Answer:
Explanation:
Given that,
The initial speed of a car, u = 0
Time, t = 18 s
Distance, d = 390 m
We need to find the acceleration of the car. Let it is a. Using the second equation of motion to find it.
or
So, the acceleration of the car is 
.
It is an imaginary transformer which has no core loss, no ohmic resistance and no leakage flux. The ideal transformer has the following important characteristic. The resistance of their primary and secondary winding becomes zero. The core of the ideal transformer has infinite permeability.
Answer:
3.626 m/s
Explanation:
v=d/t
1. -0.02/0 = 0 m/s
2. 0.86/0.2 = 4.3 m/s
3. 1.71/0.4 = 4.275 m/s
4. 2.54/0.6 = 4.23 m/s
5. 3.32/0.8 = 4.15 m/s
6. 4.08/1.0 = 4.08 m/s
7. 4.79/1.2 = 3.99 m/s
8. 5.48/1.4 = 3.91 m/s
9. 6.15/1.6 = 3.84 m/s
10. 6.76/1.8 = 3.76 m/s
11. 7.37/2.0 = 3.66 m/s
12. 7.92/2.2 = 3.6 m/s
13. 8.45/2.4 = 3.52 m/s
14. 8.96/2.6 = 3.45 m/s
the mean of these numbers is 3.626
his average velocity ks 3.626 m/s
Answer:
=0.855V
Explanation:
The induced voltage can be calculated using below expression
E =B x dA/dt
Where dA/dt = area
B= magnetic field = 6.90×10-5 T.
We were given speed of 885 km/h but we will need to convert to m/s for consistency of unit
speed = 885 km/h
speed = 885 x 10^3 m/hr
speed = 885 x 10^3/60 x60 m/s
speed = 245.8 m/s
If The aircraft wing sweep out" an area
at t= 50.4seconds then we have;
dA/dt = 50.4 x 245.8
= 123388.32m^2/s
Then from the expression above
E =B x dA/dt substitute the values of each parameters, we have
E = 6.90 x 10^-5 x 12388.32 V
E =0.855V
Hence, the average induced voltage between the tips of the wings is =0.855V
