Cars 'A' and 'C' look like they're moving at the same speed. If their tracks are parallel, then they're also moving with the same velocity.
Answer:
3.49 seconds
3.75 seconds
-43200 ft/s²
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration

Time the parachutist falls without friction is 3.19 seconds

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds


Now the initial velocity of the last half height will be the final velocity of the first half height.

Since the height are equal


Time taken to fall the first half is 2.65 seconds
Total time taken to fall is 2.65+1.1 = 3.75 seconds.
When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

Magnitude of acceleration is -43200 ft/s²
Answer: 10 m/s
Explanation: Velocity/Time
50/5= 10
:)
Answer:
FN is the forces acting on a body. When the body is at rest, the net force formula is given by, FNet = Fa + Fg.
Im in 7th and thats all I know so I hope it's enough
Answer:
1.
2.
3.The results from part 1 and 2 agree when r = R.
Explanation:
The volume charge density is given as

We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.
1. Since the cylinder is very long, Gauss’ Law can be applied.

The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is

where ‘h’ is the length of the imaginary Gaussian surface.

2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,

3. At the boundary where r = R:

As can be seen from above, two E-field values are equal as predicted.