Answer:
v = 21.25 km/h
The average velocity is 21.25km/h
Explanation:
Average velocity = total displacement/time taken
v = d/t
Given;
A car travels 50 km in 25 km /h
d1 = 50km
v1 = 25km/h
time taken = distance/velocity
t1 = d1/v1
t1 = 50/25 = 2 hours
and then travels 60km with a velocity 20 km/h
d2 = 60km
v2 = 20km/h
t2 = d2/v2 = 60/20
t2 = 3 hours
and then travels 60km with a velocity 20 km/h in the same direction
d3 = 60km
v3 = 20km/h
t3 = d3/v3 = 60/20
t3 = 3 hours
Average velocity = total displacement/total time taken
v = (d1+d2+d3)/(t1+t2+t3)
v = (50+60+60)/(2+3+3)
v = 170/8
v = 21.25 km/h
The average velocity is 21.25km/h
Answer:
Power lines
Explanation:
Generators induce that current that runs the turbines
Turbines also induce currents from water waves that is transmitted.
Transformers change A.C to D.C or vice versa.
Answer:
-6.6 km/h
Explanation:
In 7hr plane travelled 2020km;
For the first 4hr the average speed was 310km/h;
d=st, s=d/t;
Distance covered in first 4h is d = 310km/h×4h = 1240km;
See the image attached for further solution
This question is incomplete, the complete question is;
Now we will examine the electric field of a dipole. The magnitude and direction of the electric field depends on the distance and the direction. We will investigate in detail just two directions. With charges available in the simulation (all the charges are either positive or negative 1 nC increments).
how do you create a dipole with dipole moment 1 x 10-9 Cm with a direction for the dipole moment pointing to the right. Make a table below that shows the amounts of charge and the distance between the charges. There are many correct answers
Answer:
Given the data in question;
Dipole moment P = 1 × 10⁻⁹ C.m
now dipole pointing to the right;
P→
(-) ---------------->(+) 
d
so let distance between the dipoles be d
∴ P = d
Let
= 1 nC
so
P = d
1 × 10⁻⁹ = 1 × 10⁻⁹ × d
d = (1 × 10⁻⁹) / (1 × 10⁻⁹)
d = 1 m
Also Let
= 2 nC
so
P = d
1 × 10⁻⁹ = 2 × 10⁻⁹ × d
d = (1 × 10⁻⁹) / (2 × 10⁻⁹)
d = 0.5 m
Also Let
= 3 nC
so
P = d
1 × 10⁻⁹ = 3 × 10⁻⁹ × d
d = (1 × 10⁻⁹) / (3 × 10⁻⁹)
d = 0.33 m
such that;
charge distance
1 nC 1.00 m
2 nC 0.50 m
3 nc 0.33 m
4 nC 0.25 m
5 nC 0.20 m