Changing the current changes the "what" of an electromagnet
2 answers:
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Answer:
The BMX lands 5.4 m from the end of the ramp.
Explanation:
Hi there!
The position of the BMX is given by the position vector "r":
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
Where:
r = position vector at time t
x0 = initial horizontal position
v0 = initial velocity
α = jumping angle
y0 = initial vertical position
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)
Please, see the attached graphic for a better understanding of the situation. At final time, when the bicycle reaches the ground, the vector position will be "r final" (see figure). The y-component of the vector "r final" is - 2.4 m (placing the origin of the frame of reference at the jumping point). With that information, we can use the equation of the y-component of the vector "r" (see above) to calculate the time of flight. With that time, we can then obtain the x-component (rx in the figure) of the vector "r final". Then:
y = y0 + v0 · t · sin α + 1/2 · g · t²
-2.4 m = 0 m + 5.9 m/s · t · sin 40° - 1/2 · 9.8 m/s² · t²
0 = -4.9 m/s² · t² + 5.9 m/s · t · sin 40° + 2.4 m
Solving the quadratic equation:
t = 1.2 s
Now, we can calculate the x-component of the vector "r final" that is the horizontal distance traveled by the bicycle:
x = x0 + v0 · t · cos α
x = 0 m + 5.9 m/s · 1.2 s · cos 40°
x = 5.4 m
The BMX lands 5.4 m from the end of the ramp.
Have a nice day!
Answer:
acceleration are
hollow cylinder < hollow sphere < solid cylinder < solid sphere
Explanation:
To answer this question, let's analyze the problem. Let's use conservation of energy
Starting point. Highest point
Em₀ = U = m g h
Final point. To get off the ramp
Em_f = K = ½ mv² + ½ I w²
notice that we include the kinetic energy of translation and rotation
energy is conserved
Em₀ = Em_f
mgh = ½ m v² +1/2 I w²
angular and linear velocity are related
v = w r
w = v / r
we substitute
mg h = ½ v² (m + I / r²)
v² = 2 gh
v² = 2gh
this is the velocity at the bottom of the plane ,, indicate that it stops from rest, so we can use the kinematics relationship to find the acceleration in the axis ax (parallel to the plane)
v² = v₀² + 2 a L
where L is the length of the plane
v² = 2 a L
a = v² / 2L
we substitute
a =
let's use trigonometry
sin θ = h / L
we substitute
a = g sin θ \ \frac{h}{L} \ \frac{1}{1+ \frac{I}{m r^2 } }
the moment of inertia of each object is tabulated, let's find the acceleration of each object
a) Hollow cylinder
I = m r²
we look for the acerleracion
a₁ = g sin θ 1/1 + mr² / mr² =
a₁ = g sin θ ½
b) solid cylinder
I = ½ m r²
a₂ = g sin θ = g sin θ
a₂ = g sin θ ⅔
c) hollow sphere
I = 2/3 m r²
a₃ = g sin θ
a₃ = g sin θ
d) solid sphere
I = 2/5 m r²
a₄ = g sin θ
a₄ = g sin θ
We already have all the accelerations, to facilitate the comparison let's place the fractions with the same denominator (the greatest common denominator is 210)
a) a₁ = g sin θ ½ = g sin θ
b) a₂ = g sinθ ⅔ = g sin θ
c) a₃ = g sin θ = g sin θ
d) a₄ = g sin θ = g sin θ
the order of acceleration from lower to higher is
a₁ <a₃ <a₂ <a₄
acceleration are
hollow cylinder < hollow sphere < solid cylinder < solid sphere