How strong is the electric field between the plates of a 0.86 µf air-gap capacitor if they are 2.0 mm apart and each has a char
1 answer:
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Answer:
a) θ₁ = 0.487º , b) t = 0.400 s , x = 11.73 ft
Explanation:
For this exercise let's use the projectile launch relationships.
The initial height is I = 5 ft and the final height y = 4 ft
y = y₀ + t - ½ g t²
The distance to the band is x = 40 yard (3 ft / 1 yard) = 120 ft
x = v₀ₓ t
t = x / v₀ₓ
We replace
y –y₀ = v_{oy} x / v₀ₓ - ½ g x² / v₀ₓ²
v_{oy} = v₀ sin θ
v₀ₓ = vo cos θ
y –y₀ = x tan θ - ½ g x² / v₀² cos² θ
5-4 = 120 tan θ - ½ 32 120 / (300 2 cos2 θ)
1 = 120 tan θ - 0.0213 sec² θ
Let's use the trigonometry relationship
Sec² θ = 1 - tan² θ
1 = 120 tan θ - 0.0213 (1 –tan²θ)
0.0213 tan²θ + 120 tanθ -1.0213 = 0
We change variables
u = tan θ
u² + 5633.8 u - 48.03 = 0
We solve the second degree equation
u = [-5633.8 ±√(5633.8 2 + 4 48.03)] / 2
u = [- 5633.8 ± 5633.82] / 2
u₁ = 0.0085
u₂= -5633.81
u = tan θ
θ = tan⁻¹ u
For u₁
θ₁ = tan⁻¹ 0.0085
θ₁ = 0.487º
For u₂
θ₂ = -89.99º
The launch angle must be 0.487º
b) let's look for the time it takes for the arrow to arrive
x = v₀ₓ t
t = x / v₀ cos θ
t = 120 / (300 cos 0.487)
t = 0.400 s
The deer must be at a distance of
v = 20 mph (5280 ft / 1 mi) (1 h / 3600s) = 29.33 ft / s
x = v t
x = 29.33 0.4
x = 11.73 ft
1) 148 J
When lifting an object, the work done on the object is equal to its change in gravitational potential energy. Mathematically:
where
mg is the weight of the object
is the change in height
For the box in this problem,
mg = 185 N
Substituting into the equation, we find:
2) (a) 28875 J
The work done by a force applied parallel to the direction of motion of the object is given by
where
F is the magnitude of the force
d is the displacement
In this problem,
F = 825 N is the force applied by the two students together
d = 35 m is the displacement of the car
Substituting,
2) (b) 57750 J
As seen previously, the equation that gives the work done by the force is
We see that the work done is proportional to the magnitude of the force: therefore, if the force is doubled, then the work done is also doubled.
The work done previously was
W = 28875 J
Now the force is doubled, so the new work done will be
3) 4.4 J
In this case, the force acting on the ball is the force of gravity, whose magnitude is:
where
m = 0.180 kg is the mass of the ball
is the acceleration of gravity
Solving the equation,
Now we find the work done by gravity using the same formula applied before:
where d = 2.5 m is the displacement of the ball. We can apply this version of the formula since the force is parallel to the displacement. Substituting,
4) 595.2 kg
In this case, we have the work done on the box:
W = 7.0 kJ = 7000 J
And we also know the change in height of the box:
As we stated in part a), the work done on the box is equal to its change in gravitational potential energy:
Solving for m, we find
And substituting the numerical values, we find the mass of the box:
5) They do the same work
In fact, the net work done by each person on the box is equal to the change in gravitational potential energy of the box:
Where is the difference in height between the final position and the initial position of the box.
This means that the work done on the box depends only on its initial and final position, not on the path taken. The two men carry the box along different paths, however the reach at the end the same position, and they started from the same position: this means that the value of is the same for both of them, so the work they have done is exactly the same.