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2 years ago

Label the places where you decide to measure current and voltage. Try several places

Physics

1 answer:

katen-ka-za [31]2 years ago

8 0

Answer:

Explore the Intro screen of Circuit Construction Kit DC

Build a circuit that shows how to make a light bulb light up.

Figure out how to measure current and voltage.

Insert an image of your circuit with the current and voltage measured.

Imagine you’re an engineer making a string of battery powered holiday lights. If a bulb burns out current cannot flow through that bulb any longer like if the wire at the bulb has been cut. Figure out how to hook up 2 light bulbs and a battery so that when one bulb burns out or is disconnected the other stays lit.

Insert images to illustrate that your circuit works as expected.

Explain why you think it works.

Imagine that you want to make sure the battery for your string of lights will last as long as possible. A battery will last longer if it powers a circuit with low current. How could you hook up a battery and 2 light bulbs so the least amount of current flows through the battery? Use the measurement tools in the simulation to check your design.

Insert images to illustrate that your circuit works as expected.

Explain why you think it works.

Develop your understanding Part 2

Instructions: Your goal in this part is to write rules to describe how patterns of current and voltage in a circuit relate to the structure of the circuit. You will need to measure current and voltage in multiple places on several different circuits.

Examples:

Measuring Current

Measuring Voltage

“Current” is the flow of charge, measured in Amps (Coulombs/s). An ammeter measures the current past a single point in the circuit.

The current flowing through point 1 can be written as:

I1 = 0.09 A.

“Voltage” is a measure of the difference in electric potential between two points. The voltmeter measures this difference by placing the two leads (pronounced “leeds”) at two different points.

The voltage between points A and B can be written as VAB = 9 V.

Explanation:

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3 years ago

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Distance Between Two Cars A car leaves an intersection traveling west. Its position 4 sec later is 19 ft from the intersection.

faltersainse [42]

Answer:

The answer to the question is;

The rate at which the distance between the two cars is changing is equal to 14.4 ft/sec.

Explanation:

We note that the distance traveled by each car after 4 seconds is

Car A = 19 ft in the west direction.

Car B = 26 ft in the north direction

The distance between the two cars is given by the length of the hypotenuse side of a right angled triangle with the north being the y coordinate and the west being the x coordinate.

Therefore, let the distance between the two cars be s

we have

s² = x² + y²

= (19 ft)² + (26 ft)² = 1037 ft²

s = $\sqrt{1037 ft^2}$ = 32.202 ft.

The rate of change of the distance from their location 4 seconds after they commenced their journeys is given by;

Since s² = x² + y² we have

$\frac{ds^{2} }{dt} = \frac{dx^{2} }{dt} + \frac{dy^{2} }{dt}$

→ $2s\frac{ds }{dt} = 2x\frac{dx}{dt} + 2y\frac{dy }{dt}$ which gives

$s\frac{ds }{dt} = x\frac{dx}{dt} + y\frac{dy }{dt}$

We note that the speeds of the cars were given as

Car B moving north = 12 ft/sec, which is the y direction and

Car A moving west = 8 ft/sec which is the x direction.

Therefore

$\frac{dy }{dt}$ = 12 ft/sec and

$\frac{dx}{dt}$ = 8 ft/sec

$s\frac{ds }{dt} = x\frac{dx}{dt} + y\frac{dy }{dt}$ becomes

$32.202 ft.\frac{ds }{dt} = 19 ft \times 8 \frac{ft}{sec} + 26ft\times 12\frac{ft}{sec}$ = 464 ft²/sec

$\frac{ds }{dt} = \frac{464\frac{ft^{2} }{sec} }{32.202 ft.}$ = 14.409 ft/sec ≈ 14.4 ft/sec to one place of decimal.

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I truly believe it's D. iron oxide

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3 years ago

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You have been hired as a technical consultant for an early-morning cartoon series for children to make sure that the science is

katen-ka-za [31]

The initial potential energy of the wagon containing gold boxes will enable

it roll down the hill when cut loose.

The Lone Ranger and Tonto have approximately <u>5.1 seconds</u>.

Reasons:

Mass wagon and gold = 166 kg

Location of the wagon = 77 meters up the hill

Slope of the hill = 8°

Location of the rangers = 41 meters from the canyon

Mass of Lone Ranger, m₁ = 65 kg

Mass of Tonto m₂ = 66 kg

Solution;

Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m

Potential energy = m·g·h

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912

Potential energy of wagon, P.E. ≈ 17457.0912 J

By energy conservation, P.E. = K.E.

$K.E. = \mathbf{\dfrac{1}{2} \cdot m \cdot v^2}$

Where;

v = The velocity of the wagon a the bottom of the cliff

Therefore;

$\dfrac{1}{2} \times 166 \times v^2 = 17457.0912$

$v = \sqrt{\dfrac{17457.0912}{\dfrac{1}{2} \times 166} } \approx 14.5$

Velocity of the wagon, v ≈ 14.5 m/s

Momentum = Mass, m × Velocity, v

Initial momentum of wagon = m·v

Final momentum of wagon and ranger = (m + m₁ + m₂)·v'

By conservation of momentum, we have;

m·v = (m + m₁ + m₂)·v'

$\therefore v' = \mathbf{ \dfrac{m \cdot v}{(m + m_1 + m_2) }}$

Which gives;

$\therefore v' = \dfrac{166 \times 14.5}{(166 + 65 + 66) } \approx 8.1$

The velocity of the wagon after the Ranger and Tonto drop in, v' ≈ 8.1 m/s

$Time = \dfrac{Distance}{Velocity}$

$\mathrm{The \ time \ the\ Lone \ Ranger \ and \ Tonto \ have, \ t} = \dfrac{41 \, m}{8.1 \, m/s} \approx 5.1 \, s$

The Lone Range and Tonto have approximately <u>5.1 seconds</u> to grab the

gold and jump out of the wagon before the wagon heads over the cliff.

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3 years ago

Calculate the first and second velocities of the car with four washers attached to the pulley, using the formulas v1 = 0. 25 m /

castortr0y [4]

The first and second velocities of the car are 0.25/t1 and 0.50/t2 respectively.

Given data:

The value of the first distance is, d1 = 0.25 m.

The value of the second distance is, d2 = 0.50 m.

The ratio of distance covered by an object in a specific direction and the time taken to cover the distance is known as the velocity of the object. Mathematically, the expression for the velocity is,

v = d/t

Here,

d is the distance covered.

t is the average time taken to cover the distance.

Then the first velocity of the car at 0.25 m is,

v1 = d1/t1

v1 = 0.25 / t1

here, t1 is the average time for first distance.

And the second velocity of the car with four washers at the 0. 50 m mark is,

v2 = d2/t2

v2 = 0.50 /t2

here, t2 is the average time for the second distance.

Thus, we can conclude that the first and second velocities of the car are 0.25/t1 and 0.50/t2 respectively.

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2 years ago