

$c=f\lambda\\\\f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{5.2\times 10^{-7}}\\\\f=5.76\times 10^{14}\ Hz$

So, the required frequency of green light is equal to $5.76\times 10^{14}\ Hz$ .

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What is NOT a major factor affecting climate?

professor190 [17]

B Longitude

Explanation:

The line of longitude or longitude does not affects climates.

Climate is the weather condition over a place for a long period of time. It usually takes several years to delineate the climate of an area.

Longitude is the distance on the earth surface that runs form east to west.
All lines of longitude are all great circles.
Latitude are lines from north to south on the earth surface. Only the equator is a great circle.
Latitude affects climate. In fact based on the division of lines of latitude on can predict climatic patterns on earth.
Proximity to water bodies also affects climatic condition because of the effect of land and sea breeze and other climatic factors.
Elevation affects climate a whole lot. Physiography of a place determines the weather to a very large extent.

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8 0

3 years ago

Synonym of an applied force

tatuchka [14]

Answer:

Coerce, compel, constrain

Explanation:

7 0

3 years ago

Some fish have a density slightly less than that of water and must exert a force (swim) to stay submerged. What force must an 85

vichka [17]

The force require to keep grouper submerged is 8.207N.

According to Archimedes principle buoyant force of any object must equal to weight of fluid it displaced.

The expression for the force exerted to stay submerged in salt water is

F = F(b) - w(fish)

where F(b) = buoyant force

w(fish) = weight

now substitute w(b) for F(b)

→ F = Vρg - w(fish)

where V = volume of sea water

ρ = density of sea water

Now by Archimedes principle V = m(fish) / ρ(fish)

→ F = (m(fish) / ρ (fish) ) ρg - m(fish)g

F = (85 kg/1015 kg-m^-3) (1.025× 10³ kg-m^-3) (9.8 m/s^2)

- (85kg) × 9.8 m/s^2

F = 841.207N - 833N

F = 8.207 N

Hence, the force require to keep grouper submerged is 8.207N.

Learn more about Archimedes Principle here:

brainly.com/question/15076878

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7 0

2 years ago

A 100-kg tackler moving at a speed of 2.6 m/s meets head-on (and holds on to) an 92-kg halfback moving at a speed of 5.0 m/s. Pa

DIA [1.3K]

Given that,

Mass of trackler, m₁ = 100 kg

Speed of trackler, u₁ = 2.6 m/s

Mass of halfback, m₂ = 92 kg

Speed of halfback, u₂ = -5 m/s (direction is opposite)

To find,

Mutual speed immediately after the collision.

Solution,

The momentum of the system remains conserved in this case. Let v is the mutual speed after the collision. Using conservation of momentum as :

$m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{100\times 2.6+92\times (-5)}{(100+92)}\\\\V=-1.04\ m/s$

So, the mutual speed immediately after the collision is 1.04 m/s but in opposite direction.

3 0

3 years ago

By what potential difference must a proton [m_0 = 1.67E-27 kg) be accelerated to have a wavelength lambda = 4.23E-12 m? By what

Vinil7 [7]

Explanation:

1. Mass of the proton, $m_p=1.67\times 10^{-27}\ kg$

Wavelength, $\lambda_p=4.23\times 10^{-12}\ m$

We need to find the potential difference. The relationship between potential difference and wavelength is given by :

$\lambda=\dfrac{h}{\sqrt{2m_pq_pV}}$

$V=\dfrac{h^2}{2q_pm_p\lambda^2}$

$V=\dfrac{(6.62\times 10^{-34})^2}{2\times 1.6\times 10^{-19}\times 1.67\times 10^{-27}\times (4.23\times 10^{-12})^2}$

V = 45.83 volts

2. Mass of the electron, $m_p=9.1\times 10^{-31}\ kg$

Wavelength, $\lambda_p=4.23\times 10^{-12}\ m$

We need to find the potential difference. The relationship between potential difference and wavelength is given by :

$\lambda=\dfrac{h}{\sqrt{2m_eq_eV}}$

$V=\dfrac{h^2}{2q_em_e\lambda^2}$

$V=\dfrac{(6.62\times 10^{-34})^2}{2\times 1.6\times 10^{-19}\times 9.1\times 10^{-31}\times (4.23\times 10^{-12})^2}$

$V=6.92\times 10^{34}\ V$

V = 84109.27 volt

Hence, this is the required solution.

7 0

3 years ago