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kicyunya [14]
2 years ago
12

A certain quantity of a liquid has a volume of 10cm at 20°C. Calculate its volume at 50°C, if it’s cubic expansivity is 10^(-3)

K^(-1)
Physics
1 answer:
Pavel [41]2 years ago
6 0

Answer:

10.3 cm³

Explanation:

From the question given above, the following data were obtained:

Original volume (V₁) = 10 cm³

Initial temperature (θ₁) = 20 °C

Final temperature (θ₂) = 50 °C

Cubic expansivity (γ) = 10¯³ K¯¹

Final volume (V₂) =?

γ = V₂ – V₁ / V₁(θ₂ – θ₁)

10¯³ = V₂ – 10 / 10( 50 – 20)

10¯³ = V₂ – 10 / 10(30)

10¯³ = V₂ – 10 / 300

Cross multiply

10¯³ × 300 = V₂ – 10

0.3 = V₂ – 10

Collect like terms

0.3 + 10 = V₂

10.3 = V₂

V₂ = 10.3 cm³

Thus, the volume at 50 °C is 10.3 cm³

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after a circuit has been turned off, so it is important to make sure they are discharged before you touch them. Suppose a 120 mF
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The time is 110.16\times10^{-3}\ sec

Explanation:

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Using formula of discharge current

i_{0}=\dfrac{V_{0}}{R}

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We need to calculate the time

Using formula of current

i=\dfrac{V_{0}}{R}e^{\frac{-t}{RC}}

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4 0
3 years ago
Arrow_forward
garri49 [273]

Explanation:

(a) Hooke's law:

F = kx

7.50 N = k (0.0300 m)

k = 250 N/m

(b) Angular frequency:

ω = √(k/m)

ω = √((250 N/m) / (0.500 kg))

ω = 22.4 rad/s

Frequency:

f = ω / (2π)

f = 3.56 cycles/s

Period:

T = 1/f

T = 0.281 s

(c) EE = ½ kx²

EE = ½ (250 N/m) (0.0500 m)²

EE = 0.313 J

(d) A = 0.0500 m

(e) vmax = Aω

vmax = (0.0500 m) (22.4 rad/s)

vmax = 1.12 m/s

amax = Aω²

amax = (0.0500 m) (22.4 rad/s)²

amax = 25.0 m/s²

(f) x = A cos(ωt)

x = (0.0500 m) cos(22.4 rad/s × 0.500 s)

x = 0.00919 m

(g) v = dx/dt = -Aω sin(ωt)

v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)

v = -1.10 m/s

a = dv/dt = -Aω² cos(ωt)

a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)

a = -4.59 m/s²

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