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max2010maxim [7]
3 years ago
12

The distance between earth and mars is 51 G meters. What is the minimum RTT for a link between Earth and Mars if speed of light

is 3 *108 meters/second.
Physics
1 answer:
beks73 [17]3 years ago
4 0

Answer:

340 seconds = 5.667 minutes

Explanation:

As we know, S = v t or t = S / v (S = 51 x 10^9 m and v = 3 x 10^8 ms^-1)

So, t = 51 x 10^9 / 3 x 10^8 = 17 x 10^1 = 170 s

For a RTT estimation, the time span will be doubled of one way propagation for transmission and receive delay.

The over all round trip time will be = 170 x 2 = 340 seconds = 5.667 minutes

You might be interested in
A red blood cell is 7 x 10-6 m in size. This means it is 7 __________ in size. What unit completes the sentence?
aalyn [17]

If a red blood cell is 7 x 10^-^6 m in size, it means that such a cell is 7 microns or micrometer in size. The unit that completes the sentence will, thus, be microns or micrometer.

According to international system of measurement:

  • 10^{-1} m = decimeter
  • 10^{-2} m = centimeter
  • 10^{-3} m = millimeter
  • 10^{-6} m = micrometer or micron
  • 10^{-9} m = nanometer

Thus, a measurement of 7 x 10^-^6 m is the same is 7 micons or 7 micrometer.

More on metric conversion can be found here: brainly.com/question/17767575

5 0
2 years ago
A dockworker applies a constant horizontal force of 90.0 N to a block of ice on a smooth horizontal floor. The frictional force
Oliga [24]

Answer:

The mass of the ice block is equal to 70.15 kg

Explanation:

The data for this exercise are as follows:

F=90 N

insignificant friction force

x=13 m

t=4.5 s

m=?

applying the equation of rectilinear motion we have:

x = xo + vot + at^2/2

where xo = initial distance =0

vo=initial velocity = 0

a is the acceleration

therefore the equation is:

x = at^2/2

Clearing a:

a=2x/t^2=(2x13)/(4.5^2)=1.283 m/s^2

we use Newton's second law to calculate the mass of the ice block:

F=ma

m=F/a = 90/1.283=70.15 kg

6 0
3 years ago
Read 2 more answers
Describe the difference between potencial and kinetic energy
avanturin [10]
Potential energy is energy stored in an object. kinetic energy is energy of motion
8 0
3 years ago
Ask Your Teacher Suppose the roller coaster below(h1 = 36 m, h2 = 13 m, h3 = 30) passes point A with a speed of 1.00 m/s. If the
Oliga [24]

Answer:

The answer to the question is

The roller coaster will reach point B with a speed of 14.72 m/s

Explanation:

Considering both kinetic energy KE = 1/2×m×v² and potential energy PE = m×g×h

Where m = mass

g = acceleration due to gravity = 9.81 m/s²

h = starting height of the roller coaster

we have the given variables

h₁ = 36 m,

h₂ = 13 m,

h₃ = 30 m

v₁ = 1.00 m/s

Total energy at point 1 = 0.5·m·v₁² + m·g·h₁

= 0.5 m×1² + m×9.81×36

=353.66·m

Total energy at point 2 = 0.5·m·v₂² + m·g·h₂

= 0.5×m×v₂² + 9.81 × 13 × m = 0.5·m·v₂² + 127.53·m

The total energy at 1 and 2 are not equal due to the frictional force which must be considered

Total energy at point 2 = Total energy at point 1 + work done against friction

Friction work = F×d×cosθ = (\frac{1}{5} × mg)×60×cos 180 = -117.72m

0.5·m·v₂² + 127.53·m = 353.66·m -117.72m

0.5·m·v₂² = 108.41×m

v₂² = 216.82

v₂  =  14.72 m/s

The roller coaster will reach point B with a speed of 14.72 m/s

8 0
3 years ago
During the latter part of your European vacation, you are hanging out at the beach at the gold coast of Spain. As you are laying
Jlenok [28]

Well, I guess you can come close, but you can't tell exactly.

It must be presumed that the seagull was flying through the air
when it "let fly" so to speak, so the jettisoned load of ballast
of which the bird unburdened itself had some initial horizontal
velocity.

That impact velocity of 98.5 m/s is actually the resultant of
the horizontal component ... unchanged since the package
was dispatched ... and the vertical component, which grew
all the way down in accordance with the behavior of gravity.

  98.5 m/s  =  √ [ (horizontal component)² + (vertical component)² ].

The vertical component is easy; that's (9.8 m/s²) x (drop time).
Since we're looking for the altitude of launch, we can use the
formula for 'free-fall distance' as a function of acceleration and
time:

             Height = (1/2) (acceleration) (time²) .

If the impact velocity were comprised solely of its vertical
component, then the solution to the problem would be a
piece-o-cake.

                  Time = (98.5 m/s) / (9.81 m/s²) = 10.04 seconds
whence
                 Height = (1/2) (9.81) (10.04)²

                            =   (4.905 m/s²) x (100.8 sec²)  =  494.43 meters.

As noted, this solution applies only if the gull were hovering with
no horizontal velocity, taking careful aim, and with malice in its
primitive brain, launching a remote attack on the rich American.

If the gull was flying at the time ... a reasonable assumption ... then
some part of the impact velocity was a horizontal component.  That
implies that the vertical component is something less than 98.5 m/s,
and that the attack was launched from an altitude less than 494 m.   

8 0
3 years ago
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