Answer:
It would because the shape of the rocket is designed to be able to slice through the air as smooth as possible and now you may be thinking that air is already smooth but when you try to push something as large and heavy like a rocket then the shape of the rocket will be very important. The bottom of the rocket is flatter then the top so it is not designed to fly smoothly through the air. So the rocket would fall vertically downward(If it was still in one piece)because of it's shape. It is easier for the top of the rocket to go smoothly through the air then the bottom.
Explanation:
I am 90% sure this is correct but if I'm not please tell me
Answer: Electrons are transferred → Ions form → Ions are attracted to each other
Explanation:
209 km/h = 58.0 m/s
2as = v² - u²
Assuming the ball started from rest,
a = (58)² / 2 x 0.11
a = 15,290 m/s²
Answer: They create antibodies.
Answer:
a. 05cm from x axis
b. 8cm from x axis
Explanation:
If the net magnetic field is zero and the currents are in the same direction then the thanks point is between the currents i1 and i2 as show in the attachment below
a. Given that i1= 5A and i2=3A
Let assume the null point is xcm from current i1, then the null point will be (4-x)cm from current i2 since the total length is 4cm.
Now the magnetic field of the current i1 from the null point= to magnetic field of current i2 from the null point
B1=B2
μi1/2πx=μi2/2π(4-x)
i1/x=i2/(4-x)
5/x=3/(4-x)
20-5x=3x
8x=20
8x=2.5cm
since from the left of x axis is 2cm, then the null point is 2.5-2 which 0.5cm from the origin x axis.
The null point is 0.5cm from the origin x axis
b. If both current are flowing in opposite direction, the null point lies outside of the current.
Then with same analysis let assume the first current i1 is xcm from the null point and since the total length is 4cm the second current i2 will be (x-4)cm from the null point.
Also the magnetic field of the current i1 from the null point = to magnetic field of current i2 from the null point
B1=B2
μi1/2πx=μi2/2π(x-4)
i1/x=i2/(x-4)
5/x=3/(x-4)
5x-20=3x
2x=20
x=10cm.
This shows that the distance of the null point from current i1 is 10cm and the current i1 is 2cm from the x axis, then the null point is 10-2=8cm from the origin x axis.
The null point is 8cm from the x axis.
Check the attachment to see the diagram of the current and the null points
