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3 years ago

A physics student has a battery and three equal resistors. If she uses all of the

Physics

2 answers:

Soloha48 [4]3 years ago

4 0

Answer:

she can arrange the resistors and battery like a butterfly so that you know.... she can do that

Korvikt [17]3 years ago

3 0

Question: A physics student has a battery and an assortment of resistors. If she uses all of the resistors, how should she arrange them in a circuit to obtain the maximum current flow through the battery and the circuit?

Ans: she should arrange them in parallel

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A single loop of wire with an area of 0.0900 m2 is in a uniform magnetic field that has an initial value of 3.80 T, is perpendic

erica [24]

Answer:

(a) 0.0171 V

Explanation:

A = 0.09 m^2, dB/dt = 0.190 T/s

(a) According to the law of electromagntic induction

e = dФ / dt

e = A dB / dt

e = 0.09 x 0.190 = 0.0171 V

(b)

as we know

i = e / R

we can find induced current by dividing induced emf by resistance

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3 years ago

A square wave has amplitude 0 V for the low voltage and 4 V for the high voltage. Calculate the average voltage by integrating o

Margarita [4]

Answer:

V_{average} = $\frac{1}{2} V_o$ , V_{average} = 2 V

Explanation:

he average or effective voltage of a wave is the value of the wave in a period

V_average = ∫ V dt

in this case the given volage is a square wave that can be described by the function

V (t) = $\left \{ {{V=V_o \ \ \ t< \tau /2} \atop {V=0 \ \ \ \ t> \tau /2 } } \right.$

to substitute in the equation let us separate the into two pairs

V_average = $\int\limits^{1/2}_0 {V_o} \, dt + \int\limits^1_{1/2} {0} \, dt$

V_average = $V_o \ \int\limits^{1/2}_0 {} \, dt$

V_{average} = $\frac{1}{2} V_o$

we evaluate V₀ = 4 V

V_{average} = 4 / 2)

V_{average} = 2 V

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3 years ago

Pitch describes how high or low a sound is. The pitch of a sound is most dependent upon the of the sound wave.

Reil [10]

Answer: C. Frequency

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3 years ago

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It can hurt and harm the body on many different and levels

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3 years ago

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The half-life of Co-55 is 175 hours. How much of a 4000 g of Cobalt-55 sample would be left after 525 hours?

Harrizon [31]

We know that whatever amount we start with, half of it decays and forms atoms of other elements in 175 hours. So in order to figure out how much is left after 525 hours, we'll need to know how many half-lifes pass in that amount of time.

Well, (525 divided by 175) is exactly 3 half-lifes. So this will be easy.

-- After 1 half-life . . .

. . . . . 50% decays, 50% is still there.

-- After the 2nd half-life . . .

. . . . . (half of the leftover 50%) = another 25% decays, 25% is left.

-- After the 3rd half-life . . .

. . . . . (half of the leftover 25%) = another 12.5% decays, 12.5% is left.

12.5% of 4,000g = (0.125 x 4,000g) = 500 g .

============================================

Another way:

After 1 half-life, 1/2 is left.

After 2 half-lifes, 1/4 is left.

After 3 half-lifes, 1/8 is left.

1/8 of 4,000g = (4,000g/8) = 500 g .

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3 years ago