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Amiraneli [1.4K]
3 years ago
7

Calculate the total resistance in a parallel circuit made up of resistances of 2Ω, 3Ω, and 4Ω. (Hint: When converting to a decim

al from a fraction, keep the same number of significant figures in the decimal as was present in the original measurement from which the fraction was derived.)
Physics
1 answer:
natta225 [31]3 years ago
6 0

Answer:

Total resistance = 0.92Ω

Explanation:

For parallel connected resistors we have effective resistance

                 \frac{1}{R_{eff}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+.....

Here parallel circuit made up of resistances of 2Ω, 3Ω, and 4Ω.

That is

             R₁ = 2Ω

             R₂ = 3Ω

             R₃ = 4Ω

Substituting

             \frac{1}{R_{eff}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\\\\\frac{1}{R_{eff}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\\\\\frac{1}{R_{eff}}=\frac{6+4+3}{12}\\\\R_{eff}=\frac{12}{13}=0.92ohm

Total resistance = 0.92Ω

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How much force is needed to accelerate a 68 kilogram-skier at a rate of 1.2 m/sec^2?
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3 years ago
Read 2 more answers
A river flows due east at 1.60 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant v
Citrus2011 [14]

Answer:

part (a) v\ =\ 10.42\ at\ 81.17^o towards north east direction.

part (b) s = 46.60 m

Explanation:

Given,

  • velocity of the river due to east = v_r\ =\ 1.60\ m/s.
  • velocity of the boat due to the north = v_b\ =\ 10.3\ m/s.

part (a)

River is flowing due to east and the boat is moving in the north, therefore both the velocities are perpendicular to each other and,

Hence the resultant velocity i,e, the velocity of the boat relative to the shore is in the North east direction. velocities are the vector quantities, Hence the resultant velocity is the vector addition of these two velocities and the angle between both the velocities are 90^o

Let 'v' be the velocity of the boat relative to the shore.

\therefore v\ =\ \sqrt{v_r^2\ +\ v_b^2}\\\Rightarrow v\ =\ \sqrt{1.60^2\ +\ 10.3^2}\\\Rightarrow v\ =\ 10.42\ m/s.

Let \theta be the angle of the velocity of the boat relative to the shore with the horizontal axis.

Direction of the velocity of the boat relative to the shore.\therefore Tan\theta\ =\ \dfrac{v_b}{v_r}\\\Rightarrow Tan\theta\ =\ \dfrac{10.3}{1.60}\\\Rightarrow \theta\ =\ Tan^{-1}\left (\dfrac{10.3}{1.60}\ \right )\\\Rightarrow \theta\ =\ 81.17^o

part (b)

  • Width of the shore = w = 300m

total distance traveled in the north direction by the boat is equal to the product of the velocity of the boat in north direction and total time taken

Let 't' be the total time taken by the boat to cross the width of the river.\therefore w\ =\ v_bt\\\Rightarrow t\ =\ \dfrac{w}{v_b}\\\Rightarrow t\ =\ \dfrac{300}{10.3}\\\Rightarrow t\ =\ 29.12 s

Therefore the total distance traveled in the direction of downstream by the boat is equal to the product of the total time taken and the velocity of the river\therefore s\ =\ u_rt\\\Rightarrow s\ =\ 1.60\times 29.12\\\Rightarrow s\ =\ 46.60\ m

7 0
3 years ago
A 10 kg blue cart moving to the right at 25 m/s collides with a 17 kg red cart moving in the opposite direction at 16 m/s. If, a
scoray [572]

Answer:

24.8m/s

Explanation:

Given data

m1= 10kg

u1=25m/s

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u2=16m/s

v1=10m/s

v2=??

Applying the conservation of linear momentum

m1u1+m2u2=m1v1+m2v2

substitute

10*25+17*16=10*10+17*v2

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Kipish [7]

Answer:

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Explanation:

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to find out

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solution

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Δx × ΔPx  ≥ \frac{h}{4\pi }

here uncertainty momentum ΔPx = mΔVx

and uncertainty velocity = ΔVx

and h = 6.626 × 10^{-34} Js

so put here all these value in equation 1

10^{-3} × 0.0459 × ΔVx =  \frac{6.626*10^{-34}}{4\pi }

ΔVx = 1.15 × 10^{-30} m/s

and

so % of uncertainty in speed = ΔV / m

% of uncertainty in speed =  1.15 × 10^{-30}  / 55.5

% of uncertainty in speed =  2.07 × 10^{-30} %

3 0
3 years ago
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