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goldfiish [28.3K]
2 years ago
5

Consider a civilization broadcasting a signal with a power of 1.2×104 watts. The Arecibo radio telescope, which is about 300 met

ers in diameter, could detect this signal if it is coming from as far away as 110 light-years. Suppose instead that the signal is being broadcast from the other side of the Milky Way Galaxy, about 70000 light-years away.
Physics
1 answer:
Talja [164]2 years ago
6 0
The same power (1.9×10^4 watts) gets diminished by Inverse square law 
<span>(ratio of distances)² </span>
<span>= (138/70000)² </span>
<span>= 3.886506 X 10^-6 . </span>
<span>This is the diminution factor for the given sensitivity of the telescope. But at 3.8865 millionth below the sensitivity its is far below detection; it needs to be enhanced by collecting the energy over an area so many times more (by an aperture multiplied by same factor) </span>
<span>1/[3.886506 X 10^-6] = 257298.88 </span>
<span>In other words the diameter should have the ratio of square root of this </span>
<span>(70000/138). </span>
<span>This multiplied by 300m dish gives </span>
<span>300 X (70000/138) = 152173.913 m = 152.1739 km. </span>
<span>This should be the aperture of the new telescope for detecting the signal.</span>
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Ira Lisetskai [31]

Answer:

<h3>38,673.9N</h3>

Explanation:

According to newton's second law:

Force = mass * acceleration

Given

Mass = 873kg

acceleration = 44.66m/s²

Magnitude of the force is expressed as;

F = ma

F = 873 * 44.6

F = 38,673.9N

<em>Hence the magnitude of the net force exerted on the dragster during this time is 38,673.9N</em>

8 0
2 years ago
A 5000-kg freight car runs into a 10,000-kg freight car at rest. They couple upon collision and move with a speed of 2 m/s. What
Vitek1552 [10]

Answer:

C) 6 m/s

Explanation:

Given that

m₁=5000 kg

The initial velocity of 5000 kg car =u₁

m₂=10,000 kg

The initial velocity of 10000 kg car =u₂ = 0 m/s

After collision the final speed of the both car,v = 2 m/s

There is no any external force on the system that is why linear momentum will be conserved.

Linear momentum P = m v

m₁u₁ + m₂u₂ = (m₂ + m₁) v

5000 x u₁ + 10000 x 0 = (5000 + 10000) x 2

5000 x u₁ = 15000 x 2

5 x u₁ = 15 x 2

u₁ = 6 m/s

Therefore the answer is C.

C) 6 m/s

4 0
2 years ago
Calculate the net force on the right charge due to the other two. Enter a positive value if the force is directed to the right a
lbvjy [14]

Answer:

Answer:

A. - 0.017N. It acts to the left.

B. - 0.043N. It acts to the left.

C. 0.060N. It acts to the right.

Explanation:

A. For the +65μC charge, we consider it to be the origin. Hence, the two other charges are on the +x axis.

The net coulombs force on the charge is

F = [KQ(1)Q(2)]/(r^2) + [KQ(1)Q(3)]/(r^2)

Where K = Coloumbs constant =

Q(1) = charge on the leftmost side.

Q(2) = charge in the middle.

Q(3) = charge on the rightmost side.

F = [(8.988 × 10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(-95×10^-6)×(65×10^-6)]/(40^2)

F = 0.01753 - 0.03469

F = -0.017N

It has a negative sign, hence, it acts to the left.

B. For the +48μC charge, we consider it to be the origin. Hence, the leftmost charge is on the - x axis and the rightmost charge is on the +x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) + [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = -0.017 - 0.02562

F = - 0.043N

It has a negative sign, hence, it acts to the left.

C. For the -95μC charge, we consider it to be the origin. Hence, the two other charges are on the - x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) - [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(-95×10^-6)]/(40^2) - [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = +0.03469 + 0.02562

F = +0.060N

It has a positive sign, hence, it acts to the right.

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Explanation:

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True [87]

Answer:

C

Explanation:

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