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2 years ago

What is the total current in a parallel circuit containing a 12-V battery, a 2 Ω resistor, and a 4 Ω resistor?

Physics

1 answer:

deff fn [24]2 years ago

7 0

Answer:

V=IR

since the circuit is parallel both resistor have same voltage and different current value according to OHMs law.

total resistance in parallel=

1/R=1/R1+1/R2+...+1/Rn

since we have two resistor in parallel

Rt=R1R2/R1+R2

2*4/2+4=4/3 ohms

I=V/R

12/4/3=36/4=9Amp

I=12/2=6amp

I=12/4=3amp

total current

I=6+3

9amp

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A hiker determines the length of a lake by listening for the echo of her shout reflected by a cliff at the far end of the lake.

Nataliya [291]

To solve the exercise it is necessary to take into account the definition of speed as a function of distance and time, and the speed of air in the sound, as well

$v=\frac{d}{t}$

Where,

V= Velocity

d= distance

t = time

Re-arrange the equation to find the distance we have,

d=vt

Replacing with our values

$d= (343)(3.7)$

$d= 1269.1m$

It is understood that the sound comes and goes across the entire lake therefore, the length of the lake is half the distance found, that is

$L_{lake} = \frac{d}{2}$

$L_{lake} = \frac{1269.1}{2}$

$L_{lake} = 634.55m$

Therefore the length of the lake is 634,55m

8 0

3 years ago

A 1.0 kg piece of copper with a specific heat of cCu=390J/(kg⋅K) is placed in 1.0 kg of water with a specific heat of cw=4190J/(

Valentin [98]

Answer:

The temperature change of the copper is greater than the temperature change of the water.

Explanation:

deltaQ = mc(deltaT)

Where,

delta T = change in the temperature

m =mass

c = heat capacity

$\frac{(deltaT)_{Cu}}{(deltaT)_{w}}=\frac{4190J/kg.K}{390J/kg.K}\\ \\(deltaT)_{Cu}=10.74(deltaT)_{w}$

The temperature change in the copper is nearly 11 times the temperature change in the water.

So, the correct option is,

The temperature change of the copper is greater than the temperature change of the water.

Hope this helps!

7 0

3 years ago

You have a ball in your hand above the ground at rest which you then drop. What energy is involved in each part of this situatio

Juliette [100K]

When you hold the ball, you have potential energy, when you drop the ball, you have kinetic.

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3 years ago

A sled with a mass of 20 kg slides along frictionless ice at 4.5 m/s. It then crosses a rough patch of snow that exerts a fricti

uysha [10]

Answer:The sled slides 16.875m before rest.

Explanation:

$a=\frac{F}{m} =\frac{12N}{20kg}$

a=0.6 m/s²

$Vf=0=Vi-a.t$

$t=\frac{Vi}{a} =t=\frac{4.5m/s}{0.6 m/s2} =t=7.5seg$

$d= Vi.t - \frac{a.t^{2}}{2}$

$d= 4.5 * 7.5 - \frac{0.6*7.5^{2} }{2} \\\\d=16.875m$

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