Just the the moon seems to shine and the two below it.

Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 3.3 cm . Two of

sweet [91]

Answer:

$1.44\times 10^{-3}$ N

Explanation:

We are given that three charged particle are placed at each corner of equilateral triangle.

$q_1=-8.2 nC,q_2=-16.4 nC,q_3=8.0nC$

$q_1=-8.2\times 10^{-9} C$

$q_2=-16.4\times 10^{-9} C$

$q_3=8.0\times 10^{-9} C$

Side of equilateral triangle =3.3 cm= $\frac{3.3}{100}=0.033m$

We know that each angle of equilateral angle= $60^{\circ}$

Net force=F = $\sum\frac{kQq }{d^2}$

Where k= $9\times 10^9 Nm^2/C^2$

If we bisect the angle at $q_3$ then we have 30 degrees from there to either charge.

Direction of vertical force due to charge $q_1$ and $q_2$

Therefore, force will be added

Vertical force= $9\times 10^9\times q_3(q_1+q_2)\frac{cos30}{(0.033)^2})$

Vertical net force= $9\times 10^9\times 8\times 10^{-9}(-8.2-16.4)\times 10^{-9}\times 10^6\times\frac{\sqrt3}{2\times 1089}$

Vertical force = $9\times 8(-24.6)\times 10^{-9}\times 10^6\times 1.732\times \frac{1}{2178}$

Vertical force= $-1.41\times 10^{-3}N$ (towards $q_1}$

Horizontal component are opposite in direction then will b subtracted

Horizontal force= $9\times 10^9\times 8\times 10^{-9}(-8.2+16.4) \times 10^{-9}\times \frac{sin30}{(0.033)^2}$

Horizontal force= $0.27\times 10^-3}$ N(towards $q_2$

Net electric force acting on particle 3 due to particle = $\sqrt{F^2_x+F^2_y}$

Net force= $\sqrt{(-1.41\times 10^{-3})^2+(0.27\times 10^{-3})^2}$

Net force= $1.44\times 10^{-3}$ N

3 0

2 years ago

Pulsed dye lasers emit light of wavelength 585 nm in 0.45 ms pulses to remove skin blemishes such as birthmarks. The beam is usu

koban [17]

Answer:

a) E₀ = 2.125 eV, b) # photon2 = 9.2 10¹⁵ photons / mm²

Explanation:

a) To calculate the energy of a photon we use Planck's education

E = h f

And the ratio of the speed of light

c = λ f

We replace

E = h c /λ

Let's calculate

E₀ = 6.63 10⁻³⁴ 3 10⁸/585 10⁻⁹

E₀ = 3.40 10⁻¹⁹ J

Let's reduce

E₀ = 3.4 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)

E₀ = 2.125 eV

b) Let's look for the energy in each pulse

P = E / t

E = P t

E = 20.0 0.45 10⁻³

E = 9 10⁻³ J

let's use a ratio of proportions (rule of three) if we have the energy of a photon (E₀), to have the energy of 9 10⁻³ J

# photon = 9 10⁻³ /3.40 10⁻¹⁹

# photon = 2.65 10¹⁶ photons

Let's calculate the areas

Focus area

A₁ = π r²

A₁ = π (3.4/2)²

A₁ = 9,079 mm²2

Area requested for calculation r = 1 mm

A₂ = π 1²

A₂ = 3.1459 mm²

Let's use another rule of three. If we have 2.65 106 photons in an area A1 how many photons in an area A2

# photon2 = 2.65 10¹⁶ 3.1459 / 9.079

# photon2 = 9.2 10¹⁵ photons / mm²

8 0

2 years ago

True or False. Electromagnetic waves travel fastest through a vacuum.

olya-2409 [2.1K]

This
is
true
Electromagnetic waves travel fastest through vacuum

4 0

3 years ago

A 215-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope a

Misha Larkins [42]

Answer:

303.9481875 N

Explanation:

t = Time taken = 2 seconds

F = Force

r = Radius = 1.5 m

I = Moment of Inertia

$\alpha$ = Angular Acceleration

Torque

$\tau=F\times r$

$\tau=I\times \alpha$

$\\\Rightarrow F\times r=I\times \alpha\\\Rightarrow F=\frac{I\times \alpha}{r}$

Angular velocity

$\omega=rev/s\times 2\pi\\\Rightarrow \omega=0.6\times 2\pi\\\Rightarrow \omega=3.76991\ rad/s$

Angular acceleration

$\alpha=\frac{\omega}{t}\\\Rightarrow \alpha=\frac{3.76991}{2}\\\Rightarrow \alpha=1.88495\ rad/s^2$

$I=\frac{1}{2}mr^2\\\Rightarrow I=\frac{1}{2}215\times 1.5^2\\\Rightarrow I=241.875\ kgm^2$

$F=\frac{I\times \alpha}{r}\\\Rightarrow F=\frac{241.875\times 1.88495}{1.5}\\\Rightarrow F=303.9481875\ N$

The magnitude of the force to stop the merry-go-round is 303.9481875 N

3 0

3 years ago

A 195 g glider is moving at 5.3 m/s on a frictionless air track. It then collides with a stationary 295 g glider.

lord [1]

Answer:For a perfectly elastic collision, the final velocities of the carts will each be 1/2 the velocity of the initial velocity of the moving cart. For a perfectly inelastic collision, the final velocity of the cart system will be 1/2 the initial velocity of the moving cart.

Explanation:

3 0

2 years ago