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Fynjy0 [20]
3 years ago
7

NASA’s Tracking and Data Relay Satellite (TDRS) System constellation resides at geosynchronous orbit (35,000km) altitude. If a t

echnician at the Goddard Spaceflight Center in Maryland initiates a transmission to the Johnson Spaceflight Center in Houston over TDRS, how long will it be until JSC detects the transmission (one-way latency)? You may assume there is negligible processing delay on the satellite, and that c = 3x108 m/sec.
Physics
1 answer:
Leya [2.2K]3 years ago
6 0

Answer:

35,000 km = 35,000,000 m = 3.5 E107 m

t = S / v = 3.5 * 10E7 / 3.0 E10E8 = .117 sec

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24) a plane takes off at 7:00 pm and lands at 1:00 am after travelling a distance of 4818km. what is the minimum number of times
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Answer:

Twice

Step-by-Step Explanation:

Time between 7:00 PM and 1:00 AM: 6 hours

Distance: 4818km

Since the distance is 4818km, and the time is 6 hours, you divide 4818 by 6.

803.0000015999 km/h.

The average speed is 803 km/h

Which considering the ideal case scenario if the plane starts at 0 reaches the speed of 803 and the end reduces its speed from 803 to 0. This means we have come across the value of 800 at least twice. Hence, the plane was travelling at a speed of 800 km/h at least 2 times.

7 0
1 year ago
I need help on putting this diagram in order.
morpeh [17]
In what type of order are you supposed to put it in?
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3 years ago
Our milky way galaxy is 100000 lyly in diameter. a spaceship crossing the galaxy measures the galaxy's diameter to be a mere 1.
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The speed of the spaceship relative to the galaxy is 0.99999995c.

A light-year measures distance rather than time (as the name might imply). A light-year is a distance a light beam travels in one year on Earth, which is roughly 6 trillion miles (9.7 trillion kilometers). One light-year equals 5,878,625,370,000 miles. Light moves at a speed of 670,616,629 mph (1,079,252,849 km/h) in a vacuum.We multiply this speed by the number of hours in a year to calculate the distance of a light-year (8,766).

The Milky way galaxy is 100,000 light years in diameter.

The galaxy's diameter is a mere 1. 0 ly.

We know that ;

L = L_0 \sqrt{1-\frac{v^2}{c^2} }

L = 1 light year

L₀ = 100,000 light year

1 = 100,000 \sqrt{1-\frac{v^2}{c^2} }

1 = 100,000 \sqrt{1-\frac{v^2}{(3*10^8)^2} }

\frac{1}{100,000}  = \sqrt{1-\frac{v^2}{c^2} }

v = 0.999999995 c

Therefore, the speed of the spaceship relative to the galaxy is 0.99999995c.

Learn more about a light year here:

brainly.com/question/17423632

#SPJ4

5 0
1 year ago
The path of the sun on the celestial sphere is called.
Ede4ka [16]

Answer:

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The path the sun, moon, and planets take across the sky as seen from Earth. It defines the plane of the Earth's orbit around the sun. The name "Ecliptic" comes from the fact that eclipses take place along this line.

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It holds the atoms together (aka your last option)
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