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ivanzaharov [21]
2 years ago
13

For part 2 I was told that The circuit below will accommodate up to three resistors in series. Explore the circuit. It is a seri

es circuit with ammeters reading amps hooked in series and volt meters hooked in parallel. The circuit has one resistor and two spaces to accommodate resistors. Enter an R to add resistor. Enter a X to remove a resistor. By default, the cells read "enter R for Resistor" so I did that and tried 3 volts( the pics below) then I was asked Complete each of the following parts:(a) Draw the circuit diagram for the three resistors in series that you assembled in Part 2. Label the voltage and the resistances of each resistor.(b) For the three resistors in series, how did the measured total voltage, current across each resistor, total current, and voltage drop across each resistor compare to the values that you calculated?(c) How did the equivalent resistance, voltage drop across each resistor, total current, and total voltage change as you added resistors in series to the circuit?The resistors value (ohms) is 3Power output volts ( 15,10,5) I put three different ones but if I only need to use one then 15 This is 11th grade physics btw

Physics
1 answer:
Amiraneli [1.4K]2 years ago
5 0

(a) Draw the circuit diagram for the three resistors in series that you assembled in Part 2. Label the voltage and the resistances of each resistor

The diagram will look like the following:

As all the resistors are equal, their voltage drop will also be equal, which will be a third of the input voltage

(b) For the three resistors in series, how did the measured total voltage, current across each resistor, total current, and voltage drop across each resistor compare to the values that you calculated?

The value of the measured voltage on each resistor can be seen as to be exactly as calculated, which is 1/3 of the voltage input. The total current can be calculated as:

I=\frac{V}{R+R+R}=\frac{15}{3*3}=\frac{15}{9}\approx1.7A

Which is the one seen on the amperimeter

(c) How did the equivalent resistance, voltage drop across each resistor, total current, and total voltage change as you added resistors in series to the circuit?

The voltage drop on each resistor will decrease once you add more resistors, as there will be more total resistance, and thus less current. As the resistance of each resistor is constant, the relation V=RI given by Ohm's law will be reduced.

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The air pressure inside a car tire is
Valentin [98]

Answer:

0.137m²

Explanation:

Pressure = Force/Area

Given

Force = 41,500N

Pressure = 3.00atm

since 1atm = 101325.00 N/m²

3atm = 3(101325.00)

3atm = 303,975N/m²

Pressure = 303,975N/m²

Get the area

Area = Force/Pressure

Area = 41500/303,975

Area = 0.137m²'

Hence the surface area of the  inside of the tire is 0.137m²

7 0
3 years ago
Fluid flows at 2.0 m/s through a pipe of diameter 3.0 cm. What is the volume flow rate of the fluid in m^3/
fenix001 [56]

Answer:

Volume flow rate = 1.41 \times 10^{-3}m^3/s

Explanation:

The volume flow rate through a channel can be gotten by multiplying its area and its velocity.

The channel under consideration is a circular channel. Hence, the cross-sectional area can be calculated by using the relation for calculating the area of a circle

Cross-sectional area = \pi \times d^{2}/4

A = \pi \times (3\times 10^-2)^2  /4 = 7.07 \times 10 ^-4 m^2

volume flow rate= A =  7.07 \times 10 ^-4 m^2 X 2.0 = 1.41 \times 10^{-3}m^3/s

Volume flow rate = 1.41 \times 10^{-3}m^3/s

7 0
4 years ago
Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.
SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

               (b) m1 = 915kg;

                   m2 = 605kg;

                   m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

F_{pull}=m_{T}.a

m_{T}=\frac{F_{pull}}{a}

m_{T}=\frac{3615}{1.516}

m_{T}=2384.5

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

<u>For </u>m_{1}<u>:</u>

The only force acting On the m_{1} box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

m_{1} = \frac{F_{12}}{a}

m_{1} = \frac{1387}{1.516}

m_{1} = 915kg

<u>For </u>m_{2}<u>:</u>

There are two forces acting on m_{2}: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

m_{2} = \frac{F_{23}-F_{12}}{a}

m_{2} = \frac{2304-1387}{1.516}

m_{2} = 605kg

<u>For </u>m_{3}<u>:</u>

m_{3} = m_{T} - (m_{1}+m_{2})

m_{3} = 2384.5-1520.0

m_{3} = 864.5kg

8 0
3 years ago
calculate the diameter of a silver wire of length 75cm , which is extended by 1.85mm when a 10kg mass is suspended from it's end
sdas [7]

Answer:0.8\ mm

Explanation:

Given

length of wire l=75\ cm

change in length \Delta l=1.85\ mm

mass of wire m=10\ kg

Young's modulus for silver E=7.9\times 10^{10}\ N/m^2

load on wire F=mg

F=10\times 9.8=98\ kg

change in length is given by

\Delta l=\dfrac{Pl}{AE}

Where A=area of cross-section

A=\dfrac{Pl}{\Delta lE}

A=\dfrac{98\times 0.75}{1.85\times 10^{-3}\times 7.9\times 10^{10}}

A=\dfrac{73.5}{14.615\times 10^{7}}

A=5.029\times 10^{-7}\ m^2

also wire is the shape of cylinder so cross-section is given by

A=\dfrac{\pi d^2}{4}=5.029\times 10^{-7}\ m^2

\Rightarrow d^2=\dfrac{5.029\times 10^{-7}\times 4}{\pi }

\Rightarrow d^2=64.02\times 10^{-8}

\Rightarrow d=8\times 10^{-4}\ m

\Rightarrow d=0.8\ mm

4 0
3 years ago
A rocket is fired with an initial VELOCITY OF 100m/s at an angle of 55° above the horizontal, It explodes On the mountain Side 1
GuDViN [60]

Answer

688.32m and 277.44m

Explanation :

⠀

\large{\maltese{\textsf{\underline{To find :-}}}}

The X and Y coordinates of the rocket relative of firing

⠀

⠀

\large{\maltese{\textsf{\underline{Given :-}}}} \\ \\ \sf velocity (v_i) = 100m{s}^{-1} \\ \sf angle ({\theta}_{1}) = 55.0{\degree} \\ \sf time (t) = 12s

⠀

⠀

\Large{\maltese{\textsf{\underline{\underline{Step by Step Solution:-}}}}}

⠀

⠀

<u>The</u><u> </u><u>horizontal</u><u> </u><u>range</u><u> </u><u>of</u><u> </u><u>projectile</u><u> </u><u>at</u><u> </u><u>x</u><u>.</u><u> </u>

⠀

\sf \large{x = v_{xi}} \times t \\ \\ \sf \large{x = v_i \times \cos {\theta}_{i} \times t}

⠀

⠀

\large\textsf{\underline{Now substituting the required values}}

⠀

⠀

\sf x = 300 \times \cos 55{\degree} \times 12 \\ \\ \sf x = 100 \times 0.5756 \times 12 \\ \\ {\underline{\boxed{\bold{ x = 688.32m}}}}

⠀

⠀

The vertical position of projectile at y.

⠀

⠀

\sf \large  y = v_{yi} \times t -  (\frac{1}{2}  \times g  \times {t}^{2}) \\  \\   \sf  \large y = v_i \times  \cos \theta  \times t -  \frac{1}{2} g {t}^{2}

⠀

⠀

\textsf{ \large {\underline{Now substituting the required values}}  }

⠀

⠀

\sf y = 100 \times  \cos55{ \degree} \times 12 -  \frac{1}{2}   \times 9.80 \times  {12}^{2} \\  \\  \sf  y = 100 \times 0.8192 \times 12 - 0.5 \times 9.8 \times 144 \\  \\  \sf y = 983.04 - 705.6 \\  \\  \underline{ \boxed{ \bold{y = 277.44m}}}

⠀

⠀

⠀

<h3><u>Henceforth</u><u>,</u><u> </u><u>the</u><u> </u><u>distance</u><u> </u><u>at</u><u> </u><u>horizon</u><u> </u><u>is</u><u> </u><u>6</u><u>8</u><u>8</u><u>.</u><u>3</u><u>2</u><u>m</u><u> </u><u>and</u><u> </u><u>at</u><u> </u><u>vertical</u><u> </u><u>is</u><u> </u><u>2</u><u>7</u><u>7</u><u>.</u><u>4</u><u>4</u><u>m</u><u>.</u></h3>

8 0
2 years ago
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