Answer:
A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 6.10s later.
Explanation:
A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 6.0 s later. What was the rocket’s acceleration?
Answer:
it could be either or because it doesnt just depend on the height but it also depends on the pressure but then again the question didnt ask anything about the pressure so the answer should be true
Explanation:
recoils and must be tethered or he's gone.
According to Boyle’s law, For a fixed amount of an ideal gas kept at a fixed temperature, P (pressure) and V (volume) are inversely proportional.
Therefore,
![P_{1} V_{1} =P_{2} V_{2}](https://tex.z-dn.net/?f=P_%7B1%7D%20V_%7B1%7D%20%3DP_%7B2%7D%20V_%7B2%7D)
Given
,
and
.
Thus,
![V_{2} = \frac{P_{1}\times V_{1} }{ P_{2} } = \frac{1\times 15}{0.75} =20 m^3](https://tex.z-dn.net/?f=V_%7B2%7D%20%3D%20%5Cfrac%7BP_%7B1%7D%5Ctimes%20V_%7B1%7D%20%20%7D%7B%20P_%7B2%7D%20%7D%20%3D%20%5Cfrac%7B1%5Ctimes%2015%7D%7B0.75%7D%20%3D20%20m%5E3)
We have all the charges for q1, q2, and q3.
Since k = 8.988x10^2, and N=m^2/c^2
F(1) = F (2on1) + F (3on1)
F(2on1) = k |q1 q2| / r(the distance between the two)^2
k^ | 3x10^-6 x -5 x 10^-6 | / (.2m)^2
F(2on1) = 3.37 N
Since F1 is 7N,
F(1) = F (2on1) + F (3on1)
7N = 3.37 N + F (3on1)
Since it wil be going in the negative direction,
-7N = 3.37 N + F (3on1)
F(3on1) = -10.37N
F(3on1) = k |q1 q3| / r(the distance between the two)^2
r^2 x F(3on1) = k |q1 q3|
r = sqrt of k |q1 q3| / F(3on1)
= .144 m (distance between q1 and q3)
0 - .144m
So it's located in -.144m
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