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3 years ago

calculate the resistance of a wire 150cm long and diameter 2.0mm constructed from an alloy of resistivity 44*10-⁸Ωm

Physics

1 answer:

Ghella [55]3 years ago

8 0

Answer:

R = 0.21 Ω

Explanation:

the formula:

R = r x l/A

R = (44 x 10-⁸ Ωm) x 1.5 / (π x (1 x 10-³ m)²)

R = 6.6 x 10-⁷ / 3.14 x 10-⁶

R = 0.21 Ω

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4 years ago

A force acts on a 9.90 kg mobile object that moves from an initial position of to a final position of in 5.40 s. Find (a) the wo

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Time =5.40 s

Suppose the force is (2.00i + 9.00j + 5.30k) N, initial position is (2.70i - 2.90j + 5.50k) m and final position is (-4.10i + 3.30j + 5.40k) m.

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Using formula of displacement

$s=r_{2}-r_{1}$

Where, $r_{1}$ = initial position

$r_{2}$ = final position

Put the value into the formula

$s= (-4.10i + 3.30j + 5.40k)-(2.70i - 2.90j + 5.50k)$

$s= -6.80i+6.20j-0.1k$

(a). We need to calculate the work done on the object

Using formula of work done

$W=F\cdot s$

Put the value into the formula

$W=(2.00i + 9.00j + 5.30k)\cdot (-6.80i+6.20j-0.1k)$

$W=-13.6+55.8-0.53$

$W=41.67\ J$

(b). We need to calculate the average power due to the force during that interval

Using formula of power

$P=\dfrac{W}{t}$

Where, P = power

W = work

t = time

Put the value into the formula

$P=\dfrac{41.67}{5.40}$

$P=7.71\ Watt$

(c). We need to calculate the angle between vectors

Using formula of angle

$\theta=\cos^{-1}(\dfrac{r_{1}r_{2}}{|r_{1}||r_{2}|})$

Put the value into the formula

$\theta=\cos^{-1}\dfrac{(-4.10i + 3.30j + 5.40k)\cdot(2.70i - 2.90j + 5.50k)}{7.54\times6.778})$

$\theta=79.7^{\circ}$

Hence, (a). The work done on the object by the force in the 5.40 s interval is 41.67 J.

(b). The average power due to the force during that interval is 7.71 Watt.

(c). The angle between vectors is 79.7°

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4 years ago

calculate the resistance of a wire 150cm long and diameter 2.0mm constructed from an alloy of resistivity 44*10-⁸Ωm​

calculate the resistance of a wire 150cm long and diameter 2.0mm constructed from an alloy of resistivity 44*10-⁸Ωm