Coders play an important role in

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

melomori [17]

Answer:

aaaaa

Explanation:

M = Mass of the Earth

m = Mass of satellite

r = Radius of satellite

G = Gravitational constant

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

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

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

$r_b=2.44948r$

8 0

3 years ago

Chris races his Audi north down a road for 1000 meters in 20 seconds, what is his velocity?

liq [111]

The answer is 1000/20. Or that’s what I’m guessing. Lol

5 0

4 years ago

Read 2 more answers

When you exert 75 N on a jack to lift a 6000 N car, what is the jack’s actual mechanical advantage? Show your work.

professor190 [17]

Answer:

80

Explanation:

the mechanical advantage is the ratio of the load to the effort so it doesn't have units.to calculate it you use the formula

mechanical advantage=load/effort

in this case the load is 6000N and the effort is 75N

Ma=6000/75

=80

I hope this helps

3 0

3 years ago

You charge an initially uncharged 65.7-mf capacitor through a 39.1-Ï resistor by means of a 9.00-v battery having negligible int

uysha [10]

In a RC-circuit, with the capacitor initially uncharged, when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
$Q(t) = Q_0 (1-e^{-t/\tau})$
where t is the time, $Q_0 = CV$ is the maximum charge on the capacitor at voltage V, and $\tau = RC$ is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.

1) Using $R=39.1 \Omega$ and $C= 65.7 mF=65.7\cdot 10^{-3}F$ , the time constant of the circuit is:
$\tau = RC=(39.1 \Omega)(65.7 \cdot 10^{-3}F)=2.57 s$

2) To find the charge on the capacitor at time $t=1.95 \tau$ , we must find before the maximum charge on the capacitor, which is
$Q_0 = CV=(65.7 \cdot 10^{-3}F)(9 V)=0.59 C$
And then, the charge at time $t=1.95 \tau$ is equal to
$Q(1.95 \tau) = Q_0 (1-e^{-t/\tau})=(0.59 C)(1-e^{-1.95})=0.51 C$

3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be $Q_0 = 0.59 C$ . We can see this also from the previous formule, by using $t=\infty$ :
$Q(t) = Q_0 (1-e^{-\infty})=Q_0(1-0) = 0.59 C$

4 0

3 years ago

If someone were monitoring your vital signs (like your heart rate, oxygen content in your blood, blood pressure, etc.) every hou

Lilit [14]

Answer:

Accuracy

Explanation:

I think accuracy is more important. When it comes to vital organs in the body, the exactness of getting the measurement is paramount. Accuracy deals with getting very close, almost exact you may say, to a known standard. Precision on the other hand, deals with how easy a measurement can be retaken, reproduced or remade, irrespective of how far or close they are from the accepted norm.

From this, we can agree that precision neglects the most important factor, closeness or say, exactness. Precision isn't bothered by it. And while that can be excused in a few instances, it certainly can not be permitted when it comes to life, or organs of the body

7 0

3 years ago