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Helen [10]
2 years ago
5

The three forces acting on a hot-air balloon that is moving vertically are its weight, the force due to air resistance and the u

pthrust force.
The hot-air balloon descends vertically at constant speed. The force of air resistance on the balloon is F.
Which weight of material must be released from the balloon so that it ascends vertically at the same constant speed?
Physics
1 answer:
Aleksandr [31]2 years ago
3 0
Let
upthrust = T
weight = W = mg
Air resistance = F

When balloon is descending, air resistance acts upwards (positive)
By Newton's first law, the net force on the balloon is zero, or
T+F-W=0......................(1)

Let w=weight of material dumped so that balloon now travels upwards at constant speed.
Air resistance acts against motion, namely downwards.
The Newton's equation now reads
T-F-(W-w)=0................(2)

Subtract (2) from (1)
T+F-W - (T-F-(W-w)) = 0
Solve for w
w=2F, or
the WEIGHT of material to be released equals twice the resistance of air.
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B. 4 m/s

Explanation:

v=d/t

Running for 300 m at 3 m/s takes 100 seconds and running at 300 m at 6 m/s takes 50 seconds. 100 s + 50 s = 150 s (total time). Total distance is 600 m, so 600 m/ 150 s = 4 m/s.

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A spring balance consists of a pan that hangs from a spring. A damping force Fd = −bv is applied to the balance so that when an
Citrus2011 [14]

Answer:

b ≈ 64 Kg/s

Explanation:

Given

Fd = −bv

m = 2.5 kg

y = 6.0 cm = 0.06 m

g = 9.81 m/s²

The object in the pan comes to rest in the minimum time without overshoot. this means that damping is critical (b² = 4*k*m).

m is given and we find k from the equilibrium extension of 6.0 cm (0.06 m):

∑Fy = 0 (↑)

k*y - W = 0    ⇒   k*y - m*g = 0   ⇒   k = m*g / y

⇒   k = (2.5 kg)*(9.81 m/s²) / (0.06 m)

⇒   k = 408.75 N/m

Hence, if

b² = 4*k*m    ⇒     b = √(4*k*m) = 2*√(k*m)

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2 years ago
Do as physics instructor fred cauthen does and place a tennis ball close to and above the top of a basketball. drop the balls to
Ksivusya [100]

Answer:

after shock

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Explanation:

When the two balls fall they acquire the same speed since they are accelerated by the same force, their weight and the acceleration of the acceleration of gravity. When reaching the floor the mechanical energy of the system is conserved.

Upon reaching the floor, the first ball (basketball) collides with the floor, this process is very fast, at the end of the process the basketball comes out with a velicad up and collides with the much lighter tennis ball that is still descending .

we assume that the shocks are elastic, when solving the momentary and kinetic energy findings, we find the velocities after each shock

     

In this clash the tennis ball acquires a high kinetic speed with an upward direction that makes a very high height high. Again this shock is very fast and the tennis ball almost does not move.

Here we must separate the system, creating a system for the conservation of the energy of the basketball ball and another system for the tennis ball only, the conservation of energy should be applied to each system independently

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As in the elastic shock the final speed of the tennis ball is approximately 2 vo, we can calculate the maximum height

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To reconcile this with the conservation of energy we must calculate the energy for the tennis ball at two points, the first when the crash with the tennis ball ends and at the end point at its maximum height.

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Zanzabum

the correct choices are

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and

C. An ideal solution of two volatile liquids can exist over a range of pressures that are limited by the pressure for which only a trace of liquid remains, and the pressure for which only a trace of gas remains

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