Define static electricity and ohms law
1 answer:
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Answer:
The ball will drop 0.881 m by the time it reaches the catcher.
Explanation:
The position of the ball at time "t" is described by the position vector "r":
r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)
Where:
x0 = initial horizontal position.
v0x = initial horizontal velocity.
t = time.
y0 = initial vertical position.
v0y = initial vertical velocity.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
When the ball reaches the catcher, the position vector will be "r final" (see attached figure).
The x-component of the vector "r final", "rx final", will be 17.0 m. We have to find the y-component.
Using the equation of the x-component of the position vector, we can calculate the time it takes the ball to reach the catcher (notice that the frame of reference is located at the throwing point so that x0 and y0 = 0):
x = x0 + v0x · t
17.0 m = 0 m + 40.1 m/s · t
t = 17.0 m/ 40. 1 m/s = 0.424 s
With this time, we can calculate the y-component of the vector "r final", the drop of the ball:
y = y0 + v0y · t + 1/2 · g · t²
Initially, there is no vertical velocity, then, v0y = 0.
y = 1/2 · g · t²
y = -1/2 · 9.8 m/s² · (0.424 s)²
y = -0.881 m
The ball will drop 0.881 m by the time it reaches the catcher.
Answer:
(a) aₓ = 1.33 m/s² and aᵧ = -0.770 m/s²
(b) x = 0.665 m and y = 4.62 m
(c) 3.61 s
Explanation:
(a) There are two ways we can solve this. The first way is to sum the forces in the x and y direction, then use the relation tan 30° = -aᵧ/aₓ, where aᵧ is the acceleration in the +y direction (up) and aₓ is the acceleration in the +x direction (right).
The second way is to sum the forces in the parallel and perpendicular directions to find the acceleration parallel to the incline, a. Then, use the relations aᵧ = -a sin 30° and aₓ = a cos 30°.
Let's try the first method. Sum of forces in the +y direction:
∑F = ma
N cos 30° + Nμ sin 30° − mg = maᵧ
N cos 30° + Nμ sin 30° − mg = -maₓ tan 30°
Sum of forces in the +x direction:
∑F = ma
N sin 30° − Nμ cos 30° = maₓ
Substituting:
N cos 30° + Nμ sin 30° − mg = -(N sin 30° − Nμ cos 30°) tan 30°
N cos 30° + Nμ sin 30° − mg = -N sin 30° tan 30° + Nμ sin 30°
N cos 30° − mg = -N sin 30° tan 30°
N (cos 30° + sin 30° tan 30°) = mg
N = mg / (cos 30° + sin 30° tan 30°)
N = (10 kg) (10 m/s²) / (cos 30° + sin 30° tan 30°)
N = 86.6 N
Now, solving for the accelerations:
N sin 30° − Nμ cos 30° = maₓ
aₓ = N (sin 30° − μ cos 30°) / m
aₓ = (86.6 N) (sin 30° − 0.4 cos 30°) / 10 kg
aₓ = 1.33 m/s²
N cos 30° + Nμ sin 30° − mg = maᵧ
aᵧ = N (cos 30° + μ sin 30°) / m − g
aᵧ = (86.6 N) (cos 30° + 0.4 sin 30°) / 10 kg − 10 m/s²
aᵧ = -0.770 m/s²
Now let's try the second method.
Sum of forces in the perpendicular direction:
∑F = ma
N − mg cos 30° = 0
N = mg cos 30°
Sum of forces in the parallel direction:
∑F = ma
mg sin 30° − Nμ = ma
mg sin 30° − mgμ cos 30° = ma
a = g (sin 30° − μ cos 30°)
a = (10 m/s²) (sin 30° − 0.4 cos 30°)
a = 1.536 m/s²
Solving for the accelerations:
aₓ = a cos 30°
aₓ = 1.33 m/s²
aᵧ = -a sin 30°
aᵧ = -0.770 m/s²
As you can see, the second method is faster and easier, but both methods will give you the same answer.
(b) In the x direction:
Given:
x₀ = 0 m
v₀ = 0 m/s
aₓ = 1.33 m/s²
t = 1 s
Find: x
x = x₀ + v₀ t + ½ at²
x = 0 m + (0 m/s) (1 s) + ½ (1.33 m/s²) (1 s)²
x = 0.665 m
In the y direction:
Given:
y₀ = 5 m
v₀ = 0 m/s
aᵧ = -0.770 m/s²
t = 1 s
Find: y
y = y₀ + v₀ t + ½ at²
y = 5 m + (0 m/s) (1 s) + ½ (-0.770 m/s²) (1 s)²
y = 4.62 m
(c) In the y direction:
Given:
y₀ = 5 m
y = 0 m
v₀ = 0 m/s
aᵧ = -0.770 m/s²
Find: t
y = y₀ + v₀ t + ½ at²
0 m = 5 m + (0 m/s) t + ½ (-0.770 m/s²) t²
t = 3.61 s
