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A toy rocket launcher can project a toy rocket at a speed as high as 35.0 m/s.
1 answer:
Answer:
(a) 62.5 m
(b) 7.14 s
Explanation:
initial speed, u = 35 m/s
g = 9.8 m/s^2
(a) Let the rocket raises upto height h and at maximum height the speed is zero.
Use third equation of motion
h = 62.5 m
Thus, the rocket goes upto a height of 62.5 m.
(b) Let the rocket takes time t to reach to maximum height.
By use of first equation of motion
v = u + at
0 = 35 - 9.8 t
t = 3.57 s
The total time spent by the rocket in air = 2 t = 2 x 3.57 = 7.14 second.
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Refer to the diagram shown.
Because the surface is frictionless, the resistive for, R, is zero.
Let m = the mass of the object.
Let a = acceleration due to the applied force.
Therefore
12.7 N = (m kg)*(a m/s²)
a = 12.7/m m/s²
The object travels 16.1 m in 2.5 s, starting from rest. Therefore
16.1 N = (1/2)*(12.7/m m/s²)*(2.5 s)² = 39.6875/m N
m = 16.1/39.6875 = 0.4057 kg
For freefall, let g = acceleration due to gravity.
The time to fall from 10.3 m is 2.88 s, therefore
10.3 m = (1/2)*(g m/s²)*(2.88 s)² = 4.1472g m
g = 10.3/4.1472 = 2.484 m/s²
Answer:
The gravitational acceleration on the planet is 2.5 m/s² (nearest tenth)
Because the surface is frictionless, the resistive for, R, is zero.
Let m = the mass of the object.
Let a = acceleration due to the applied force.
Therefore
12.7 N = (m kg)*(a m/s²)
a = 12.7/m m/s²
The object travels 16.1 m in 2.5 s, starting from rest. Therefore
16.1 N = (1/2)*(12.7/m m/s²)*(2.5 s)² = 39.6875/m N
m = 16.1/39.6875 = 0.4057 kg
For freefall, let g = acceleration due to gravity.
The time to fall from 10.3 m is 2.88 s, therefore
10.3 m = (1/2)*(g m/s²)*(2.88 s)² = 4.1472g m
g = 10.3/4.1472 = 2.484 m/s²
Answer:
The gravitational acceleration on the planet is 2.5 m/s² (nearest tenth)
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