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statuscvo [17]
3 years ago
5

A student has designed a circuit with one battery, two light bulbs, and two switches, pictured to the left. She would like for e

ach switch to control one light bulb, so that when one switch is open and the other is closed, one of the light bulbs will remain lit. From the image, we can see that neither light bulb is lit, even though one of the switches is closed. What is the error the student made? Your explanation should include a discussion of the flow of electrons in the circuit.

Physics
2 answers:
Gnoma [55]3 years ago
5 0

Answer:

The bulb is not powered because the negative end of the battery is not connected to the electric current,making it dysfunctional. The electrons have to flow through the positive and negative ends of the battery in order for it to work.

Hope this helps.

Misha Larkins [42]3 years ago
5 0
A.Switch the circuits A hopes this helps you out
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Rick shoots a basketball at an angle of 35' from the horizontal. It leaves his hands 7 feet from the ground with a velocity of 2
Korvikt [17]

Given:

The angle of projection of the basketball, θ=35°

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To find:

The parametric equations describing the shot.

Explanation:

The range, x of the basketball is given by,

x=v\cos\theta t

On substituting the known values,

\begin{gathered} x=20\times\cos35\degree\times t \\ \implies x=16.4t \end{gathered}

The change in the height, y of the basketball is given by,

y=-v\sin\theta t+\frac{1}{2}gt^2

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On substituting the known values,

\begin{gathered} y=-20\times\sin35\degree\times t+\frac{1}{2}\times32\times t^2 \\ \implies y=-11.5t+16t^2 \end{gathered}

Final answer:

The parametric equations describing the shot are

\begin{gathered} \begin{equation*} x=16.4t \end{equation*} \\ \begin{equation*} y=-11.5t+16t^2 \end{equation*} \end{gathered}

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