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tatyana61 [14]
3 years ago
11

A 5 kg block of copper (specific heat capacity = 385 J/kg°C) at 20°C is given

Physics
1 answer:
monitta3 years ago
6 0

Answer:

44^{\circ}C

Explanation:

When heat energy is supplied to a substance, its temperature increases according to the equation:

Q=mC\Delta T

where

Q is the energy supplied

m is the mass of the substance

C is the specific heat capacity

\Delta T is the increase in temperature

For the block of copper in this problem:

m = 5 kg

C = 385 J/kg°C

Q = 46,200 J

Therefore, the increase in temperature is

\Delta T=\frac{Q}{mC}=\frac{46,200}{(5)(385)}=24^{\circ}C

And since the initial temperature is 20°C, the final temperature will be

T_f=20^{\circ}C+\Delta T=44^{\circ}C

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A pole AB of length 10.0m and weight 600N has its center of gravity 4.0m from the end A, and lies on horizontal ground .Calculat
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Answer:

The force required to begin to lift the pole from the end 'A' is 240 N

Explanation:

The given parameters for the pole AB are;

The length of the pole, l = 10.0 m

The weight of the pole, W = 600 N ↓

The distance of the center of gravity of the pole from the side 'A' = 4.0 m

Let 'F_A' represent the force required to begin to lift the pole from the end 'A' and let a force applied in the upwards direction be positive

For equilibrium, the sum of moment about the point 'B' = 0, therefore, taking moment about 'B', we have

F_A × 10.0 m - W × 4.0 m = 0

∴ F_A × 10.0 m = W × 4.0 m = 600 N × 4.0 m

F_A × 10.0 m = 600 N × 4.0 m

∴  F_A = 600 N × 4.0 m/(10.0 m) = 240 N

The force required to begin to lift the pole from the end 'A', F_A = 240 N.

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What is the wavelength, in nm, of a photon with energy
lara31 [8.8K]

Answer:

(a)  λ = 4136 nm → infrared

(b) λ = 413.6 nm → visible light

(c) λ = 41.36 nm → ultraviolet

Explanation:

The wavelength of infrared is on the range of 700 nm to 1000000 nm

The wavelength of visible light is between 400 nm and 700 nm

The wavelength of ultraviolet ray on the range of 10 nm to 400 nm

The wavelength of photon is given by;

E = hf

f is the frequency of the wave = c / λ

E = h\frac{c}{\lambda}\\\\ \lambda = \frac{hc}{E}

Where;

c is the speed of light = 3 x 10⁸ m/s

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

(a) 0.3 eV = 0.3 x 1.602 x 10⁻¹⁹ J

\lambda = \frac{(6.626 * 10^{-34})(3*10^8)}{(0.3)*(1.602*10^{-19})}\\\\\lambda = 4.136 *10^{-6} \ m

λ = 4136 x 10⁻⁹ m

λ = 4136 nm → infrared

(b) 3.0 eV

\lambda = \frac{(6.626 * 10^{-34})(3*10^8)}{(3)*(1.602*10^{-19})}\\\\\lambda = 4.136 *10^{-7} \ m

λ = 413.6 x 10⁻⁹ m

λ = 413.6 nm →visible light

(c) 30 eV

\lambda = \frac{(6.626 * 10^{-34})(3*10^8)}{(30)*(1.602*10^{-19})}\\\\\lambda = 4.136 *10^{-8} \ m

λ = 41.36 x 10⁻⁹ m

λ = 41.36 nm →ultraviolet

5 0
3 years ago
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