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sashaice [31]
3 years ago
9

Microwaves travel at the speed of light, 300,000 km/s. What is the wavelength of a microwave received at 1000 gigahertz?

Physics
1 answer:
n200080 [17]3 years ago
7 0

Answer:

3 x 10⁻⁴m

Explanation:

Given parameters:

Speed of light  = 300000km/s

Frequency  = 1000gigahz  = 1000 x 10⁹Hz

Unknown:

Wavelength = ?

Solution:

To solve this problem, we use the expression below:

    V  = F ∧  

V is the speed

F is the frequency

∧ is the wavelength

      300000 = 1000 x 10⁹ x   ∧  

Wavelength  = 3 x 10⁻⁷km = 3 x 10⁻⁴m

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You place a point charge q = -4.00 nC a distance of 9.00 cm from an infinitely long, thin wire that has linear charge density 3.
valentinak56 [21]

Answer:

F=6\times 10^{-7}\ N

Explanation:

Given:

  • quantity of point charge, q=-4\times 10^{-9}\ C
  • radial distance from the linear charge, r=0.09\ m
  • linear charge density, \lambda=3\times 10^{-9}\ C.m^{-1}

<u>We know that the electric field by the linear charge  is given as:</u>

E=\frac{\lambda}{2\pi.\epsilon_0.r}

E=\frac{1}{2}\times 9\times 10^9\times \frac{3\times10^{-9}}{0.09}

E=150\ N.C^{-1}

<u>Now the force on the given charge can be given as:</u>

F=E.q

F=150\times 4\times 10^{-9}

F=6\times 10^{-7}\ N

3 0
4 years ago
Solids diffuse because the particles cannot move.
Nastasia [14]

Solids cannot diffuse.

4 0
3 years ago
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A sealed test tube traps 25.0 cm3 of air at a pressure of 1.00 atm and temperature of 18°C. The test tube’s stopper has a diamet
puteri [66]

Answer:

180° C

Explanation:

First we start by finding the area of the stopper.

A = πd²/4, where d = 1.5 cm = 0.015 m

A = 3.142 * 0.015² * ¼

A = 1.767*10^-4 m²

Next we find the force on the stopper

F = (P - P•)A, where

F = 10 N

P = pressure inside the tube,

P• = 1 atm

10 = (P - 101325) * 1.767*10^-4

P - 101325 = 10/1.767*10^-4

P - 101325 = 56593

P = 56593 + 101325

P = 157918 Pascal

Now, remember, in an ideal gas,

P1V1/T1 = P2V2/T2, where V is constant, then we have

P1/T1 = P2/T2, and when we substitute the values, we have

101325/(273 + 18) = 157918/ T2

101325/291 = 157918/ T2

T2 = (157918 * 291)/101325

T2 = 453 K

T2 = 453 - 273 = 180° C

3 0
4 years ago
A 2,000 kg car starts from rest and coasts down from the top of a 5.00 m long driveway that is sloped at an angel of 20o with th
ehidna [41]

Answer:

The speed at the bottom of the driveway is3.67m/s.

Explanation:

Height,h= 5sin20°= 1.71m

Potential energy PE=mgh= 2000×9.8×1.71

PE= 33516J

KE= PE- Fk ×d

0.5mv^2= 33516 - (4000×5)

0.5×2000v^2= 33516 - 20000

1000v^2= 13516

v^2= 13516/1000

v =sqrt 13.516

v =3.67m/s

5 0
3 years ago
A thin taut string is fixed at both ends and stretched along the horizontal x-axis with its left end at x = 0. It is vibrating i
Fofino [41]

Answer:

(a) Wavelength is 0.436 m

(b) Length is 0.872 m

(c) 11.518 m/s

Solution:

As per the question:

The eqn of the displacement is given by:

y(x, t) = (1.22 cm)sin[14.4 m^{- 1}x]cos[(166\ rad/s)t]          (1)

n = 4

Now,

We know the standard eqn is given by:

y = AsinKxcos\omega t           (2)

Now, on comparing eqn (1) and (2):

A = 1.22 cm

K = 14.4 m^{- 1}

\omega = 166\ rad/s

where

A = Amplitude

K = Propagation constant

\omega = angular velocity

Now, to calculate the string's wavelength,

(a) K = \frac{2\pi}{\lambda}

where

K = propagation vector

\lambda = \frac{2\pi}{K}

\lambda = \frac{2\pi}{14.4} = 0.436\ m

(b) The length of the string is given by:

l = \frac{n\lambda}{2}

l = \frac{4\times 0.436}{2} = 0.872\ m

(c)  Now, we first find the frequency of the wave:

\omega = 2\pi f

f = \frac{\omega}{2\pi}

f = \frac{2\pi}{166} = 26.42\ Hz

Now,

Speed of the wave is given by:

v = f\lambda

v = 26.419\times 0.436 = 11.518\ m/s

4 0
3 years ago
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