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mel-nik [20]
2 years ago
15

What is the value of R2 in this parallel circuit? (5 stars)

Physics
1 answer:
o-na [289]2 years ago
4 0

Answer:

20 Ω

Explanation:

Voltage, current, and resistance are related by Ohm's law:

V = IR

40 V = (4 A) R

R = 10 Ω

The total resistance of the circuit is 10 Ω.

Resistors in parallel have a total resistance of:

1/R = 1/R₁ + 1/R₂

1 / (10 Ω) = 1 / (20 Ω) + 1/R₂

R₂ = 20 Ω

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In order get the block up to a speed of 3.58 m/s in 1.77 s, it must undergo an acceleration <em>a</em> of

<em>a</em> = (3.58 m/s) / (1.77 s) ≈ 2.02 m/s²

When the spring is getting pulled, Newton's second law tells us

• the net vertical force is

∑ <em>F</em> = <em>n</em> - <em>mg</em> = 0

where ∑ <em>F</em> is the net force, <em>n</em> is the magnitude of the normal force, and <em>mg</em> is the weight of the block - it follows that <em>n</em> = <em>mg</em> ; and

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∑ <em>F</em> = <em>F</em> - <em>f</em> = <em>ma</em>

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where <em>k</em> is the spring constant and <em>µ</em> is the coefficient of kinetic friction.

When the block is being pulled at a constant speed, Newton's second law says

• the net vertical force is still

∑ <em>F</em> = <em>n</em> - <em>mg</em> = 0

so that <em>n</em> = <em>mg</em> again; and

• the net horizontal force is

∑ <em>F</em> = <em>F</em> - <em>f</em> = 0

This time, <em>F</em> stretches the spring by <em>y</em> = 0.0544 m, so we have

<em>F</em> - <em>f</em> = <em>ky</em> - <em>µmg</em> = 0

Solve the equations in boldface for <em>k</em> and <em>µ</em> :

<em>kx</em> - <em>µmg</em> = <em>ma</em>

<em>ky</em> - <em>µmg</em> = 0

==>   <em>k</em> (<em>x</em> - <em>y</em>) = <em>ma</em>

==>   <em>k</em> = <em>ma</em> / (<em>x</em> - <em>y</em>)

==>   <em>k</em> = (17.6 kg) (2.02 m/s²) / (0.250 m - 0.0544 m) ≈ 182 N/m

Then

<em>ky</em> - <em>µmg</em> = 0

==>   <em>µ</em> = <em>ky </em>/ (<em>mg</em>)

==>   <em>µ</em> = (182 N/m) (0.0544 m) / ((17.6 kg) <em>g</em>) ≈ 0.0574

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