Answer:
3054.4 km/h
Explanation:
Using the conservation of momentum
momentum before separation = 5M × 2980 Km/h where M represent the mass of the module while 4 M represent the mass of the motor
initial momentum = 14900 M km/h
let v be the new speed of the motor so that the
new momentum = 4Mv and the new momentum of the module = M ( v + 94 km/h )
total momentum = 4Mv + Mv + 93 M = 5 Mv + 93M
initial momentum = final momentum
14900 M km/h = 5 Mv + 93M
14900 km/h = 5v + 93
14900 - 93 = 5v
v = 2961.4 km/h
the speed of the module = 2961.4 + 93 = 3054.4 km/h
The greater the temperature, the greater the volume - this is Charles's law, said by Jacques Charles, a French inventor, scientist, and mathematician.
Answer:
we can see that this time period is independent of the mass of the child so answer would be same if the child mass is different
Explanation:
Natural frequency of a simple pendulum of L length is given as
so the time period of the oscillation is given as
so we will have
also from above formula we can see that this time period is independent of the mass of the child so answer would be same if the child mass is different
** Missing information: The vertical distance from surface of liquid to bottom of the object is sought in this question, with the condition that the object is at equilibrium **
Ans: The vertical distance = y = M/(ρA)
Explanation:Support the vertical distance = y
Object's density = M/(A*h) (since A*h = volume)
By applying the condition,
(M/(Ah))/ρ = y/h
M/(ρAh) = y/h
y = M/(ρA)
This is a perfect opportunity to stuff all that data into the general equation for the height of an object that has some initial height, and some initial velocity, when it is dropped into free fall.
H(t) = (H₀) + (v₀ T) + (1/2 a T²)
Height at any time 'T' after the drop =
(initial height) +
(initial velocity) x (T) +
(1/2) x (acceleration) x (T²) .
For the balloon problem ...
-- We have both directions involved here, so we have to define them:
Upward = the positive direction
Initial height = +150 m
Initial velocity = + 3 m/s
Downward = the negative direction
Acceleration (of gravity) = -9.8 m/s²
Height when the bag hits the ground = 0 .
H(t) = (H₀) + (v₀ T) + (1/2 a T²)
0 = (150m) + (3m/s T) + (1/2 x -9.8 m/s² x T²)
-4.9 T² + 3T + 150 = 0
Use the quadratic equation:
T = (-1/9.8) [ -3 plus or minus √(9 + 2940) ]
= (-1/9.8) [ -3 plus or minus 54.305 ]
= (-1/9.8) [ 51.305 or -57.305 ]
T = -5.235 seconds or 5.847 seconds .
(The first solution means that the path of the sandbag is part of
the same path that it would have had if it were launched from the
ground 5.235 seconds before it was actually dropped from balloon
while ascending.)
Concerning the maximum height ... I don't know right now any other
easy way to do that part without differentiating the big equation.
So I hope you've been introduced to a little bit of calculus.
H(t) = (H₀) + (v₀ T) + (1/2 a T²)
H'(t) = v₀ + a T
The extremes of 'H' (height) correspond to points where h'(t) = 0 .
Set v₀ + a T = 0
+3 - 9.8 T = 0
Add 9.8 to each side: 3 = 9.8 T
Divide each side by 9.8 : T = 0.306 second
That's the time after the drop when the bag reaches its max altitude.
Oh gosh ! I could have found that without differentiating.
- The bag is released while moving UP at 3 m/s .
- Gravity adds 9.8 m/s of downward speed to that every second.
So the bag reaches the top of its arc, runs out of gas, and starts
falling, after
(3 / 9.8) = 0.306 second .
At the beginning of that time, it's moving up at 3 m/s.
At the end of that time, it's moving with zero vertical speed).
Average speed during that 0.306 second = (1/2) (3 + 0) = 1.5 m/s .
Distance climbed during that time = (average speed) x (time)
= (1.5 m/s) x (0.306 sec)
= 0.459 meter (hardly any at all)
But it was already up there at 150 m when it was released.
It climbs an additional 0.459 meter, topping out at 150.459 m,
then turns and begins to plummet earthward, where it plummets
to its ultimate final 'plop' precisely 5.847 seconds after its release.
We can only hope and pray that there's nobody standing at
Ground Zero at the instant of the plop.
I would indeed be remiss if were to neglect, in conclusion,
to express my profound gratitude for the bounty of 5 points
that I shall reap from this work. The moldy crust and tepid
cloudy water have been delicious, and will not soon be forgotten.
