A cyclist and his bicycle have a combined mass of 85 kg. If he is moving at

A beam of light has a wavelength of 650 nm in vacuum. (a) What is the speed of this light in a liquid whose index of refraction

Lady_Fox [76]

Answer:

The speed of this light and wavelength in a liquid are $2.04\times10^{8}\ m/s$ and 442 nm.

Explanation:

Given that,

Wavelength = 650 nm

Index refraction = 1.47

(a). We need to calculate the speed

Using formula of speed

$n = \dfrac{c}{v}$

Where, n = refraction index

c = speed of light in vacuum

v = speed of light in medium

Put the value into the formula

$1.47=\dfrac{3\times10^{8}}{v}$

$v=\dfrac{3\times10^{8}}{1.47}$

$v= 2.04\times10^{8}\ m/s$

(b). We need to calculate the wavelength

Using formula of wavelength

$n=\dfrac{\lambda_{0}}{\lambda}$

$\lambda=\dfrac{\lambda_{0}}{n}$

Where, $\lambda_{0}$ = wavelength in vacuum

$\lambda$ = wavelength in medium

Put the value into the formula

$\lambda=\dfrac{650\times10^{-9}}{1.47}$

$\lambda=442\times10^{-9}\ m$

Hence, The speed of this light and wavelength in a liquid are $2.04\times10^{8}\ m/s$ and 442 nm.

3 0

3 years ago

There are 2 concentric cylinders. These cylinders are very long with length L. The inner cylinder has a radius R1 and is a solid

OLga [1]

To get the charge along the inner cylinder, we use Gauss Law
E = d R1/2εo
For the outer cylinder the charge can be calculated using
E = d R2^2/2εoR1
where d is the charge density
Use these two equations to get the charge in between the cylinders and the capacitance between them.

5 0

3 years ago

Two parallel 3.0-meter long wires conduct current. The current in the top wire is 12.5 A and flows to the right. The top wire fe

Aleksandr [31]

Complete question:

Two parallel 3.0-meter long wires conduct current. The current in the top wire is 12.5 A and flows to the right. The top wire feels a repulsive force of 2.4 x 10^-4 N created by the interaction of the 12.5 A current and the magnetic field created by the bottom current (I). Find the magnitude and direction of the bottom current, if the distance between the two wires is 40cm.

Answer:

The bottom current is 12.8 A to the right.

Explanation:

Given;

length of the wires, L = 3.0 m

current in the top wire, I₁ = 12.5 A

repulsive force between the two wires, F = 2.4 x 10⁻⁴ N

distance between the two wires, r = 40 cm = 0.4 m

The repulsive force between the two wires is given by;

$F = \frac{\mu_oI_1I_2L}{2\pi r}\\\\I_{2} = \frac{2F\pi r}{\mu_oI_1L}$

Where;

I₂ is the bottom current

The direction of the bottom current must be in the same direction as the top current since the force between the two wires is repulsive.

$I_{2} = \frac{2F\pi r}{\mu_oI_1L}\\\\I_{2} = \frac{2(2.4*10^{-4})(\pi)(0.4)}{(4\pi*10^{-7})(12.5)(3)}\\\\I_{2} = 12.8 \ A$

Therefore, the bottom current is 12.8 A to the right.

3 0

2 years ago

Betty is six years old and often asks her parents many questions. Her parents respond to her questions however trivial they may

Brums [2.3K]

This behavior helps Betty in <u>intellectual </u>development.

5 0

2 years ago

Read 2 more answers

What is the definition of work when net force is parallel to the distance?

goldfiish [28.3K]

Work in general is given by W=F·d where F is the force vector and d is the displacement vector. The dot symbol is the dot product which is a measure of how parallel two vectors are. It can be replaced by the cosine of the angle between the two vectors and the vectors replaced by their magnitudes. If F and d are parallel then the angle is zero and the cosine is unity. So in this case work can be defined as the product of the magnitudes of the force and distance:
W=Fd

6 0

3 years ago