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docker41 [41]
3 years ago
12

two objects are thrown from the top of a tall building and experience no appreciable air resistance. one is thrown up and the ot

her is thrown down, both with the same initial speed. what are their speeds when they hit the street?
Physics
2 answers:
NNADVOKAT [17]3 years ago
5 0

Answer:

Final speeds are equal to each other.

Explanation:

Both objects experiment a change in their velocities because of gravity. Let model each object by following equations:

Object A - Throw Up

v = -\sqrt{v_{o}^{2}+2\cdot g \cdot h}

Object B - Throw Down

v = -\sqrt{v_{o}^{2}+2\cdot g \cdot h}

Where h is the height difference between the top of the building and the street. Let consider upward speed positive.

Final speeds are equal to each other.

Iteru [2.4K]3 years ago
3 0

Answer:

The two objects are traveling at the same speed.

Explanation:

Neglecting air resistance, an object that is thrown up from the top of a tall building has the same speed as the second object thrown down from the top of the same tall building since the initial speed is the same.

The object thrown up is not traveling faster neither is the object thrown down traveling faster.  

Therefore, the two objects will have the same speed when they hit the ground but their time of landing might be different.

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A capacitor with an initial potential difference of 185 V is discharged through a resistor when a switch between them is closed
GrogVix [38]

Answer:

  • a. \tau =  2.1161 s
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Explanation:

<h3>a.</h3>

The equation for the voltage V of  discharging capacitor in an RC circuit at time t is:

V(t) = V_0 e^{(- \frac{t}{\tau}) }

where V_0 is the initial voltage, and \tau is the time constant.

For our problem, we know

V_0 = 185 \ V

and

V(10 \ s) = V_0 e^{(- \frac{10 \ s}{\tau}) } = 1.64 \ V

So

185 \ V \ e^{(- \frac{10 \ s}{\tau}) } = 1.64 \ V

e^{(- \frac{10 \ s}{\tau}) } = \frac{1.64 \ V}{ 185 \ V }

ln (e^{(- \frac{10 \ s}{\tau}) } ) = ln (\frac{1.64 \ V}{ 185 \ V })

- \frac{10 \ s}{\tau}  = ln (\frac{1.64 \ V}{ 185 \ V })

\tau =  \frac{-10 \s}{ln (\frac{1.64 \ V}{ 185 \ V }) }

This gives us

\tau =  2.1161 s

and this is the time constant.

<h3>b.</h3>

At t = 18.8 s we got:

V(18.8 \ s) = 185 \ V  \ e^{(- \frac{18.8 \ s}{2.1161 s}) }

V(18.8 \ s) = 185 \ V \ e^{(- \frac{18.8 \ s}{2.1161 s}) }

V(18.8 \ s) = 0.0256 \ V

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