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Rus_ich [418]
3 years ago
10

The total resistance of a 15-ohm, an 65-ohm and a 35-ohm resistor connected in parallel ​

Physics
1 answer:
yanalaym [24]3 years ago
3 0

Answer:

I think its 9.0397 Ohms

Explanation:

take the reciprocal of all the resistances: 1/15, 1/65, 1/35

then add them: = 151/1365

then reciprocal the answer: =1365/151

And chuck it on a calculator: =9.04 Ohms

I think this is right but I'm not entirely sure. Tell me if I'm right by the way!

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The muscular system and the skeletal system of the human body work together to
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 A. Allow movement.  
Muscles connect to your skeleton and they contract and move the skeleton along. <span>They help the process of movement happen in a smoother manner.</span>
4 0
3 years ago
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A block of mass 57.1 kg rests on a slope having an angle of elevation of 28.3°. If pushing downhill on the block with a force ju
stira [4]

Answer:

The coefficient is 0.90

Explanation:

Drawing a diagram makes thing easier, we will assume that the acceleration tends to zero because it start barely moving.

-F_s+mg*sin(\theta)+F=0\\F_s=57.1kg*9.8m/s^2*sin(28.3)+177N\\F_s=442N\\F_s=\µ*N\\N=m*g*cos(\theta)\\N=57.1*9.8*cos(28.3)=493N\\\\\µ=\frac{442N}{493N}=0.90

3 0
3 years ago
In a long straight wire, what current is required to exert a 1.0μN force on a 1.0μC charge moving at 1.5×106m/s parallel to the
Bas_tet [7]

Answer:

Current in the wire is given as

i = 1.17 \times 10^{-7} A

Explanation:

magnetic field due to long current carrying wire is given as

B = \frac{\mu_0 i}{2\pi r}

so we have magnetic force on moving charge is given as

F = qvB

so we have

F = (1\times 10^{-6})(1.5 \times 10^6)(\frac{\mu_0 i}{2\pi (0.35)})

so we have

1\times 10^{-6} = 1.5 \times \frac{2 i}{0.35}

i = 1.17 \times 10^{-7} A

6 0
3 years ago
Iron filings scattered around a magnet will be most strongly drawn toward the _____.
victus00 [196]
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8 0
3 years ago
Read 2 more answers
Suppose an electron is trapped within a small region and the uncertainty in its position is 24. 0 x 10-15 m. what is the minimum
sergeinik [125]

The minimum uncertainty in the electron's momentum is

Δp = 2.2822 \times 10 {}^{ - 20} kg/ms

Given:

Uncertainty in position (ΔX)

= 24 \times 10 {}^{ - 15}m

planck's constant (h)

= 6.26 \times 10 {}^{ - 34} js

To find:

uncertainty in momentum (Δp)

Δx.Δph/4π

24 \times 10 {}^{ - 15} .Δp =  \frac{6.26 \times 10 {}^{ - 34} }{4 \times  \frac{22}{7} }

24 \times 10 {}^{ - 15}. Δp =  \frac{6.26 \times 10 {}^{ - 34} \times 7 }{8}

Δp =  \frac{43.82 \times 10 {}^{ - 34} }{8 \times 24 \times 10 {}^{ - 15} }

Δp =  \frac{43.82 \times 10 {}^{ - 34}  \times 10 {}^{ 15} }{192}

Δp =  \frac{4382 \times 10 {}^{ - 21} }{192}

Δp =22.822 \times 10 {}^{ - 21}

Δp = 2.2822  \times 10 {}^{ - 20} kg/ms

learn more about electron's momentum from here: brainly.com/question/28203580

#SPJ4

3 0
2 years ago
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