3 years ago

State the objects in the universe that give out their own light

Physics

1 answer:

Valentin [98]3 years ago

3 0

All of those are stars.

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2 years ago

A diffraction pattern forms when light passes through a single slit. The wavelength of the light is 691 nm. Determine the angle

expeople1 [14]

Explanation:

Given that,

Wavelength of the light, $\lambda=691\ nm=691\times 10^{-9}\ m$

(a) Slit width, $a=3.8\times 10^{-4}\ m$

The angle that locates the first dark fringe is given by :

$sin\theta=\dfrac{\lambda}{a}$

$sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-4}}$

$\theta=0.104^{\circ}$

(b) Slit width, $a=3.8\times 10^{-6}\ m$

The angle that locates the first dark fringe is given by :

$sin\theta=\dfrac{\lambda}{a}$

$sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-6}}$

$\theta=10.47^{\circ}$

Hence, this is the required solution.

7 0

3 years ago

In the diagram, q1 = +6.60*10^-9 C and q2 = +3.10*10^-9 C. Find the magnitude of the total electric field at point P. pls help?

kirza4 [7]

Answer:

|E(t)| = 1258,46 [N/C]

α = 67,5⁰ (angle with respect x-axis)

Explanation:

E(t) Electric Field is a vector, so we need to determine module and direction

E(t) = E(q₁) + E (q₂) Where E(q₁) and E (q₂) are electric fields due to electric charge q₁ and q₂ respectively.

E(q₁) = K * q₁/ (d₁)² K = 9 *10⁹ [N*m²/C²] d₁ = 0,350 m

E(q₁) = 9 *10⁹ * 6,6*10⁻⁹ / 0,1225 [N*m²/C²] *C/m²

E(q₁) = 484,9 [N/C]

E(q₂) = 9 *10⁹ * 3,1*10⁻⁹ / 0,024025

E(q₂) = 1161,29

Then

|E(t)| = √ |Eq₁|² + |Eq₂|²

|E(t)| = √ ( 484,9)² +( 1161,29)²

|E(t)| = √ 235128 + 1348594,46

|E(t)| = 1258,46 [N/C]

And tanα = 1161,29/484,9 tanα = 2,395 α = 67,5⁰

The angle of the vector electric field with the x-axis

3 0

3 years ago

Read 2 more answers