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2 years ago

What does the wavelength of light determine?

Physics

1 answer:

sergejj [24]2 years ago

8 0

Answer:

The wavelength of the light determines its characteristics. For example, short wavelengths are high energy gamma-rays and x-rays, long wavelengths are low energy radio waves. The whole range of wavelengths is called the electromagnetic spectrum.

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In a mass spectrometer a particle of mass m and charge q is accelerated through a potential difference V and allowed to enter a

Ghella [55]

Answer:

m = B²qR² / 2 V

Explanation:

If v be the velocity after acceleration under potential difference of V

kinetic energy = loss of electric potential energy

1/2 m v² = Vq ,

v² = 2 Vq / m ----------------------- ( 1 )

In magnetic field , charged particle comes in circular motion in which magnetic force provides centripetal force

magnetic force = centripetal force

Bqv = mv² / R

v = BqR / m

v² = B²q²R² / m² ------------------------- (2)

from (1) and (2)

B²q²R² / m² = 2 Vq / m

m = B²q²R² / 2 Vq

m = B²qR² / 2 V

7 0

3 years ago

The scientist to first introduce the concept of inertia was

skelet666 [1.2K]

The first scientist to introduce the concept of inertia was Galileo

4 0

3 years ago

A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo

-BARSIC- [3]

Answer:

It would take $\tau(\ln 9 - \ln 8)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{8}{9} \, q_0$ .

It would take $\tau(\ln 9 - \ln 7)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{7}{9}\, q_0$ .

Note that $\ln 9 = 2\,\ln 3$ , and that $\ln 8 = 3\, \ln 2$ .

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is $\tau$ , and the initial charge of the capacitor be $q_0$ . Then at time $t$ , the charge stored in the capacitor would be:

$\displaystyle q(t) = q_0 \, e^{-t / \tau}$ .

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{8}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{8}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{8}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{8}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 8 - \ln 9$ .

$t = - \tau \, \left(\ln 8 - \ln 9\right) = \tau(\ln 9 - \ln 8)$ .

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{7}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{7}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{7}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 7 - \ln 9$ .

$t = - \tau \, \left(\ln 7 - \ln 9\right) = \tau(\ln 9 - \ln 7)$ .

7 0

3 years ago

Adhira loves to ride her bike around the neighborhood. She starts riding 1.2 miles at 30° S of E. Then, she rides another 2.0 mi

zepelin [54]

Answer:

D = 1.8677 miles , θ = 24.28º at South of West

Explanation:

This is an exercise in adding vectors, the easiest way to solve them is to decompose the vectors and add each component algebraically. Let's use trigonometry

first displacement. d = 1.2 miles to 30º south of East

cos ( 360-30) = cos (-30) = x₁ / d

sin (-30) = y₁ / d

x₁ = d cos (-30)

y₁ = d sin (-30)

x₁ = 1.2 cos (-30) = 1,039 miles

y₁ = 1.2 sin (-30) = -0.6 miles

second shift. d = 2.0 miles to 20º West of South

cos (270-20) = x₂ / d

cos (250) = y₂ / d

x₂ = 2.0 cos 250 = -0.684 miles

y₂ = 2.0 sin250 = -1.879 miles

Third displacement. d = 1.6 miles to 30º South of West

cos (180 + 30) = x₃ / d

sin (210) = y₃ / d

x₃ = 1.6 cos 210 = -1.3856 miles

y₃ = 1.6 sin 210 = -0.8 miles

Fourth displacement. d = 2.6 miles to 15º West of North

cos (90 + 15) = x₄ / d

sin (105) = y₄ / d

x₄ = 2.6 cos 105 = -0.6729 miles

y₄ = 2.6 sin 105 = 2,511 miles

having all the components we add

x-axis (West-East direction)

X = x₁ + x₂ + x₃ + x₄

X = 1.039 -0.684 - 1.3846 - 0.6729

X = -1.7025 miles

Y = y₁ + y₂ + y₃ + y₄

Y = -0.6 -1.879 -0.8 +2.511

Y = -0.768

The modulus of this displacement is we use the Pythagorean theorem

D = √ (X² + Y²)

D = √ (1.7025² + 0.768²)

D = 1.8677 miles

let's use trigonometry to find the direction

tan θ = Y / X

θ = tan⁻¹ Y / x

θ = tan⁻¹ (0.768 / 1.7025)

θ = 24.28º

as the two components are negative this angle is in the third quadrant

therefore in cardinal direction form is

θ = 24.28º at South of West

4 0

3 years ago

Why can you see your own reflected image in a mirror but not on a dry, painted wall?

kaheart [24]

A wall uses diffuse reflection while a mirror uses specular reflection. For example, when parallel light rays enter a mirror, they remain parallel when exiting the mirror, allowing you to see a reflection of the light rays. On the contrary, when incident light rays enter a wall which is painted, the rays scatter, not allowing you to see anything but a painted wall.

7 0

3 years ago