In Hooke’s law, what does k represent?
2 answers:
Answer:
d) stiffness of the spring
Explanation:
As per Hooke's law for spring we know that restoring force of the spring is proportional to the displacement of the spring from its natural length
so it is given as
here
k = spring constant which depends on the material of the spring
here we know that
k = it is stiffness constant so more stiff spring will have more stiffness constant
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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