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2 years ago

At what rate must electrons in a wire vibrate to emit microwaves with a wavelength of 1.00 mm

2 answers:

7 0

The equation to use is v=fλ where v is the speed of light, f is the frequency, i.e. the rate at which they vibrate, and λ is the wavelength in meters.

Rearrange the above equation and you get f=v/λ. So the answer is f=300000000/0.001 or 300GHz.

AnnyKZ [126]2 years ago

5 0

Answer:

f = 3×10¹¹ Hz

Explanation:

Relation ship between frequency and wavelength

"The wave speed (v) is defined as the distance traveled by a wave per unit time. If considered that the wave travels a distance of one wavelength in one period,

ν=λ/T

As we know that T = 1/f, hence we can express the above equation as,

V = f λ

The wave speed is equal to the product of its frequency and wavelength, and this implies the relationship between frequency and wavelength."

The relation between frequency and wavelength is

λ×f = c

c = speed of light = 3×10⁸ m/s

λ = 1.00 mm = 10⁻³ m

f=c/λ

f=(3×10⁸ m/s)/ 10⁻³ m

f = 3×10¹¹ Hz

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How, if at all, would the equations written in Parts C and E change if the projectile was thrown from the cliff at an angle abov

sveta [45]

Answer:

x = v₀ cos θ t , y = y₀ + v₀ sin θ t - ½ g t2

Explanation:

This is a projectile launch exercise, in this case we will write the equations for the x and y axes

Let's use trigonometry to find the components of the initial velocity

sin θ = $v_{oy}$ / v₀

cos θ = v₀ₓ / v₀

v_{y} = v_{oy} sin θ

v₀ₓ = vo cos θ

now let's write the equations of motion

X axis

x = v₀ₓ t

x = v₀ cos θ t

vₓ = v₀ cos θ

Y axis

y = y₀ + $v_{oy}$ t - ½ g t2

y = y₀ + v₀ sin θ t - ½ g t2

v_{y} = v₀ - g t

v_{y} = v₀ sin θ - gt

$v_{y}^{2}$ = v_{oy}^2 sin² θ - 2 g y

As we can see the fundamental change is that between the horizontal launch and the inclined launch, the velocity has components

7 0

2 years ago

A satellite, orbiting the earth at the equator at an altitude of 400 km, has an antenna that can be modeled as a 1.76-m-long rod

ivann1987 [24]

Answer:

The inducerd emf is 1.08 V

Solution:

As per the question:

Altitude of the satellite, H = 400 km

Length of the antenna, l = 1.76 m

Magnetic field, B = $8.0\times 10^{- 5}\ T$

Now,

When a conducting rod moves in a uniform magnetic field linearly with velocity, v, then the potential difference due to its motion is given by:

$e = - l(vec{v}\times \vec{B})$

Here, velocity v is perpendicular to the rod

Thus

e = lvB (1)

For the orbital velocity of the satellite at an altitude, H:

$v = \sqrt{\frac{Gm_{E}}{R_{E}} + H}$

where

G = Gravitational constant

$m_{e} = 5.972\times 10^{24}\ kg$ = mass of earth

$R_{E} = 6371\ km$ = radius of earth

$v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}}{6371\times 1000 + 400\times 1000} = 7670.018\ m/s$

Using this value value in eqn (1):

$e = 1.76\times 7670.018\times 8.0\times 10^{- 5} = 1.08\ V$

5 0

2 years ago

To exercise, a man attaches a 4.0 kg weight to the heel of his foot. When his leg is stretched out before him, what is the torqu

Masja [62]

Answer:

B. τ = 16 Nm

Explanation:

In order to find the torque exerted by the weight attached to the heel of man's foot, when his leg is stretched out. We use following formula:

τ = Fd

here,

τ = Torque = ?

F = Force exerted by the weight = Weight = mg

F = mg = (4 kg)(10 m/s²) = 40 N

d = distance from knee to weight = 40 cm = 0.4 m

Therefore,

τ = (40 N)(0.4 m)

8 0

2 years ago

Your 64-cm-diameter car tire is rotating at 3.5 rev/s when suddenly you press down hard on the accelerator. After traveling 200

andreyandreev [35.5K]

Answer: angular acceleration = 0.748rad/s²

Explanation: according to the question, our answer needs to be in rad/s², thus all units in rev/s will be converted to rad/s

Assuming the motion of the object is of a constant angular acceleration, then newton's laws of motion is applicable.

The formulae below is used

v² = u² + 2αθ

v = final angular speed =6rev/s = 6*2π = 12π rad/s

u =initial angular speed =3.5rev/s = 3.5 *2π = 7π rad/s

Note 1 rev = 2π rad.

α = angular acceleration.

θ = angular displacement.

Diameter = 64cm = 0.64m, radius = 64/2 = 32cm = 0.32m

The angular displacement can be gotten using the formulae below

S = rθ, where s= linear distance covered = 200m, r = radius = 0.32m

θ = S/r = 200/0.32=625 rad.

By substituting the parameter we have that

(12π)² = (7π)² + 2α(625)

1421.22 = 486.31 + 1250α

1421.22 - 486.31 = 1250α

934.91 = 1250α

α = 934.91/1250

α= 0.748 rad/s²

4 0

2 years ago

A flat uniform circular disk (radius = 2.00 m, mass = 1.00

Ostrovityanka [42]

Answer:

The resulting angular speed of the disk is 0.5 rad/s

Explanation:

Step 1: Data given

Radius of the circular disk = 2.00 meters

Mass of the circular disk = 1.00

Mass op the person = 40.0 kg

Distance from the axis = 1.25 m

tangential speed = 2.00 m/s

Step 2:

There is no external torque acting on the system so we can apply the law of conservation of angular momentum In this case the momentum is conserved.

Angular momentum of the man = Iω

⇒ With I = Inertia of the man about the axis of rotation = M*r²

⇒ I = 40 *1.25² = 62.5

⇒ with ω = Angular velocity of the man

⇒ v = 2m/s

⇒ Circumference of the circle = 2πr = 2 * 3.14 * 1.25 = 7.85m

⇒The time to describe this circle t = 2πr/ v

⇒ in 1 revolution the angle θ = 2π radians

This angle is subtended in time t = 2πr/ v

⇒ The angular speed = ω = θ/t = 2π ( v/ 2πr) = v/r = 2/1.25 = 1.6 rad/s

⇒ The angular momentum of man = I*ω = 62.5 * 1.6 = 100

Since the angular momentum is conserved, before and after the man starts running we have :

Angular momentum of disk = angular momentum of the man

⇒ with Angular momentum of disk = Idisk ωdisk

⇒ Idisk = MdiskR

⇒ with Angular momentum of disk = 100

or I(disk)*ω(disk) = 100

I(disk) = M(disk)*R ²/2 = 100*2*2 / 2 = 200

⇒ with M(disk) = the mass of the disk = 1.00 * 10² kg

⇒ with R = the radius of the disk = 2.00 m

200 ωdisk = 100

ωdisk = 100/200 = 0.5 rad/s

The resulting angular speed of the disk is 0.5 rad/s

(Since the angular speed is positive, the rotation is counterclockwise)

5 0

2 years ago