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bearhunter [10]

4 years ago

9

If no heat is lost to the surroundings, how much heat must be added to raise the temperature from 20.0 ∘C to 85.0 ∘C ?

1 answer:

andrey2020 [161]4 years ago

3 0

Answer:

Explanation:

$C_{water}$ = 4190 J/kg.K

$C_{Al}$ = 910 J/Kg. K

$m_{Al}$ = 1.50 kg

$m_{water}$ = 1.80 kg

$Q_{added} = Q_{Al} + Q_{water}$

$=m_{Al} C_{Al}$ ΔT + $m_{water} C_{water}$ ΔT

= (1.50)(910)(85.0-20)+(1.80)(4190)(85.0-20)

= 578,955 J

= 579 kJ

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Show how the alternative definition of power, found in your book, can be derived by substituting the definitions of work and spe

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Let us consider body moves a distance S due to the force F.

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A stretched string has a mass per unit length of 5.40 g/cm and a tension of 17.5 N. A sinusoidal wave on this string has an ampl

kondaur [170]

Answer:

Part a)

$y_m = 0.157 mm$

part b)

$k = 101.8 rad/m$

Part c)

$\omega = 579.3 rad/s$

Part d)

here since wave is moving in negative direction so the sign of $\omega$ must be positive

Explanation:

As we know that the speed of wave in string is given by

$v = \sqrt{\frac{T}{m/L}}$

so we have

$T = 17.5 N$

$m/L = 5.4 g/cm = 0.54 kg/m$

now we have

$v = \sqrt{\frac{17.5}{0.54}}$

$v = 5.69 m/s$

now we have

Part a)

$y_m$ = amplitude of wave

$y_m = 0.157 mm$

part b)

$k = \frac{\omega}{v}$

here we know that

$\omega = 2\pi f$

$\omega = 2\pi(92.2) = 579.3 rad/s$

so we have

$k = \frac{579.3}{5.69}$

$k = 101.8 rad/m$

Part c)

$\omega = 579.3 rad/s$

Part d)

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4 0

3 years ago

An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0

Murrr4er [49]

Complete question:

An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0 V battery. Find the value of its capacitance.

Answer:

The value of its capacitance is 1.027 x 10⁻¹² F

Explanation:

Given;

area of the plate, A = 2.9 cm² = 2.9 x 10⁻⁴ m²

separation distance of the plates, d = 2.5 mm = 2.5 x 10⁻³ m

voltage of the battery, V = 18 V

The value of its capacitance is calculated as;

$C = \frac{k\epsilon_0A}{d} \\\\C = \frac{(1)(8.85\times 10^{-12})(2.9 \times 10^{-4})}{2.5 \times 10^{-3}} \\\\C = 1.027 \times 10^{-12} \ F$

Therefore, the value of its capacitance is 1.027 x 10⁻¹² F

8 0

3 years ago