Answer:
55.8W
Explanation:
P= V^2/R
R= V^2/P
For series connection
Req= R1+ R2= V^2/310 + V^2/180
R=V^2/P= V^2/310 + V^2/180
But V^2 will cancel out
P= 1/(1/310 + 1/180)
P= 55.8W
Answer:
a. Time=25seconds
b.distance=1041.67m
Explanation:
a.The equation for 
in terms of m/s is 
after conversion.
To find when speed reaches 300km/hr=83.33m/s, we find 
and solve for 
b. From a, above we already have our t=25seconds as the time it takes before the plane is airborne.
#To find distance travelled in that time , we substitute for
in our distance equation:
Hence the distance of the plane before it gets airborne is 1041.67m
Answer:
d=9.462×10^15 meters
Explanation:
<u>Relation between distance, temps and velocity:</u>
d=v*t
t=1year*(365days/1year)*/(24hours/1day)*(3600s/1h)=31536000s
So:
1 light year=d=3*10^8m/s*3.154*10^7s=9.462×10^15 meters
Answer:
Six meters refers to the distance the baseball travels in one second.
Explanation:
6 m/s means that every second, the baseball travels 6m. So, 6 meters is the distance traveled in one second.
I pretty sure that 3 is b and 4 is A and 5 I need a full picture
