Answer:
B. ) 0.34 m
I definitely guessed and got the right answer so :))
Power delivered = (energy delivered) / (time to deliver the energy)
Power delivered = (4,000 J) / (0.5 sec)
Power delivered = 8,000 watts
I'm a little surprised to learn that Electro draws his power from the mains. This is VERY good news for Spiderman ! It means that Spiderman can always avoid tangling with Electro ... all he has to do is stay farther away from Electro than the length of Electro's extension cord.
But OK. Let's assume that Electro draws it all from the mains. Then inevitably, there must be some loss in Electro's conversion process, between the outlet and his fingertips (or wherever he shoots his bolts from).
The efficiency of Electro's internal process is
<em>(power he shoots out) / (power he draws from the mains) </em>.
So, if he delivers energy toward his target at the rate of 8,000 watts, he must draw power from the mains at the rate of
<em>(8,000 watts) / (his internal efficiency) . </em>
Answer:
![F_{net} = 1.875\,N](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%201.875%5C%2CN)
Explanation:
Asúmase que la patinadora experimenta una aceleración constante. La fuerza neta experimentada por la patinadora:
![F_{net} = (50\,kg)\cdot \left[\frac{\left(15\,\frac{m}{s}\right)^{2}-\left(0\,\frac{m}{s}\right)^{2} }{2\cdot (3000\,m)} \right]](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20%2850%5C%2Ckg%29%5Ccdot%20%5Cleft%5B%5Cfrac%7B%5Cleft%2815%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%5Cright%29%5E%7B2%7D-%5Cleft%280%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%5Cright%29%5E%7B2%7D%20%7D%7B2%5Ccdot%20%283000%5C%2Cm%29%7D%20%5Cright%5D)
![F_{net} = 1.875\,N](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%201.875%5C%2CN)
There's no such thing as "stationary in space". But if the distance
between the Earth and some stars is not changing, then (A) w<span>avelengths
measured here would match the actual wavelengths emitted from these
stars. </span><span>
</span><span>If a star is moving toward us in space, then (A) Wavelengths measured
would be shorter than the actual wavelengths emitted from that star.
</span>In order to decide what's actually happening, and how that star is moving,
the trick is: How do we know the actual wavelengths the star emitted ?