Linear momentum is the product of mass and velocity. In this case, it is simply:
How are convergent boundaries different from divergent boundaries
1 answer:
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Answer:
a) x = v₀² sin 2θ / g
b) t_total = 2 v₀ sin θ / g
c) x = 16.7 m
Explanation:
This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity
sin θ = / vo
cos θ = v₀ₓ / vo
v_{oy} = v_{o} sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 13.5 sin 32 = 7.15 m / s
v₀ₓ = 13.5 cos 32 = 11.45 m / s
a) In the x axis there is no acceleration so the velocity is constant
v₀ₓ = x / t
x = v₀ₓ t
the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
t = v_{o} sin θ / g
we substitute
x = v₀ cos θ (2 v_{o} sin θ / g)
x = v₀² /g 2 cos θ sin θ
x = v₀² sin 2θ / g
at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,
b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time
at the highest point the vertical speed is zero
v_{y} = v_{oy} - gt
v_{y} = 0
t = v_{oy} / g
t = v₀ sin θ / g
as the time to get on and off is the same the total time or flight time is
t_total = 2 t
t_total = 2 v₀ sin θ / g
c) we calculate
x = 13.5 2 sin (2 32) / 9.8
x = 16.7 m
Refer to the diagram shown below.
The basket is represented by a weightless rigid beam of length 0.78 m.
The x-coordinate is measured from the left end of the basket.
The mass at x=0 is 2*0.55 = 1.1 kg.
The weight acting at x = 0 is W₁ = 1.1*9.8 = 10.78 N
The mass near the right end is 1.8 kg.
Its weight is W₂ = 1.8*9.8 = 17.64 N
The fulcrum is in the middle of the basket, therefore its location is
x = 0.78/2 = 0.39 m.
For equilibrium, the sum of moments about the fulcrum is zero.
Therefore
(10.78 N)*(0.39 m) - (17.64 N)*(x-0.39 m) = 0
4.2042 - 17.64x + 6.8796 = 0
-17.64x = -11.0838
x = 0.6283 m
Answer: 0.63 m from the left end.
The basket is represented by a weightless rigid beam of length 0.78 m.
The x-coordinate is measured from the left end of the basket.
The mass at x=0 is 2*0.55 = 1.1 kg.
The weight acting at x = 0 is W₁ = 1.1*9.8 = 10.78 N
The mass near the right end is 1.8 kg.
Its weight is W₂ = 1.8*9.8 = 17.64 N
The fulcrum is in the middle of the basket, therefore its location is
x = 0.78/2 = 0.39 m.
For equilibrium, the sum of moments about the fulcrum is zero.
Therefore
(10.78 N)*(0.39 m) - (17.64 N)*(x-0.39 m) = 0
4.2042 - 17.64x + 6.8796 = 0
-17.64x = -11.0838
x = 0.6283 m
Answer: 0.63 m from the left end.
Answer:
N₂=20.05 rpm
Explanation:
Given that
R= 19 cm
I=0.13 kg.m²
N₁ = 24.2 rpm
ω₁= 2.5 rad/s
m= 173 g = 0.173 kg
v=1.2 m
Initial angular momentum L₁
L₁ = Iω₁ - m v r ( negative sign because bird coming opposite to motion of the wire motion)
Final linear momentum L₂
L₂= I₂ ω₂
I₂ = I + m r²
The is no any external torque that is why angular momentum will be conserve
L₁ = L₂
Iω₁ - m v r = I₂ ω₂
Iω₁ - m v r = ( I + m r²) ω₂
Now by putting the all values
Iω₁ - m v r = ( I + m r²) ω₂
0.13 x 2.5 - 0.173 x 1.2 x 0.19 = ( 0.13 + 0.173 x 0.19²) ω₂
0.325 - 0.0394 = 0.136 ω₂
ω₂ = 2.1 rad/s
N₂=20.05 rpm