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1 answer:
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Answer:
the center of mass is 316,670m bellow the release point.
Explanation:
First, we must find the distance at which the objects are in time t = 360s
We will use the formula for vertical distance in free fall
is the initial velocity, is 0 for both stones since they were just dropped.
g is acceleration of gravity and t is time ()
At the first stone has been falling for the entire 360 seconds, its position h1 is:
And at 360 seconds the second stone has been fallin for, so its position h2 is:
And finally using the equation for the center of mass:
We know that the mass of the second stone is twice the mass of the first stone so:
replacing these values in the equation for the center of mass
Finally, replacing the values we found fot h1 and h2:
the center of mass is 316,670m bellow the release point.
Answer:
speed when it reaches y = 4.00cm is
v = 14.9 g.m/s
Explanation:
given
q₁=q₂ =2.00 ×10⁻⁶
distance along x = 3.00cm= 3×10⁻²
q₃= 4×10⁻⁶C
mass= 10×10 ⁻³g
distance along y = 4×10⁻²m
r₁ = =
= 3.61cm = 0.036m
r₂ = =
= 5cm = 0.05m
electric potential V =
change in potential ΔV =
ΔV = , where
2.00μC
ΔV =
ΔV = 2 × 9×10⁹ × 2×10⁻⁶ ×
ΔV= 2.789×10⁵
= ΔV × q₃
ˣ 10×10⁻³ ×v² = 2.789×10⁵× 4 ×10⁻⁶
v² = 223.12 g.m/s
v = 14.9 g.m/s
Given data
*The mass of Bruce is m_1 = 45 kg
*The initial velocity of the Bruce is u_1 = 2 m/s
*The mass of the biff is m_2 = 90 kg
*The initial velocity of the Biff is u_2 = -7 m/s
*The final velocity of the first glider is v_2 = -1 m/s
According to the law of conservation of linear momentum, the total linear momentum of a system remains constant
Applying the law of conservation of momentum as
Substitute the known values in the above expression as
Hence, the speed of the bruce knock backwards is v_1 = -10 m/s