Answer:
vB = 15.4 m/s
Explanation:
Principle of conservation of energy:
Because there is no friction the mechanical energy is conserve
ΔE = 0
ΔE : mechanical energy change (J)
K : Kinetic energy (J)
U: Potential energy (J)
K = (1/2)mv²
U = m*g*h
Where :
m: mass (kg)
v : speed (m/s)
h : hight (m)
Ef - Ei = 0
(K+U)final - (K+U)initial =0
(K+U)final = (K+U)initial
((1/2)mv²+m*g*h)final = ((1/2)mv²+m*g*h)initial , We divided by m both sides of the equation:
((1/2)vB² + g*hB = (1/2 )vA²+ g*hA
(1/2) (vB)² + (9.8)*(14.7) = 0 + (9.8)(26.8 )
(1/2) (vB)² = (9.8)(26.8 ) - (9.8)*(14.7)
(vB)² = (2)(9.8)(26.8 - 14.7)
(vB)² = 237.16
vB = 15.4 m/s : speed of the cart at B
4% of 110 is 4.4. So the possible range of speeds is the interval from 110-4.4 till 110+4.4.
105.6 till 114.4
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.
(Example 1 )
<span>If the Voltage that furnishes the current is an ideal (no internal resistance) Voltage source. Then; </span>
<span>V/R = i </span>
<span>V/2R = i/2 If external resistance doubles, current reduced to 1/2 of original value </span>
<span>V/3R = i/3 If external resistance triples, current reduced to 1/3 of original value </span>
<span>(Example 2) </span>
<span>But if the Voltage that furnishes the current is a practical [contains an internal resistance (Ri)] Voltage source. Then the current is a function of the Voltage source`s internal resistance, which does not double nor triple, plus the external resistance which is being doubled and tripled. </span>
<span>V/(R + Ri) = i </span>
<span>V/(2R + Ri) = greater than i/2 but less than I. </span>
<span>V/(3R + Ri) = greater than i/3 but less than i/2</span>
