Mercury is added to a cylindrical container to a depth d and then the rest of the cylinder is filled with water.
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
The average dissipated power in a resistor in a ac circuit is:
where R is the resistance, and
is the root mean square current, defined as
where
is the peak value of the current. Substituting the second formula into the first one, we find
and if we re-arrange this formula and use the data of the problem, we can find the value of the peak current I0:
where R is the resistance, and
where
and if we re-arrange this formula and use the data of the problem, we can find the value of the peak current I0: