HELP ME PLSSS!!! WILL MARK BRAINLIEST!!
1 answer:
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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
Answer:
Δ KE = 249158.6 kJ
Explanation:
given data
Truck mass M = 1560 Kg
Truck initial speed, u = 28 m/s
mass of car m = 1070 Kg
initial speed of car u1 = 0 m/s
solution
first we get here final speed by using conservation of momentum that is express as
Mu = (M+m) V .......................1
put here value we get
1560 × 28 = (1560 + 1070 ) V
solve it we get
final speed V = 16.60 m/s
and
Change in kinetic energy will be here
Δ KE = .................2
put here value and we get
Δ KE =
solve it we get
Δ KE = 249158.6 kJ