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Bad White [126]
3 years ago
12

Light with a frequency of 7.18 Ã 1014 hz lies in the violet region of the visible spectrum. what is the wavelength of this frequ

ency of light? answer in units of nm.
Physics
1 answer:
andre [41]3 years ago
6 0
The frequency is 7.18 × 10^14 Hz, c= 3 × 10^8 m/s, where c is the speed of light.
But; c = λf, where λ is the wavelength of the wave and f is the frequency.
Making λ the subject, we get;
λ = c/f
  = (3 ×10^8)/ (7.18 ×10^14)
  = 4.17827 × 10^-7 m
but 1 m = 10^9 nm
Therefore;
 = (4.17827 ×10^-7) × 10^9
 = 417.827 nm
Hence, the wavelength= 417.827 nm
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Answer:

The correct option is (b).

Explanation:

Given that,

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We know that, the electric force is given by :

F=qE\\\\=1.6\times 10^{-19}\times 2\times 10^2\\\\F=3.2\times 10^{-17}\ N

So, the required force on the electron is equal to 3.2\times 10^{-17}\ N.

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3 years ago
A QL = -26 μC charge is placed on the x-axis at x = - 0.2 m. A QR = 26 μC charge is placed at x = +0.2 m. (for all answers below
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4 0
3 years ago
An engine has an energy input of 125 J, and 35 J of that energy is transformed into useful energy. What is the efficiency of the
Korolek [52]
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8 0
4 years ago
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Which of the following changes would decrease the coefficient of friction needed for this ride?
posledela
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7 0
3 years ago
If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h
raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory (u_x) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

u^2=u_x^2+u_y^2

Where, u_x\ and\ u_y are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1

Now, we know that, for a projectile motion, the maximum height is given as:

H=\frac{u_y^2}{2g}

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m

Therefore, the maximum height reached is 4.63 m.

3 0
4 years ago
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