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3 years ago

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A 2.1 cm length of tungsten filament in a small lightbulb has a resistance of 0.067 ω. find its diameter. the resistivity of the

filament is 5.6 × 10−8 ω · m. answer in units of mm.

1 answer:

kkurt [141]3 years ago

3 0

The resistance of an ohmic material is given by
$R= \frac{ \rho l}{a}$
where l is the length and a is the cross sectional area. For a circular wire, this area is that of a circle:
$R= \frac{\rho l}{ {d^2} } \frac{4}{\pi}=\frac{5.6\times10^-8 \times0.021m\times4}{\pi d^2} = \frac{1.497\times 10^-9}{d^2}$
Then solving for diameter, d:
$d^2=\frac{1.497\times 10^-9}{0.067}=2.235\times10^-8$
$d=0.1495mm$

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Two simple pendulum of slightly different length , are set off oscillating in step is a time of 20s has elasped , during which t

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Answer:

Length of longer pendulum = 99.3 cm

Length of shorter pendulum = 82.2 cm

Explanation:

Since the longer pendulum undergoes 10 oscillations in 20 s, its period T = 20 s/10 = 2 s.

From T = 2π√(l/g), the length of the pendulum. l = T²g/4π²

substituting T = 2s and g = 9.8 m/s² we have

l = T²g/4π²

= (2 s)² × 9.8 m/s² ÷ 4π²

= 39.2 m ÷ 4π²

= 0.993 m

= 99.3 cm

Now, for the shorter pendulum to be in step with the longer pendulum, it must have completed some oscillations more than the longer pendulum. Let x be the number of oscillations more in t = 20 s. Let n₁ = number of oscillations of longer pendulum and n₂ = number of oscillations of longer pendulum.

So, n₂ = n₁ + x. Also n₁ = t/T₁ and n₂ = t/T₂ where T₂ = period of shorter pendulum.

t/T₂ = t/T₁ + x

1/T₂ = 1/T₁ + x (1)

Also, the T₂ = t/n₂ = t/(n₁ + x) (2)

From (1) T₂ = T₁/(T₁ + x) (3)

equating (2) and (3) we have

t/(n₁ + x) = T₁/(T₁ + x)

substituting t = 20 s and n₁ = 10 and T₁ = 2s, we have

20 s/(10 + x) = 2/(2 + x)

10/(10 + x) = 1/(2 + x)

(10 + x)/10 = (2 + x)

(10 + x) = 10(2 + x)

10 + x = 20 + 10x

collecting like terms

10x - x = 20 - 10

9x = 10

x = 10/9

x = 1.11

x ≅ 1 oscillation

substituting x into (2)

T₂ = t/n₂ = t/(n₁ + x)

= 20/(10 + 1)

= 20/11

= 1.82 s

Since length l = T²g/4π²

substituting T = 1.82 s and g = 9.8 m/s² we have

l = T²g/4π²

= (1.82 s)² × 9.8 m/s² ÷ 4π²

= 32.46 m ÷ 4π²

= 0.822 m

= 82.2 cm

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Answer:

Part a)

$T = 42 N$

Part b)

$v_f = 11.8 m/s$

Part c)

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Part a)

While bucket is falling downwards we have force equation of the bucket given as

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so we have

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now we have

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$a = \frac{mg}{(\frac{M}{2} + m)}$

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Part b)

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Part c)

now in order to find the time of fall we can use another equation

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$11.8 - 0 = 7 t$

$t = 1.7 s$

Part d)

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so here we can use force balance

$F = T + Mg$

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