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gtnhenbr [62]
3 years ago
7

On average, how many stars would we have to search before we would expect to hear a signal? assume there are 500 billion stars i

n the galaxy.
Physics
1 answer:
Keith_Richards [23]3 years ago
7 0

We would have to search at least 5,000,000,000 (5 billion) stars before we would expect to hear a signal.

To find out the number of stars that we will need to search to find a signal, we need to use the following formula:

  • total of stars/civilizations
  • 500,000,000,000 (500 billion) stars / 100 civilization = 5,000,000,000 (5 billion)

This shows it is expected to find a civilization every 5 billion stars, and therefore it is necessary to search at least 5 billion stars before hearing a signal from any civilization.

Note: This question is incomplete; here is the complete question.

On average, how many stars would we have to search before we would expect to hear a signal? Assume there are 500 billion stars in the galaxy.

Assuming 100 civilizations existed.

Learn more about stars in: brainly.com/question/2166533

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Review:
Tresset [83]

Answer:

1. 1. A quantity is completely described by magnitude alone. A quantity Is completely described by a magnitude with a direction.

[a]. scalar, vector

b. vector, scalar

2.2. Speed is a velocity is a quantity and quantity.

a. scalar, vector

[b]. vector, scalar

3 0
2 years ago
28. Identify whether the following objects are in
Marianna [84]

Answer:

a. A baseball after it has been  hit - not in free fall

b. A rock that is thrown in the  air - not in free fall

c. The moon - free-fall

d. A paper airplane - not in free fall

e. A bird flying - not in free fall

Explanation:

  1. The free-fall is defined as the falling of an object due to the action of gravity. The object is not experiencing any other force neglecting the air resistance.
  2. If an object is in free-fall, the direction of its motion is directed towards the center of the earth. It does not have a horizontal component of velocity.
  3. If the body is under free-fall, but a centripetal force acts on it where it is equal to the gravitational force at that point. The object will have two components of velocity along the tangential line, perpendicular to the radius of the orbit.

a. A baseball after it has been  hit - not in free fall according to point 1 & 2.

b. A rock that is thrown in the  air - not in free fall according to point 1.

c. The moon - free-fall according to point 3.

d. A paper airplane - not in free fall according to point 1 & 2.

e. A bird flying - not in free fall according to point 1 & 2.

7 0
3 years ago
The surface of the Sun has a temperature of about 5 800 K. If the radius of the Sun is 7 × 108 m, determine the power output of
Nimfa-mama [501]

Answer:

a. 3.95\times10^{26}W

Explanation:

T = temperature of the surface of sun = 5800 K

r = Radius of the Sun = 7 x 10⁸ m

A = Surface area of the Sun

Surface area of the sun is given as

A = 4\pi r^{2} \\A = 4(3.14) (7\times10^{8})^{2}\\A = 6.2\times10^{18} m^{2}

e = Emissivity = 1

\sigma = Stefan's constant = 5.67 x 10⁻⁸ Wm⁻²K⁻⁴

Using Stefan's law, Power output of the sun is given as

P = \sigma e AT^{4} \\P = (5.67\times10^{-8}) (1) (6.2\times10^{18}) (5800)^{4}\\P = 3.95\times10^{26} W

4 0
3 years ago
Objects 1 and 2 attract each other with a electrostatic force of 72.0 units. If the distance separating objects 1 and 2 is chang
qaws [65]

Answer:

288.0 units; that is the electrostatic force of attraction become quadruple of its initial value.

Explanation:

If all other parameters are constant,

Electrostatic Force of attraction ∝ (1/r²)

F = (k/r²) = 72.0

If r₁ = r/2, what happens to F₁

F₁ = (k/r₁²) = k/(r/2)² = (4k/r²) = 4F = 4 × 72 = 288.0 units

5 0
4 years ago
Read 2 more answers
A 2.45-kg frictionless block is attached to an ideal spring with force constant 355 N/m. Initially the spring is neither stretch
ANTONII [103]

Answer:

A.    A = 0.913 m

B.    amax = 132.24m/s^2

C.    Fmax = 324.01N

Explanation:

When the block is moving at the equilibrium point , its velocity is maximum.

A. To find the amplitude of the motion you use the following formula for the maximum velocity:

v_{max}=A\omega          (1)

vmax = maximum velocity = 11.0 m/s

A: amplitude of the motion = ?

w: angular frequency = ?

Then, you have to calculate the angular frequency of the motion, by using the following formula:

\omega=\sqrt{\frac{k}{m}}           (2)

k: spring constant = 355 N/m

m: mass of the object = 2.54 kg

\omega = \sqrt{\frac{355N/m}{2.45kg}}=12.03\frac{rad}{s}

Next, you solve the equation (1) for A and replace the values of vmax and w:

A=\frac{v}{\omega}=\frac{11.0m/s}{12.03rad/s}=0.913m

The amplitude of the motion is 0.913m

B. The maximum acceleration of the block is given by:

a_{max}=A\omega^2 = (0.913m)(12.03rad/s)^2=132.24\frac{m}{s^2}

The maximum acceleration is 132.24 m/s^2

C. The maximum force is calculated by using the second Newton law and the maximum acceleration:

F_{max}=ma_{max}=(2.45kg)(132.24m/s^2)=324.01N

It is also possible to calculate the maximum force by using:

Fmax = k*A = (355N/m)(0.913m) = 324.01N

The maximum force exertedbu the spring on the object is 324.01 N

4 0
3 years ago
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