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Maru [420]
3 years ago
13

Once juno reaches jupiter, what is the minimum amount of time it takes for the transmitted signals to travel from the spacecraft

to earth?
Physics
2 answers:
arsen [322]3 years ago
5 0

speed of light = juno to earth/time

time= juno/3x10^8. if dist in metres answer is in seconds.

probably not long

Komok [63]3 years ago
4 0

In very very very round figures . . .

-- Jupiter is about 5.2 times as far from the sun as the earth is.

-- So when Jupiter and the EARTH are aligned in both orbits, Jupiter is about

(4.2) x (150 million kilometers) = 630 million kilometers

Time = (distance) / (speed)

The speed of light and radio is 300,000 km/second

Time = (630 million / 300 thousand)

<em>Time = 2,100 seconds</em>

That's 35 minutes.

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Answer:

If they both start at the same height then most likely that means that each has an intical velocity  and a net force sooooo ball A 19.6

Explanation:

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A sound wave travels at 379 m/sec and has a wavelength of 8 meters. Calculate its frequency and period.
olga nikolaevna [1]

\qquad\qquad\huge\underline{{\sf Answer}}

Wavelength, Wavespeed and frequency depends on each other in the following way :

\qquad \sf  \dashrightarrow \: frequency =  \dfrac{wave \:  \: velocity}{wavelength}

\qquad \sf  \dashrightarrow \: f =  \dfrac{389}{8}

\qquad \sf  \dashrightarrow \: f \approx 48.62 \:  \: hertz

And we know the Reciprocal relationship between frequency and period ~

So, let's find it's period •

\sf{\qquad \sf  \dashrightarrow \: t = \dfrac{8}{389}}

\sf{\qquad \sf  \dashrightarrow \: t \approx 0.02 sec }

6 0
2 years ago
Two loudspeakers emit sound waves along the x-axis. A listener in front of both speakers hears a maximum sound intensity when sp
yKpoI14uk [10]

Answer

given,

difference between the two consecutive maximum

λ = 0.870 - 0.540

λ = 0.33 m

speed of sound = 340 m/s

b) frequency of the sound

v = f x λ

340 = f x 0.33

f =\dfrac{340}{0.33}

    f = 1030.3 Hz

a) phase difference

  the expression of phase difference is given by

   \phi = \dfrac{2\pi}{\lambda}\ \delta

   \delta = \Delta x - \lambda

   \delta = 0.540 - 0.33

   \delta = 0.21\ m

now,

   \phi = \dfrac{2\pi}{\lambda}\ \times 0.21

   \phi = \dfrac{2\pi}{0.33}\ \times 0.21

   \phi = 3.99 rad

8 0
3 years ago
A 99.1-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk =
Stels [109]

Answer:

628.022466 N

8.61 m/s

Explanation:

m = Mass

\mu = Coefficient of friction

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

F_f=\mu mg\\\Rightarrow F_f=0.646\times 99.1\times 9.81\\\Rightarrow F_f=628.022466\ N

Magnitude of frictional force is 628.022466 N

F=ma\\\Rightarrow a=\frac{F_f}{m}\\\Rightarrow a=\frac{628.022466}{99.1}\\\Rightarrow a=6.33726\ m/s^2

v=u+at\\\Rightarrow 0=u-6.33726\times 1.36\\\Rightarrow u=8.61\ m/s

Initial speed of the player is 8.61 m/s

4 0
3 years ago
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