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3 years ago

Light waves have constructive interference because they are emitted _____.

1 answer:

3 0

Constructive Interference occurs when the crust of one way falls on the crust of other and the trough of one wave falls on the trough of other. This way the points of the wave re-enforce each other resulting in an increased amplitude of the wave.

Since the crust and troughs of two waves are at the same place at the same time, the waves are said to be in phase. So the correct answer to the question is option A.

Light waves have constructive interference because they are emitted <span>In Phase</span>

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In physics, why would an equation like y=mx+b be plotted as a straight line rather than as a parabola, as is done in math?

notka56 [123]

Answer:

Explanation:

In Both Physics and Math

y=mx+b is plotted as straight line where

m=slope of line

b=intercept on Y-axis

whereas Equation of parabola is something like this

$y^2=4ax$

$x^2=4ay$

Math is a tool to solve Physics problems so equations are same in math and physics

3 0

2 years ago

A solid conducting sphere of radius 2.00 cm has a charge of 6.88 μC. A conducting spherical shell of inner radius 4.00 cm and ou

zepelin [54]

Explanation:

Given that,

Radius R= 2.00

Charge = 6.88 μC

Inner radius = 4.00 cm

Outer radius = 5.00 cm

Charge = -2.96 μC

We need to calculate the electric field

Using formula of electric field

$E=\dfrac{kq}{r^2}$

(a). For, r = 1.00 cm

Here, r<R

So, E = 0

The electric field does not exist inside the sphere.

(b). For, r = 3.00 cm

Here, r >R

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times6.88\times10^{-6}}{(3.00\times10^{-2})^2}$

$E=6.88\times10^{7}\ N/C$

The electric field outside the solid conducting sphere and the direction is towards sphere.

(c). For, r = 4.50 cm

Here, r lies between R₁ and R₂.

So, E = 0

The electric field does not exist inside the conducting material

(d). For, r = 7.00 cm

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times(-2.96\times10^{-6})}{(7.00\times10^{-2})^2}$

$E=5.43\times10^{6}\ N/C$

The electric field outside the solid conducting sphere and direction is away of solid sphere.

Hence, This is the required solution.

6 0

2 years ago

What is the change in electric potential energy when the dipole moment of a molecule changes its orientation with respect to e⃗

AleksandrR [38]

Electric potential energy of a dipole is given as

$U = - P.E$

$U = - PE cos\theta$

so change in the potential energy is given by

$U_f - U_i = (-PEcos\theta_2)- (-PEcos\theta_1)$

$\Delta U = PE (cos\theta_1 - cos\theta_2)$

here initially it was parallel to electric field while finally it is perpendicular to the electric field

so we have

$\Delta U = PE(cos 0 - cos 90)$

$\Delta U = PE$

so above is the change in potential energy of dipole

8 0

2 years ago

If a golf ball is swinging on a pendulum on the planet mars (g=3.711 m/s²) and it takes 7 seconds to complete one oscillation, w

sertanlavr [38]

T = 2*pi*Sqrt (L/g)

T = Period = Time to complete one oscillation, L = Length of the pendulum, g = gravitational acceleration.

Then,

L = {T/(2*pi)}^2*g = {7/(2*pi)}^2*3.711 = 4.606 m

4 0

3 years ago

Saturn has an orbital period of 29.46 years. In two or more complete sentences, explain how to calculate the average distance fr

soldi70 [24.7K]

$Given:\\T=29.46y\approx 9.29\cdot 10^8s\\M_S\approx2.0\cdot 10^{30}kg\\G=6.67\cdot 10^{-11} \frac{m^3}{kg\cdot s^2} \\\\Find:\\R=?\\\\Solution:\\\\F_g= G\frac{mM_s}{R^2} \\\\F_c= \frac{mv^2}{R} \\\\F_g=F_c\\\\G\frac{mM_s}{R^2}=\frac{mv^2}{R} \Rightarrow G\frac{M_s}{R^2}=\frac{v^2}{R}\\\\v=\omega r\\\\G\frac{M_s}{R^2}= \frac{\omega^2R^2}{R}\Rightarrow G\frac{M_s}{R^2}=\omega^2R \\\\\omega= \frac{2 \pi }{T} \\\\G\frac{M_s}{R^3}= \frac{4 \pi ^2}{T^2}$

$GM_ST^2=4 \pi ^2R^3\Rightarrow R= \sqrt[3]{ \frac{GM_ST^2}{4 \pi ^2} }\\\\\\R= \sqrt[3]{ \frac{6.67\cdot 10^{-11} \frac{m^3}{kg\cdot s^2}\cdot2.0\cdot 10^{30}kg( 9.29\cdot 10^8s)^2}{4\cdot 3.14^2} } \approx 1.42\cdot 10^{12}m$

3 0

3 years ago