Answer:
Explanation:
40 divided by 10 then which would equal 4. Add the 1.0 , 2 ,and 15 together. Then multply the 60 by 18.0 after you are done dividing the answer is 3 with a remainder of 6.
Answer:
The magnitude of the electric field at a point equidistant from the lines is 
Explanation:
Given that,
Positive charge = 24.00 μC/m
Distance = 4.10 m
We need to calculate the angle
Using formula of angle



We need to calculate the magnitude of the electric field at a point equidistant from the lines
Using formula of electric field

Put the value into the formula



Hence, The magnitude of the electric field at a point equidistant from the lines is 
Using p=v * i
p=250 * 0.8=200w = 0.2kw
power consumed in a day=0.2 *8=1.6 kwh
for one month=1.6 * 30 =48kwh
monthly bills= 48 *3 = Rs 144
Answer:
1.) 400s
2.) 1.875 m/s
3.) 1.125 m/s
Explanation:
Given that you traveled 150.meters south at a speed of 1.50m/s,
Time = distance/ speed
Substitute speed and distance into the formula
Time = 150/1.5
Time = 100 s
and then traveled 600. Meters north at a rate of 2.00 m/s.
Time = 600 / 2
Time = 300 s
1.) Total time = 100 + 300
Total time = 400 s
2.) The average speed will be total distance travelled over total time.
Total distance travelled = 150 + 600
Total distance travelled = 750 m
Substitute all the parameters into the formula
Average speed = 750/400
Average speed = 1.875 m/s
3.) Average velocity will be displacement over total time
Displacement = 600 - 150
Displacement = 450 m
Average velocity = 450/400
Average velocity = 1.125 m/s