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Serga [27]
3 years ago
15

In 1974, the Arecibo Radio telescope in Puerto Rico was used to transmit a signal to M13, a star cluster about 25,000 light-year

s away. How long will it take the message to reach M13, and how far has the message travelled so far (in light-years)?
Physics
1 answer:
Arte-miy333 [17]3 years ago
6 0

Answer:

The Arecibo message is a 1974 interstellar radio message carrying basic information about humanity and Earth sent to globular star cluster M13 in the hope that extraterrestrial intelligence might receive and decipher it. The message was broadcast into space a single time via frequency modulated radio waves at a ceremony to mark the remodeling of the Arecibo radio telescope in Puerto Rico on 16 November 1974.The message was aimed at the current location of M13 some 25,000 light years away because M13 was a large and close collection of stars that was available in the sky at the time and place of the ceremony.The message consisted of 1,679 binary digits, approximately 210 bytes, transmitted at a frequency of 2,380 MHz and modulated by shifting the frequency by 10 Hz, with a power of 450 kW. The "ones" and "zeros" were transmitted by frequency shifting at the rate of 10 bits per second. The total broadcast was less than three minutes.

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Describe how we know the plates once formed a supercontinent and how we know this.
Lady_Fox [76]

Answer:

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2 years ago
Recall that photosynthetic rates remain relatively constant in regions near the equator. Imagine that tropical environments pers
solniwko [45]

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a straight line sloping upward (atmospheric CO2 levels would not seasonally oscillate, but would have increased over time)

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8 0
3 years ago
A horizontal force of 12N is applied to 1.5kg of block which rests on a horizontal surface. If the coefficient friction is 0.3,
Rama09 [41]

Answer:

7 m/s²

Explanation:

From the question given above, the following data were:

Force applied (Fₐ) = 15 N

Mass (m) of block = 1.5 Kg

Acceleration due to gravity (g) = 10 m/s²

Coefficient of friction (μ) = 0.3

Acceleration (a) of block =?

Next, we shall determine the frictional force. This can be obtained as follow:

Mass (m) of block = 1.5 Kg

Acceleration due to gravity (g) = 10 m/s²

Coefficient of friction (μ) = 0.3

Normal reaction (R) = mg = 1.5 × 10 = 15 N

Frictional force (Fբ) =?

Fբ = μR

Fբ = 0.3 × 15

Fբ = 4.5 N

Next, we shall determine the net force acting on the block. This can be obtained as follow:

Force applied (Fₐ) = 15 N

Frictional force (Fբ) = 4.5 N

Net force (Fₙ) =.?

Fₙ = Fₐ – Fբ

Fₙ = 15 – 4.5

Fₙ = 10.5 N

Finally, we shall determine the acceleration produced in the block. This can be obtained as follow:

Mass (m) of block = 1.5 Kg

Net force (Fₙ) = 10.5 N

Acceleration (a) of block =?

Fₙ = ma

10.5 = 1.5 × a

Divide both side by a

a = 10.5 / 1.5

a = 7 m/s²

Therefore, the acceleration produced in the block is 7 m/s²

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