I can guarantee you that it is not
C.<span>the angle that the incident ray makes with a line drawn perpendicular to the reflecting surface I hope this somewhat helps</span>
Answer:
three point charge positioned one x-axis if the charge and corresponding positions are +32Mc x=0 +20Mc x=40cm - 60Mc x=60cm find force 32Mc
Explanation:
The standard wave format for any wave is transverse wave
The EMF of the battery includes the force to to drive across its internal resistance. the total resistance:
R = internal resistance r + resistance connected rv
R = r + rv
Now find the current:
V 1= IR
I = R / V1
find the voltage at the battery terminal (which is net of internal resistance) using
V 2= IR
So the voltage at the terminal is:
V = V2 - V1
This is the potential difference vmeter measured by the voltmeter.
Answer:
Explanation:
from the question we have the following:
distance between Sacramento and los angles = 400 miles
speed of car A = 60 mph
start time of car A = 11 am
speed of car B = 75 mph
start time of car B = 12 pm
distance of Fresno from Los Angeles = 150 miles
- To start off let's allow car A to travel for one hour (from 11 am to 12 pm), during which it would have covered a distance of 60 miles.
- Now the time would be 12 pm and the distance between the two cars would be 400 - 60 (distance traveled by car A within 11 am to 12 pm) = 340 miles
- From 12 pm to the time both cars will meet, the distance covered by car A + distance covered by car B would be equal to 340 miles. Therefore
- Distance covered by car A = speed x time(t) = 60 x t = 60t
- Distance covered by car B = speed x time(t) = 75 x t = 75t
- 60t + 75t = 340 miles
- 135t = 340
- t = 2.51 hours
- Recall that at their meeting point, the distance covered by car B = 75t = 75 x 2.62 = 188.89 miles
- Since Fresno is 150 miles from Los Angeles, car B which is 188.89 miles from Los Angeles at their meeting point would be 188.89 - 150 = 38.89 miles from Fresno
- 38.89 miles would also be the distance of car A from Fresno since that is their meeting point.
