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Mariana [72]
3 years ago
6

Two satellites revolve around the Earth. Satellite A has mass m and has an orbit of radius r. Satellite B has mass 6m and an orb

it of unknown radius rb. The forces of gravitational attraction between each satellite and the Earth is the same. Find rb.
Physics
1 answer:
melomori [17]3 years ago
8 0

Answer:

aaaaa

Explanation:

M = Mass of the Earth

m = Mass of satellite

r = Radius of satellite

G = Gravitational constant

F=G\frac{Mm}{r^2}

F=G\frac{M6m}{r_b^2}

G\frac{Mm}{r^2}=G\frac{M6m}{r_b^2}\\\Rightarrow \frac{1}{r^2}=\frac{6}{r_b^2}\\\Rightarrow \frac{r_b^2}{r^2}=6\\\Rightarrow \frac{r_b}{r}=\sqrt{6}\\\Rightarrow r_b=2.44948r

r_b=2.44948r

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Carbon-14 has a half-life of 5,730 years. if the age of an object older than 50,000 years cannot be determined by radiocarbon da
Aleonysh [2.5K]
<h3><u>Answer;</u></h3>

Carbon-14 levels in a sample are undetectable after approximately 9 half lives

<h3><u>Explanation;</u></h3>
  • <em><u>The half life of Carbon-14 is 5,730 years . Half life is the time taken by a radioactive material to decay by half of its original mass.  Therefore, it  would take a time of 5730 years for a sample of 100 g of carbon-14 to decay to 50 grams</u></em>
  • <em><u>A period of 50,000 years, is equivalent to; </u></em>

<em><u>  50,000÷5,730 </u></em>

<em><u>= 8.73 half lives</u></em>

<em>Which is approximately equal to 9 half lives.</em>

  • Therefore, if the age of an object older than 50,000 years cannot be determined by radiocarbon dating, then <em><u>Carbon-14 levels in a sample are undetectable after approximately 9 half lives</u></em>.
6 0
3 years ago
Read 2 more answers
A sound wave has a frequency of 425 Hz. What is the period of this wave
Oduvanchick [21]

Answer: The time period of the given wave with frequency of 425 Hertz is 0.0023 seconds.

Explanation:

Frequency of the wave = 425 Hertz =425 sec^{-1}

\text{Time period}=\frac{1}{Frequency}

\text{Time period}==\frac{1}{425 sec^{-1}}=0.0023 sec

The time period of the given wave with frequency of 425 Hertz is 0.0023 seconds.

5 0
3 years ago
Over a time interval of 1.71 years, the velocity of a planet orbiting a distant star reverses direction, changing from +17.3 km/
lutik1710 [3]

Answer:

a)40100m/s

b)-4.348x10^- m/s^2

Explanation:

to calculate the change in the planet's velocity we have to rest the speeds

ΔV=-22.8-17.3=-40.1km/s=40100m/s

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t  (1)\\\\

{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\

X=Xo+ VoT+0.5at^{2}    (3)\\

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve

for this problem we have to convert the time interval ins seconds, we know that a year has 53926560s

t=1.71years=53926560*1.71=92214417.6

then we can use the ecuation number 1 to calculate the aceleration

Vf=-22.8km/s

Vo=17.3km/s

Vf=Vo+at

a=(vf-vo)/t

a=(-22.8-17.3)/92214417.6

a=-4.348x10^-7 km/s^2=-4.348x10^- m/s^2

4 0
3 years ago
46 points :)
IgorC [24]

Answer:

Its is dividing by 2

Explanation:

It starts with 100 them it goes to 50, 25, 12.5 so its a cycle of dividing by 2

3 0
3 years ago
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kolezko [41]

Answer:

1800J

Explanation:

Given parameters:

Weight of the book  = 20N

Total distance covered  = 45m + 15m + 30m  = 90m

Unknown:

Total work performed on the books  = ?

Solution:

To solve this problem we must understand that work done is the force applied to move a body through a certain distance.

So;

    Work done  = Force x distance

  Work done  = 20 x 90  = 1800J

8 0
3 years ago
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