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irina1246 [14]
3 years ago
8

A 5 kg box drops a distance of 10 m to the ground. If 70% of the initial potential energy goes into increasing the internal ener

gy of the box, determine the magnitude of the increase.
Physics
1 answer:
Elina [12.6K]3 years ago
7 0

Answer:

Explanation:

From the given information:

The initial PE (PE)_i = m×g×h

= 5 kg × 9.81 m/s² × 10 m

= 490.5 J

The change in Potential energy P.E of the box is:

ΔP.E = P.E_f -P.E_i

ΔP.E = 0 - (PE)_i

ΔP.E = -P.E_i

If we take a look at conservation of total energy for determining the change in the internal energy of the box;

\Delta P.E + \Delta K.E + \Delta U = 0

\Delta U = -\Delta P.E - \Delta K.E

this can be re-written as:

\Delta U =- (-\Delta P.E_i) - \Delta K.E

Here, K.E = 0

Also, 70% goes into raising the internal energy for the box;

Thus,

\Delta U =(70\%) \Delta P.E_i-0

\Delta U =(0.70) (490.5)

ΔU = 343.35  J

Thus, the magnitude of the increase is = 343.35 J

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Answer:

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Option 5.

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For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electro
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Answer:

V_d = 1.75 × 10⁻⁴ m/s

Explanation:

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Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

Radius of wire = 0.625 mm

Current, I = 3A

Area of the wire, A = \frac{\pi d^2}{4} = A = \frac{\pi 0.625^2}{4}

Now,

The current density, J is given as

J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}= 2444619.925 A/mm²

now, the electron density, n = \frac{\rho}{M}N_A

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n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3

Now,

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V_d=\frac{J}{ne}

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V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e} = 1.75 × 10⁻⁴ m/s

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