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3 years ago

Which resistors in the circuit must have the same amount of charge passing through each second?

Physics

1 answer:

Nutka1998 [239]3 years ago

5 0

Answer:

Option B) C and D

Explanation:

The amount of charge passing through a resistor each second corresponds to the definition of current, so the problem is asking "which resistors have the same amount of current".

Let's keep in mind that:

- When two resistors are in series (=they are connected into the same branch of the circuit), the current flowing through each resistor is the same

- When two resistors are in parallel (=they are connected into different branches), the potential difference across each resistor is the same

Looking at the previous definitions, we have to find the two resistors in the circuit that are connected in series. We see that resistors C and D are in the same branch, so they are in series: therefore, the current flowing through resistor C and D is the same.

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How much pressure is on the bottom of a pot that holds 20N of soup? the surface area of the pot is 0.05m2

fomenos

Answer:

400

Explanation:

Formula used in solution:

$P = \frac{F}{A}$

$P - pressure,\: F - force,\: A - area$

The given information:

$F = 20N$

$A = 0.05m^{2}$

Solution

$P = \frac{F}{A} = \frac{20}{0.05} = 20 * 20 = 400$

Answer:

$Pressure = 400 \: Pascals$

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3 years ago

All objects, regardless of their mass, fall with the same rate of acceleration on

tamaranim1 [39]

Answer:

A -TRUE

Explanation:

The mass, size, and shape of the object are not a factor in describing the motion of the object. So all objects, regardless of size or shape or weight, free fall with the same acceleration.

4 0

2 years ago

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In the equation Ca(s) + 2H2O(l) -->Ca(OH)2(aq) + H2(g) list each element used as a reactant, and tell how many atoms of each

Mekhanik [1.2K]

The reactants are Calcium, Hydrogen and Oxygen.

We need 1 atom of Calcium, 4 atoms of Hydrogen and 2 atoms of Oxygen.

Hope this helps.

r3t40

3 0

3 years ago

One string of a certain musical instrument is 70.0 cm long and has a mass of 8.79 g . It is being played in a room where the spe

Svetach [21]

To solve this problem we will apply the concepts of linear mass density, and the expression of the wavelength with which we can find the frequency of the string. With these values it will be possible to find the voltage value. Later we will apply concepts related to harmonic waves in order to find the fundamental frequency.

The linear mass density is given as,

$\mu = \frac{m}{l}$

$\mu = \frac{8.79*10^{-3}}{70*10^{-2}}$

$\mu = 0.01255kg/m$

The expression for the wavelength of the standing wave for the second overtone is

$\lambda = \frac{2}{3} l$

Replacing we have

$\lambda = \frac{2}{3} (70*10^{-2})$

$\lambda = 0.466m$

The frequency of the sound wave is

$f_s = \frac{v}{\lambda_s}$

$f_s = \frac{344}{0.768}$

$f_s = 448Hz$

Now the velocity of the wave would be

$v = f_s \lambda$

$v = (448)(0.466)$

$v = 208.768m/s$

The expression that relates the velocity of the wave, tension on the string and linear mass density is

$v = \sqrt{\frac{T}{\mu}}$

$v^2 = \frac{T}{\mu}$

$T= \mu v^2$

$T = (0.01255kg/m)(208.768m/s)^2$

$T = 547N$

The tension in the string is 547N

PART B) The relation between the fundamental frequency and the $n^{th}$ harmonic frequency is

$f_n = nf_1$

Overtone is the resonant frequency above the fundamental frequency. The second overtone is the second resonant frequency after the fundamental frequency. Therefore

$n=3$

Then,

$f_3 = 3f_1$

Rearranging to find the fundamental frequency

$f_1 = \frac{f_3}{3}$

$f_1 = \frac{448Hz}{3}$

$f_1 = 149.9Hz$

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3 years ago

One runner in a relay race has a 1.50 s lead and is running at a constant speed of 3.25 m/s. The runner has 30.0 m to run before

yarga [219]

The second runner must run 3.3m/s. If the leading runner is 1.5 seconds ahead and there are 30m left, the second runner would need to run slightly faster than the lead in order to finish at the same time. To calculate this I did 30/1.5 which gave me 0.05. I added this onto the speed of the lead runner to get 3.3m/s :)

6 0

3 years ago

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