meg goes swimming on a hot afternoon. when she comes out of the pool, her foot senses that the pavement is unbearably hot. suppo
1 answer:
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Answer:
(a) The package lands 682 meters horizontally ahead from the point the package was dropped from the plane
(b) The horizontal component = 39.0 m/s
The vertical component = 171.55 m/s
(c) The angle of impact is 77.19°
Explanation:
The parameters given are;
Velocity of the plane, vₓ = 39.0 m/s
Height of the plane above the ground, h = 1.50 × 10² m = 1,500 m
(a) The time, t, before the package hits the ground when dropped from the plane is given by the relation;
h = u·t + 1/2×g×t²
Where:
g = Acceleration due to gravity = 9.81 m/s²
u = Initial vertical velocity = 0 m/s
Hence;
1500 = 0×t + 1/2 × 9.81 × t² = 4.905·t²
∴ t = √(1500/4.905) = 17.49 s
The horizontal distance the package travels before landing = 17.49 × 39 ≈ 682 m
The package lands 682 meters horizontally ahead from the point the package was dropped from the plane
(b) The vertical velocity, , of the package just before landing is given by the relation;
² = u² + 2·g·h
u = 0 m/s
∴ ² = 0 + 2×9.81×1500 = 29430 m²/s²
= √29430 = 171.55 m/s
Hence the horizontal component = 39.0 m/s
The vertical component = 171.55 m/s
(c) The angle of impact, θ, is given as follows;
∴ θ = tan⁻¹(4.4) = 77.19°.
Answer:
a)
b)
Step-by-step explanation.
In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).
Once we drew the points, we can start calculating the forces:
which yields:
(I will assume the positions are in meters)
Next, we can make use of the force formula:
so we substitute the values:
which yields:
Now we can find its components:
And we can now write them together for the first force, so we get:
We continue with the next force. The procedure is the same so we get:
which yields:
Next, we can make use of the force formula:
which yields:
Now we can find its components:
And we can now write them together for the second, so we get:
We continue with the next force. The procedure is the same so we get:
which yields:
Next, we can make use of the force formula:
which yields:
Now we can find its components:
And we can now write them together for the third force, so we get:
So in order to find the resultant force, we need to add the forces together:
so we get:
So when adding the problem together we get that:
which is the answer to part a), now let's take a look at part b).
b)
Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:
which yields:
and now we take the formula for the electric field which is:
so we go ahead and substitute:
