Answer:
distance = 6.1022 x10^16[m]
Explanation:
To solve this problem we must use the formula of the average speed which relates distance to time, so we have
v = distance / time
where:
v = velocity = 3 x 10^8 [m/s]
distance = x [meters]
time = 6.45 [light years]
Now we have to convert from light-years to seconds in order to get the distance in meters.
![t = 6.45 [light-years]*365[\frac{days}{1light-year}]*24[\frac{hr}{1day}] *60[\frac{min}{1hr}]*60[\frac{seg}{1min} ] =203407200 [s]](https://tex.z-dn.net/?f=t%20%3D%206.45%20%5Blight-years%5D%2A365%5B%5Cfrac%7Bdays%7D%7B1light-year%7D%5D%2A24%5B%5Cfrac%7Bhr%7D%7B1day%7D%5D%20%2A60%5B%5Cfrac%7Bmin%7D%7B1hr%7D%5D%2A60%5B%5Cfrac%7Bseg%7D%7B1min%7D%20%5D%20%3D203407200%20%5Bs%5D)
Now using the formula:
distance = v * time
distance = (3*10^8)*203407200
distance = 6.1022 x10^16[m]
you would have to look up the formula and know what time and how fast and how far
Type of energy was used is kinetic energy.
Answer:
I = 0.451 amp
Explanation:
given,
C = 8.0 µF
V = 2 V
resistor connected between two terminal = 6 Ω
current flowing through resistor = 13 µsec
Q = CV
Q = 8 x 2
Q = 16 µC
for an RC discharge circuit
V = V_0e^{-\dfrac{t}{RC}}
I = \dfrac{-Q_0}{RC}e^{-\dfrac{t}{RC}}
t = 13 µsec
I = \dfrac{-16}{6\times 2}e^{-\dfrac{13}{6\times 2}}
I = 0.451 amp
neglecting -ve sign just to show direction.