A current of 3A passes through a 2.05cm diameter copper wire. if the wire has 8.5x1028 free electrons per cubic meter. How many
2 answers:
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Answer:
Explanation:
The two cars are under an uniform linear motion. So, the distance traveled by them is given by:
is the same for both cars when the second one catches up with the first. If we take as reference point the initial position of the second car, we have:
We have . Thus, solving for t:
The motion of the package can be described as the motion of a projectile,
given that it has an horizontal velocity and it is acted on by gravity.
- a)
= 70·i
- b) The package will reach ground in approximately <u>12.77 seconds</u>.
- c) The speed of the package as it lands is approximately <u>145.51 m/s</u>.
- d) The path of the package based on a stationary frame of reference is <u>parabolic</u>
- e) The path of the package as seen from the plane is <u>directly vertical</u> downwards
Reasons:
Velocity of the aircraft = 70 m/s
Direction of flight of the aircraft = Eastward
Height from which the aircraft drops the package, h = 800 m
a) The initial velocity of the package, = 70·i
b) The time it will take the package to reach the ground, <em>t</em>, is given by the formula;
Where;
g = The acceleration due to gravity ≈ 9.81 m/s²
Therefore;
Which gives;
The time it will take the package to reach the ground, t ≈ <u>12.77 seconds</u>
c) The vertical velocity just before the package reaches the ground, , is given as follows;
= 2·g·h
Therefore;
= √(2·g·h)
Which gives;
= √(2 × 9.81 × 800) ≈ 125.28
≈ 125.28 m/s
Which gives; = 70·i - 125.28·j
Therefore, |v| = √(70² + (-125.28)²) ≈ 143.51
The speed of the package as it lands, |v| ≈ <u>143.51 m/s</u>
d) The motion of the package that includes both horizontal and vertical motion is a projectile motion.
Therefore;
The path of the package is the path of a projectile, which is a <u>parabolic shape</u>.
e) As seen by someone on the aeroplane, the horizontal velocity will be
zero, therefore, the package will appear as accelerating <u>directly vertical</u>
downwards.
Learn more about projectile motion here: