What is the answer, friends?

The reflective surface of a CD consists of spirals of equally spaced grooves. If you shine a laser pointer on a CD, each groove

Ipatiy [6.2K]

Answer:

d = 1.55 * 10⁻⁶ m

Explanation:

To calculate the distance between the adjacent grooves of the CD, use the formula, $d = \frac{m \lambda}{sin(A_{m}) }$ ..........(1)

The fringe number, m = 1 since it is a first order maximum

The wavelength of the green laser pointer, $\lambda$ = 532 nm = 532 * 10⁻⁹ m

Distance between the central maximum and the first order maximum = 1.1 m

Distance between the screen and the CD = 3 m

$A_{m}$ = Angle between the incident light and the diffracted light

From the setup shown in the attachment, it is a right angled triangle in which

$sin(A_{m}) = \frac{opposite}{Hypotenuse} \\sin(A_{m}) =\frac{1.1}{\sqrt{1.1^{2}+3^{2}}}$

$sin(A_{m} ) = 0.344\\A_{m} = sin^{-1} 0.344\\A_{m} = 20.14^{0}$

Putting all appropriate values into equation (1)

$d = \frac{1* 532*10^{-9} }{0.344 }\\d = 0.00000155 m\\d = 1.55 * 10^{-6} m$

3 0

3 years ago

if a stone is projected at an angle of 50 degrees to the horizontal with an initial velocity of 50m/s, what is the vertical comp

Vilka [71]

Answer:

38.3 m/s

Explanation:

To find vertical component of initial velocity, you'd have to use sine ratio:

$\displaystyle{\sin \theta = \dfrac{u_y}{u}}$

$\displaystyle{u_y}$ is vertical component of initial velocity and $\displaystyle{u}$ is initial velocity given which is 50 m/s.

A stone is projected at an angle of 50 degrees so $\displaystyle{\theta}$ = 50°. Substitute in the formula:

$\displaystyle{\sin 50^{\circ} = \dfrac{u_y}{50}}\\\\\displaystyle{50 \sin 50^{\circ} = u_y}\\\\\displaystyle{u_y = 38.3 \ \, \sf{m/s}}$

Therefore, the vertical component of initial velocity is approximately 38.3 m/s

(The picture is also attached for visual reference!)

3 0

2 years ago

Which of the following could be the source of resistance in a household electric circuit?

Tpy6a [65]

D. all of these

all of these use electricity

Hope I helped!

8 0

3 years ago

Read 2 more answers

On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini

irinina [24]

Answer:

$v_a=0.8176 m/s$

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being $m_a$ and $m_b$ the masses of pucks a and b respectively, the initial momentum of the system is

$M_1=m_av_a+m_bv_b$

Since b is initially at rest

$M_1=m_av_a$

After the collision and being $v'_a$ and $v'_b$ the respective velocities, the total momentum is

$M_2=m_av'_a+m_bv'_b$

Both momentums are equal, thus

$m_av_a=m_av'_a+m_bv'_b$

Solving for $v_a$

$v_a=\frac{m_av'_a+m_bv'_b}{m_a}$

$v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}$

$v_a=0.8176 m/s$

The initial kinetic energy can be found as (provided puck b is at rest)

$K_1=\frac{1}{2}m_av_a^2$

$K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J$

The final kinetic energy is

$K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2$

$K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J$

The change of kinetic energy is

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

3 0

3 years ago

A block of mass, m, sits on the ground. A student pulls up on

kakasveta [241]

Answer a

Explanation: a

3 0

3 years ago