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Helen [10]
4 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]4 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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A 15kg ball accelerates at a rate of 3m/s/s. What force was required?
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Answer:

<h2>45 N</h2>

Explanation:

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3 0
3 years ago
A proton has a speed of 3.50 Ã 105 m/s when at a point where the potential is +100 V. Later, itâs at a point where the potential
ruslelena [56]

Answer:

(a). The change in the protons electric potential is 0.639 kV.

(b). The change in the potential energy of the proton is 1.022\times10^{-16}\ J

(c). The work done on the proton is -8\times10^{-18}\ J.

Explanation:

Given that,

Speed v= 3.50\times10^{5}\ m/s

Initial potential V=100 V

Final potential = 150 V

(a). We need to calculate the change in the protons electric potential

Potential energy of the proton is

U=qV=eV

Using conservation of energy

K_{i}+U_{i}=K_{f}+U_{f}

\dfrac{1}{2}mv_{i}^2+eV_{i}=\dfrac{1}{2}mv_{f}^2+eV_{f}

]\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e(V_{f}-V_{i})

\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e\Delta V

\Delta V=\dfrac{m(v_{i}^2-v_{f}^2)}{2e}

Put the value into the formula

\Delta V=\dfrac{1.67\times10^{-27}(3.50\times10^{5}-0)^2}{2\times1.6\times10^{-19}}

\Delta V=639.2=0.639\ kV

(b). We need to calculate the change in the potential energy of the proton

Using formula of potential energy

\Delta U=q\Delta V

Put the value into the formula

\Delta U=1.6\times10^{-19}\times639.2

\Delta U=1.022\times10^{-16}\ J

(c). We need to calculate the work done on the proton

Using formula of work done

\Delta U=-W

W=q(V_{2}-V_{1})

W=-1.6\times10^{-19}(150-100)

W=-8\times10^{-18}\ J

Hence, (a). The change in the protons electric potential is 0.639 kV.

(b). The change in the potential energy of the proton is 1.022\times10^{-16}\ J

(c). The work done on the proton is -8\times10^{-18}\ J.

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Answer:

B. develop hypotheses that will be subjected to empirical test.

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