Efficiency of a machine = (useful output energy) / (total input energy)
Efficiency = (500 J output work) / (Carlotta's 2,000 J input work)
Efficiency = (500 J) / (2,000 J)
Efficiency = 0.25
<em>Efficiency = 25%</em>
Answer:
The correct answer is B.
The astronaut will know due to the light from the explosion.
Explanation:
Sound and vibrations require a medium such as air to travel through. Space, there is no air. Only a vacuum. So sound and vibrations are unable to travel. Light requires no medium to travel. It can go through a vacuum.
Therefore the Astronaut will see a bright flash of light as it travels from the explosion to outer space. It is also important to note that light can travel very far because nothing else interacts with its wave particles and as such, it cannot be impeded.
Cheers!
Answer:
Hemoglobin bound to oxygen absorbs blue-green light, which means that it reflects red-orange light into our eyes, appearing red. That's why blood turns bright cherry red when oxygen binds to its iron. Without oxygen connected, blood is a darker red color.
Hope this helps :)
Answer:
R=4.22*10⁴km
Explanation:
The tangential speed 
of the geosynchronous satellite is given by:
Because 
is the circumference length (the distance traveled) and T is the period (the interval of time).
Now, we know that the centripetal force of an object undergoing uniform circular motion is given by:
If we substitute the expression for in this formula, we get:
Since the centripetal force is the gravitational force 
between the satellite and the Earth, we know that:
![F_g=\frac{GMm}{R^{2}}\\\\\implies \frac{GMm}{R^{2}}=\frac{4m\pi ^{2}R}{T^{2}}\\\\R^{3}=\frac{GMT^{2}}{4\pi^{2}} \\\\R=\sqrt[3]{\frac{GMT^{2}}{4\pi^{2}} }](https://tex.z-dn.net/?f=F_g%3D%5Cfrac%7BGMm%7D%7BR%5E%7B2%7D%7D%5C%5C%5C%5C%5Cimplies%20%5Cfrac%7BGMm%7D%7BR%5E%7B2%7D%7D%3D%5Cfrac%7B4m%5Cpi%20%5E%7B2%7DR%7D%7BT%5E%7B2%7D%7D%5C%5C%5C%5CR%5E%7B3%7D%3D%5Cfrac%7BGMT%5E%7B2%7D%7D%7B4%5Cpi%5E%7B2%7D%7D%20%5C%5C%5C%5CR%3D%5Csqrt%5B3%5D%7B%5Cfrac%7BGMT%5E%7B2%7D%7D%7B4%5Cpi%5E%7B2%7D%7D%20%7D)
Where G is the gravitational constant (
) and M is the mass of the Earth (
). Since the period of the geosynchronous satellite is 24 hours (equivalent to 86400 seconds), we finally can compute the radius of the satellite:
![R=\sqrt[3]{\frac{(6.67*10^{-11}Nm^{2}/kg^{2})(5.97*10^{24}kg)(86400s)^{2}}{4\pi^{2}}}\\\\R=4.22*10^{7}m=4.22*10^{4}km](https://tex.z-dn.net/?f=R%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%286.67%2A10%5E%7B-11%7DNm%5E%7B2%7D%2Fkg%5E%7B2%7D%29%285.97%2A10%5E%7B24%7Dkg%29%2886400s%29%5E%7B2%7D%7D%7B4%5Cpi%5E%7B2%7D%7D%7D%5C%5C%5C%5CR%3D4.22%2A10%5E%7B7%7Dm%3D4.22%2A10%5E%7B4%7Dkm)
This means that the radius of the orbit of a geosynchronous satellite that circles the earth is 4.22*10⁴km.
(a) The maximum potential difference across the resistor is 339.41 V.
(b) The maximum current through the resistor is 0.23 A.
(c) The rms current through the resistor is 0.16 A.
(d) The average power dissipated by the resistor is 38.4 W.
<h3>Maximum potential difference</h3>
Vrms = 0.7071V₀
where;
V₀ = Vrms/0.7071
V₀ = 240/0.7071
V₀ = 339.41 V
<h3> rms current through the resistor </h3>
I(rms) = V(rms)/R
I(rms) = (240)/(1,540)
I(rms) = 0.16 A
<h3>maximum current through the resistor </h3>
I₀ = I(rms)/0.7071
I₀ = (0.16)/0.7071
I₀ = 0.23 A
<h3> Average power dissipated by the resistor</h3>
P = I(rms) x V(rms)
P = 0.16 x 240
P = 38.4 W
Learn more about maximum current here: brainly.com/question/14562756
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