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3 years ago

14

Conversations with astronauts on the lunar surface were characterized by a kind of echo in which the earthbound person’s voice w

as so loud in the astronaut’s space helmet that it was picked up by the astronaut’s microphone and transmitted back to Earth. It is reasonable to assume that the echo time equals the time necessary for the radio wave to travel from the Earth to the Moon and back (that is, neglecting any time delays in the electronic equipment). Calculate the distance from Earth to the Moon given that the echo time was 2.56 s and that radio waves travel at the speed of light (3.00×108 m/s) .

1 answer:

abruzzese [7]3 years ago

8 0

Answer:

3.85*10^8m

Explanation:

We know that

Speed v = distance/time t

So distance = v x t

So = 3*10^8 x 2.56s

= distatance = 7.7*10^8m

However this is double the distance from the earth to the moon, so to get distance from earth to the moon we divide by 2

7.7*10^8/2= 3.85*10^8m

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Una pelota de tenis de 100 g de masa lleva una rapidez de 20 m/s. Al ser golpeada por una raqueta, se mueve en sentido contrario

REY [17]

Answer:

Explanation:

El impulso aplicado a la pelota produce una variación en su momento lineal.

J = m (V -Vo)

Conviene elegir positivo el sentido de la velocidad final.

J = 0,100 kg [40 - (- 20)] m/s = 6 kg m/s

Saludos Herminio

7 0

3 years ago

Greg throws a 2.8-kg pumpkin horizontally off the top of the school roof in order to hit Mr. H's car. The car has parked a dista

Igoryamba

Answer:

The horizontal velocity is $v = 9.2 m/s$

Explanation:

From the question we are told that

The mass of the pumpkin is $m = 2.8 \ kg$

The distance of the the car from the building's base is $d = 13.4 \ m$

The height of the roof is $h = 10.4 \ m$

The height is mathematically represented as

$h = \frac{1}{2} gt^2$

Where g is the acceleration due to gravity which has a value of $g =9.8 \ m/s^2$

substituting values

$10.4= 0.5 * 9.8 * t$

making the time taken the subject of the formula

$t = \frac{10.4}{0.5 * 9.8 }$

$t = 1.457 \ s$

The speed at which the pumpkin move horizontally can be represented mathematically as

$v = \frac{d}{t}$

substituting values

$v =\frac{13.4}{1.457}$

$v = 9.2 m/s$

7 0

3 years ago

A 6.00-mH solenoid is connected in series with a 5.0-μF capacitor and an AC source. The solenoid has internal resistance 3.0 Ω w

son4ous [18]

Answer:

5773.50269 Hz

23 A

Explanation:

$L$ = Inductance = 6 mH

$C$ = Capacitance = 5 μF

$R$ = Resistance = 3 Ω

$\epsilon$ = Maximum emf = 69 V

Resonant angular frequency is given by

$\omega=\dfrac{1}{\sqrt{LC}}\\\Rightarrow \omega=\dfrac{1}{\sqrt{6\times 10^{-3}\times 5\times 10^{-6}}}\\\Rightarrow \omega=5773.50269\ Hz$

The resonant angular frequency is 5773.50269 Hz

Current is given by

$I=\dfrac{\epsilon}{R}\\\Rightarrow I=\dfrac{69}{3}\\\Rightarrow I=23\ A$

The current amplitude at the resonant angular frequency is 23 A

7 0

3 years ago

Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.

DiKsa [7]

Answer:0.061

Explanation:

Given

$T_C=300 k$

Temperature of soup $T_H=340 K$

heat capacity of soup $c_v=33 J/K$

Here Temperature of soup is constantly decreasing

suppose T is the temperature of soup at any instant

efficiency is given by

$\eta =\frac{dW}{Q}=1-\frac{T_C}{T}$

$dW=Q(1-\frac{T_C}{T})$

$dW=c_v(1-\frac{T_C}{T})dT$

integrating From $T_H$ to $T_C$

$\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT$

$W=\int_{T_C}^{T_H}33\cdot (1-\frac{300}{T})dT$

$W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}$

$W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]$

Now heat lost by soup is given by

$Q=c_v(T_C-T_H)$

Fraction of the total heat that is lost by the soup can be turned is given by

$=\frac{W}{Q}$

$=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}$

$=\frac{T_C-T_H-T_C\ln (\frac{T_C}{T_H})}{T_C-T_H}$

$=\frac{300-340-300\ln (\frac{300}{340})}{300-340}$

$=\frac{-40+37.548}{-40}$

$=0.061$

4 0

3 years ago

Calculate the power of convex lens of focal length 20cm

maxonik [38]

D = 1/f, where D is the power in diopters and f is the focal length in meters.

D=1/20

<u>D=0.05</u>

6 0

3 years ago