What does a star color tell you about the amount of energy it emits?

Light from a helium-neon laser (λ=633nm) passes through a circular aperture and is observed on a screen 4.0 m behind the apertur

devlian [24]

Answer:

d = 0.247 mm

Explanation:

given,

λ = 633 nm

distance from the hole to the screen = L = 4 m

width of the central maximum = 2.5 cm

2 y = 0.025 m

y = 0.0125 m

For circular aperture

$sin \theta = 1.22\dfrac{\lambda}{d}$

using small angle approximation

$\theta = \dfrac{y}{D}$

now,

$\dfrac{y}{D} = 1.22\dfrac{\lambda}{d}$

$y = 1.22\dfrac{\lambda\ D}{d}$

$d = 1.22\dfrac{\lambda\ D}{y}$

$d = 1.22\dfrac{633\times 10^{-9}\times 4}{0.0125}$

d =0.247 x 10⁻³ m

d = 0.247 mm

the diameter of the hole is equal to 0.247 mm

5 0

2 years ago

Which statements about Earth’s core help explain Earth’s magnetic field? Check all that apply.

Stella [2.4K]

All the statements can be used to explain about Earth's magnetic field.

Earth's magnetic field is almost same to that of the bar magnet. The magnetic field of the earth occurs from its core. It has an inner core which is solid in nature. in the outer core, there is a motion of the liquid due to heat flow from the inner core to the core mantle boundary which proves that the outer core is liquid and moving. The Earth's core consists of iron and nickel as the Eddy currents in the iron and nickel liquid helps generating the Earth's magnetic field.

Hence, all the three statements are true and can be used to explain the Earth's magnetic field.

5 0

3 years ago

Read 2 more answers

How tall would a tower need to be if the period of a pendulum were 30. seconds

Virty [35]

We use the following expression

T = 2*pi *sqrt(l/g)

Where T is the period of the pendulum

l is the length of the pendulum

and g the acceleration of gravity

We solve for l

l = [T/2*pi]² *g = [30s/2*pi]²* 9.8 [m/s²] = 223.413 m

The tower would need to be at least 223.413 m high

4 0

2 years ago

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a un

Kamila [148]

Answer:

a) 6738.27 J

b) 61.908 J

c) $\frac{4492.18}{v_{car} ^{2} }$

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

$I$ = $\frac{1}{2}$ $mr^{2}$

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

$I$ = $\frac{1}{2}$ $*11*1.1^{2}$ = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = $Iw^{2}$ = 6.655 x $31.82^{2}$ = 6738.27 J

b) second flywheel has

radius = 2.8 m

mass = 16 kg

moment of inertia is

$I$ = $\frac{1}{2}$ $mr^{2}$ = $\frac{1}{2}$ $*16*2.8^{2}$ = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = $Iw$ = 6.655 x 31.82 = 211.76 kg-m^2-rad/s