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3 years ago

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The metal with the highest melting temperature is tungsten which melts at around 3400 K. What is the wavelength of the peak of t

he blackbody distribution for this temperature. Is this visible, infrared or ultraviolet?

1 answer:

alekssr [168]3 years ago

7 0

Explanation:

It is given that, the metal with the highest melting temperature is tungsten which melts at around 3400 K, T = 3400 K

We need to find the wavelength of the peak of the black body distribution for this temperature. It can be calculated using Wein's displacement law as :

$\lambda\times T=k$

k is the constant, $k=2.89\times 10^{-3}\ m.k$

$\lambda=\dfrac{k}{T}$

$\lambda=\dfrac{2.89\times 10^{-3}\ m.k}{3400\ k}$

$\lambda=8.5\times 10^{-7}\ m$

or

$\lambda=850\ nm$

The wavelength of infrared is from 700 nm to 1 mm. So, the lies in infrared region of the spectrum. Hence, this is the required solution.

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Answer:

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Explanation:

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2 years ago

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 V. It is used to charge two storage batt

Natali [406]

Complete Question

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 $\Omega$ . It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300 $\Omega$ . If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?

Answer:

a

The additional resistance is $R_z = 4.4 \Omega$

b

The rate at which internal energy increase at the supply is $Z_1 = 32 W$

c

The rate at which internal energy increase in the battery is $Z_1 = 32 W$

d

The rate at which internal energy increase in the added series resistance is $Z_3 = 70.4 W$

e

the increase rate of the chemically energy in the battery is $C = 48 W$

Explanation:

From the question we are told that

The open circuit voltage is $V = 40.0V$

The internal resistance is $R = 2 \Omega$

The emf of each battery is $e = 6.00 V$

The internal resistance of the battery is $r = 0.300V$

The charging current is $I = 4.00 \ A$

Let assume the the additional resistance to to added to the circuit is $R_z$

So this implies that

The total resistance in the circuit is

$R_T = R + 2r +R_z$

Substituting values

$R_T = 2.6 +R_z$

And the difference in potential in the circuit is

$E = V -2e$

=> $E = 40 - (2 * 6)$

$E = 28 V$

Now according to ohm's law

$I = \frac{E}{R_T}$

Substituting values

$4 = \frac{28}{R_z + 2.6}$

Making $R_z$ the subject of the formula

So $R_z = \frac{28 - 10.4}{4}$

$R_z = 4.4 \Omega$

The increase rate of internal energy at the supply is mathematically represented as

$Z_1 = I^2 R$

Substituting values

$Z_1 = 4^2 * 2$

$Z_1 = 32 W$

The increase rate of internal energy at the batteries is mathematically represented as

$Z_2 = I^2 r$

Substituting values

$Z_2 = 4^2 * 2 * 0.3$

$Z_2 = 9.6 \ W$

The increase rate of internal energy at the added series resistance is mathematically represented as

$Z_3 = I^2 R_z$

Substituting values

$Z_3 = 4^2 * 4.4$

$Z_3 = 70.4 W$

Generally the increase rate of the chemically energy in the battery is mathematically represented as

$C = 2 * e * I$

Substituting values

$C = 2 * 6 * 4$

$C = 48 W$

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3 years ago

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guapka [62]

Answer:

C

Explanation:

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then, as it boils it remains at constant temp ( boiling point)

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