Driving home from school one day, you spot a ball rolling out into the street (Figure 5-21). You brake for 1.20 s, slowing your
950-kg car from 16.0 m/s to 9.50 m/s.
a) What was the average force exerted on your car during braking?
b) How far did you travel while braking?
a) What was the average force exerted on your car during braking?
b) How far did you travel while braking?
1 answer:
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Answer:
Ф_cube /Ф_sphere = 3 /π
Explanation:
The electrical flow is
Ф = E A
where E is the electric field and A is the surface area
Let's shut down the electric field with Gauss's law
Фi = ∫ E .dA = / ε₀
the Gaussian surface is a sphere so its area is
A = 4 π r²
the charge inside is
q_{int} = Q
we substitute
E 4π r² = Q /ε₀
E = 1 / 4πε₀ Q / r²
To calculate the flow on the two surfaces
* Sphere
Ф = E A
Ф = 1 / 4πε₀ Q / r² (4π r²)
Ф_sphere = Q /ε₀
* Cube
Let's find the side value of the cube inscribed inside the sphere.
In this case the radius of the sphere is half the diagonal of the cube
r = d / 2
We look for the diagonal with the Pythagorean theorem
d² = L² + L² = 2 L²
d = √2 L
we substitute
r = √2 / 2 L
r = L / √2
L = √2 r
now we can calculate the area of the cube that has 6 faces
A = 6 L²
A = 6 (√2 r)²
A = 12 r²
the flow is
Ф = E A
Ф = 1 / 4πε₀ Q/r² (12r²)
Ф_cubo = 3 /πε₀ Q
the relationship of these two flows is
Ф_cube /Ф_sphere = 3 /π
Answer:
47.48 m
Explanation:
The horizontal range of an object can be estimated using the equation below:
For the maximum horizontal range to occur, the angle is 45°. Then, the horizontal range will be:
=
In addition, the velocity (v) can be estimated using the equation below:
v = gt/2
Therefore, the maximum horizontal range is:
= (9.81*4.4/2)^2 / 9.81 = 47.48 m