18. How much force would it take to push another, larger friend who has a mass of 70 Kg to accelerate

An advanced computer sends information to its various parts via infrared light pulses traveling through silicon fibers (n = 3.50

sasho [114]

Answer:

d = 6.43 cm

Explanation:

Given:

- Speed resistance coefficient in silicon n = 3.50

- Memory takes processing time t_p = 0.50 ns

- Information is to be obtained within T = 2.0 ns

Find:

- What is the maximum distance the memory unit can be from the central processing unit?

Solution:

- The amount of time taken for information pulse to travel to memory unit:

t_m = T - t_p

t_m = 2.0 - 0.5 = 1.5 ns

- We will use a basic relationship for distance traveled with respect to speed of light and time:

d = V*t_m

- Where speed of light in silicon medium is given by:

V = c / n

- Hence, d = c*t_m / n

-Evaluate: d = 3*10^8*1.5*10^-9 / 3.50

d = 0.129 m 12.9 cm

- The above is the distance for pulse going to and fro the memory and central unit. So the distance between the two is actually d / 2 = 6.43 cm

8 0

3 years ago

What is the average speed of a bird that flies 1.55 km in 30 minutes

Alexxandr [17]

Answer:

the average speed in annual work out is 46.5

5 0

3 years ago

A plane electromagnetic wave, with wavelength 4.1 m, travels in vacuum in the positive direction of an x axis. The electric fiel

marusya05 [52]

(a) $7.32\cdot 10^7 Hz$

The frequency of an electromagnetic waves is given by:

$f=\frac{c}{\lambda}$

where

$c=3.0\cdot 10^8 m/s$ is the speed of light

$\lambda=4.1 m$ is the wavelength of the wave in the problem

Substituting into the equation, we find

$f=\frac{3.0\cdot 10^8 m/s}{4.1 m}=7.32\cdot 10^7 Hz$

(b) $4.60\cdot 10^8 rad/s$

The angular frequency of a wave is given by

$\omega = 2\pi f$

where

f is the frequency

For this wave,

$f=7.32\cdot 10^7 Hz$

So the angular frequency is

$\omega=2\pi(7.32\cdot 10^7 Hz)=4.60\cdot 10^8 rad/s$

(c) $1.53 m^{-1}$

The angular wave number of a wave is given by

$k=\frac{2\pi}{\lambda}$

where

$\lambda$ is the wavelength of the wave

For this wave, we have

$\lambda=4.1 m$

so the angular wave number is

$k=\frac{2\pi}{4.1 m}=1.53 m^{-1}$

(d) $1.03\cdot 10^{-6}T$

For an electromagnetic wave,

$E=cB$

where

E is the magnitude of the electric field component

c is the speed of light

B is the magnitude of the magnetic field component

For this wave,

E = 310 V/m

So we can re-arrange the equation to find B:

$B=\frac{E}{c}=\frac{310 V/m}{3\cdot 10^8 m/s}=1.03\cdot 10^{-6}T$

(e) z-axis

In an electromagnetic wave, the electric field and the magnetic field oscillate perpendicular to each other, and they both oscillate perpendicular to the direction of propagation of the wave. Therefore, we have:

- direction of propagation of the wave --> positive x axis

- direction of oscillation of electric field --> y axis

- direction of oscillation of magnetic field --> perpendicular to both, so it must be z-axis

(f) 127.5 W/m^2

The time-averaged rate of energy flow of an electromagnetic wave is given by:

$I=\frac{E^2}{2\mu_0 c}$

where we have

E = 310 V/m is the amplitude of the electric field

$\mu_0$ is the vacuum permeability

c is the speed of light

Substituting into the formula,

$I=\frac{(310 V/m)^2}{2(4\pi\cdot 10^{-7} H/m) (3\cdot 10^8 m/s)}=127.5 W/m^2$

(g) $1.53\cdot 10^{-8} kg m/s$

For a surface that totally absorbs the wave, the rate at which momentum is transferred to the surface given by

$\frac{dp}{dt}=\frac{A}{c}$

where the <S> is the magnitude of the Poynting vector, given by

$=\frac{EB}{\mu_0}=\frac{(310 V/m)(1.03\cdot 10^{-6} T)}{4\pi \cdot 10^{-7}H/m}=254.2 W/m^2$

and where the surface is

A = 1.8 m^2

Substituting, we find

$\frac{dp}{dt}=\frac{(254.2 W/m^2)(1.8 m^2)}{3\cdot 10^8 m/s}=1.53\cdot 10^{-8} kg m/s$

(h) $8.47\cdot 10^{-7} N/m^2$

For a surface that totally absorbs the wave, the radiation pressure is given by

$p=\frac{}{c}$

where we have

$=254.2 W/m^2$

$c=3\cdot 10^8 m/s$

Substituting, we find

$p=\frac{254.2 W/m^2}{3\cdot 10^8 m/s}=8.47\cdot 10^{-7} N/m^2$

8 0

3 years ago

At the north magnetic pole the earth’s magnetic field is vertical and has a strength of 0.62 gauss. The earth’s field at the sur

Anika [276]

Answer:

A) Dipole moment; m = 8.02 x 10^(22) J/T

B) I = 3.51 x 10^(9) A

Explanation:

The components of a magnetic field of a dipole are;

B_r = (μ_o•m/2πr³).cosθ

B_θ = (μ_o•m/4πr³).sin θ

B_Φ = 0

Let's make m the subject in the B_r equation ;

m = (2πr³•B_r)/(μ_o•cosθ)

Where;

B_r is magnetic field = 0.62 Gauss = 6.2 x 10^(-5) T

μ_o is the magnetic constant and has a value of 4π × 10^(−7) H/m

m is magnetic moment.

r is equal to radius of earth =6.371 x 10^(6)m

Thus, if we set θ = 0,we can solve for m as below;

m = (2π(6.371 x 10^(6))³•6.2 x 10^(-5) )/(4π × 10^(−7)•cos0)

Thus, m = 8.02 x 10^(22) J/T

B) Now, to find the current, let's use the expression for the magnetic field on the z-axis of the current ring.

B_z = (μ_o•Ib²/(2(z² + b²/2)^(3/2)))

So, let's set z = R and b = R/2

Thus, we now have;

B_z = (μ_o•I)/(5^(3/2)•R)

Making I the subject, we have;

I = [(5^(3/2)•R)•B_z]/μ_o

Plugging in the relevant values, we have;

I = [(5^(3/2) x 6.371 x 10^(6)) x 6.2 x 10^(-5)]/(4π × 10^(−7))

I = 3.51 x 10^(9) A

4 0

3 years ago

a cheetah starting from rest accelerates at a rate of 2.5m/s^2 to reach a speed of 45 m/s.how fare does the cheetah travel

Helen [10]

Answer:

405 m

Explanation:

Given:

v₀ = 0 m/s

a = 2.5 m/s²

v = 45 m/s

Find: Δx

v² = v₀² + 2aΔx

(45 m/s)² = (0 m/s)² + 2 (2.5 m/s²) Δx

Δx = 405 m

7 0

3 years ago