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2 years ago

What resistance must be connected in parallel with a 633-Ω resistor to produce an equivalent resistance of 205 Ω?

Physics

1 answer:

alukav5142 [94]2 years ago

5 0

Answer:

303 Ω

Explanation:

Given

Represent the resistors with R1, R2 and RT

R1 = 633

RT = 205

Required

Determine R2

Since it's a parallel connection, it can be solved using.

1/Rt = 1/R1 + 1/R2

Substitute values for R1 and RT

1/205 = 1/633 + 1/R2

Collect Like Terms

1/R2 = 1/205 - 1/633

Take LCM

1/R2 = (633 - 205)/(205 * 633)

1/R2 = 428/129765

Take reciprocal of both sides

R2 = 129765/428

R2 = 303 --- approximated

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I think the answer is 120 minutes hope this helps

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2 years ago

Piaget used a pendulum apparatus to assess whether children had reached the _____ stage of cognitive development.

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Answer:

formal operational

Explanation:

The Formal operational thinking has tested experiments using the pendulum. The method involved three factors (variables) the length of the string, the weight of the object and the force or impulse that is printed on it. The task was to find out which factor is most important to determine the swing speed of the pendulum. Children who have already reached the formal operational stage solve the task systematically, testing one variable at a time.

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2 years ago

Three small spheres, having masses m1 = 1 kg, m2 = 3 kg, and m3 = 4 kg, are held fixed on the x axis in deep space where the eff

Vlad [161]

Answer: $F_{net} = 0.7411 N$ towards the mass $m_{3}$

Explanation: We know that the gravitational force is a long range force which is always attractive in nature.

Given that:

mass $m_{1} = 1 kg$

mass $m_{2} = 3kg$

mass $m_{3} = 4kg$

The masses are positioned on X-axis at the following points:

Position of mass $m_{1}$ $x_{1} = 0$

Position of mass $m_{2}$ $x_{2} = 3$

Position of mass $m_{3}$ $x_{3} = 6$

Mathematically:

Gravitational force on mass $m_{2}$ due to mass $m_{1}$ is given by

$F_{21} = G \frac{m_{1}.m_{2}}{(r_{21})^2}$ ...................(1)

where: $(r_{21})^2$ = the radial distance between masses $m_{2}$ & $m_{1}$ =3

Similarly, gravitational force on mass $m_{2}$ due to mass $m_{3}$ is given by

$F_{23} = G \frac{m_{3}.m_{2}}{(r_{23})^2}$ ............................(2)

where: $(r_{23})^2$ = the radial distance between masses $m_{2}$ & $m_{3}$ =3

Now, put the respective values in the above equations.

$F_{21} = 6.67 \times 10^{-11 }\times \frac{1\times 3}{3^2}$

$F_{21} = 2.2233\times 10^{-11} N$

Again,

$F_{23} = 6.67 \times 10^{-11 }\times \frac{1\times 4}{3^2}$

$F_{23} = 2.9644\times 10^{-11} N$

∵Mass $m_{2}$ is in the middle of the masses $m_{3}$ & $m_{1}$ therefore the forces $F_{23}$ & $F_{21}$ will attract them in radially opposite direction.

∴ $F_{net} = F_{23} -F_{21} \\\\F_{net} = 2.9644-2.2233\\\\F_{net} = 0.7411 N$ towards the mass $m_{3}$

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2 years ago

Read 2 more answers

2. A kg of iron and a kilo of lead, both at 1000 degrees Celsius, are kept on ice. It is discovered after some time that the iro

Sunny_sXe [5.5K]

The specific heat capacity of iron is greater than that of lead hence the iron will melt the ice more than the lead.

<h3>What is the specific heat capacity?</h3>

The specific heat capacity refers to the amount of heat that is required to raise the temperature of 1Kg of a body by 1 K.

We know that the specific heat capacity of iron is greater than that of lead hence the iron will melt the ice more than the lead.

Learn more about heat capacity:brainly.com/question/13499849

#SPJ1

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1 year ago

The electric potential difference across the membrane of a body cell is 0.070 V (higher on the outside than on the inside). The

12345 [234]

Answer:

The electric field is $8.75 \times 10^{6}~v~m^{-1}$ and the ditection is from outer to inner side of the membrane.

Explanation:

We know the electric field ( $\vec{E}$ ) is given by $\vec{E} = - \nabla V$ , 'V' being the potential.

In 1-D, it can be written as

$E=\dfrac{V}{d}$

where 'd' is the separation of space in between the potential difference is created.

Given, $V = 0.070~V~$ and the thickness of the cell membrane is $d = 8.0 \times 10^{-9}~m$ .

Therefore the created electric field through the cell membrane is

$E = \dfrac{0.07~V}{8 \times 10^{-9}~m} = 8.75 \times 10^{6}~m~s^{-1}$

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2 years ago