As an object moves, the distance it travels increases with time. Agree Disagree
2 answers:
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Answer:
a) The fragment speeds of 0.3 kg is 33.3 m / s on the y axis
0.7 kg is 109.4 ms on the x axis
b) Y = 109.3 m
Explanation:
This is a moment and projectile launch exercise.
a) Let's start by finding the initial velocity of the projectile
sin 40 = voy / v₀
= v₀ sin 40
= 50.0 sin40
= 32.14 m / s
cos 40 = v₀ₓ / V₀
v₀ₓ = v₀ cos 40
v₀ₓ = 50.0 cos 40
v₀ₓ = 38.3 m / s
Let us define the system as the projectile formed t all fragments, for this system the moment is conserved in each axis
Let's write the amounts
Initial mass of the projectile M = 2.0 kg
Fragment mass 1 m₁ = 1.0 kg and its velocity is vₓ = 0 and = -10.0 m / s
Fragment mass 2 m₂ = 0.7 kg moves in the x direction
Fragment mass 3 m₃ = 0.3 kg moves up (y axis)
Moment before the break
X axis
p₀ₓ = m v₀ₓ
Y Axis y
= 0
After the break
X axis
= m₂ v₂
Axis y
= m₁ v₁ + m₃ v₃
Let's write the conservation of the moment and calculate
Y Axis
0 = m₁ v₁ + m₃ v₃
Let's clear the speed of fragment 3
v₃ = - m₁ v₁ / m₃
v₃ = - (-10) 1 / 0.3
v₃ = 33.3 m / s
X axis
M v₀ₓ = m₂ v₂
v₂ = v₀ₓ M / m₂
v₂ = 38.3 2 / 0.7
v₂ = 109.4 m / s
The fragment speeds of 0.3 kg is 33.3 m / s on the y axis
0.7 kg is 109.4 ms on the x axis
b) The speed of the fragment is 33.3 m / s and has a starting height of where the fragmentation occurred, let's calculate with kinematics
² =
² - 2 gy
0 = ²-2gy
y = ² / 2g
y = 32.14² / 2 9.8
y = 52.7 m
This is the height where the break occurs, which is the initial height for body movement of 0.3 kg
² =
² - 2 g y₂
0 = ² - 2 g y₂
y₂ = ² / 2g
y₂ = 33.3²/2 9.8
y₂ = 56.58 m
Total body height is
Y = y + y₂
Y = 52.7 + 56.58
Y = 109.3 m
Answer:
The force applied on one wheel during braking = 6.8 lb
Explanation:
Area of the piston (A) = 0.4
Force applied on the piston(F) = 6.4 lb
Pressure on the piston (P) =
⇒ P =
⇒ P = 16
This is the pressure inside the cylinder.
Let force applied on the brake pad =
Area of the brake pad ()= 1.7
Thus the pressure on the brake pad () =
When brake is applied on the vehicle the pressure on the piston is equal to pressure on the brake pad.
⇒ P =
⇒ 16 =
⇒ = 16 ×
Put the value of we get
⇒ = 16 × 1.7
⇒ = 27.2 lb
This the total force applied during braking.
The force applied on one wheel = =
= 6.8 lb
⇒ The force applied on one wheel during braking.