To solve this problem it is necessary to apply the concepts related to the Kinetic Energy and the Energy Produced by the heat loss. In mathematical terms kinetic energy can be described as:

$KE = \frac{1}{2} mv^2$

Where,

m = Mass

v = Velocity

Replacing we have that the Total Kinetic Energy is

$KE = \frac{1}{2} mv^2$

$KE = \frac{1}{2} (5*10^{-3})(300)^2$

$KE = 225J$

On the other hand the required Energy to heat up t melting point is

$Q_1 = mC_p \Delta T$

$Q_2 = L_f m$

Where,

m = Mass

$C_p =$ Specific Heat

$\Delta T =$ Change at temperature

$L_f =$ Latent heat of fussion

Heat required to heat up to melting point,

$Q = Q_1+Q_2$

$Q = mC_p \Delta T+L_f m$

$Q = 5*0.128*(327-20) + 5*24.7$

$Q = 310J$

The energy required to melt is larger than the kinetic energy. Therefore the heat of fusion of lead would be 327 ° C: The melting point of lead.

4 0

3 years ago

A wave x meters long has a speed of y meters per second. The frequency of the wave is

Sergio [31]

The correct answer is (b.) y/x hertz. That is because the formula to get the frequency is f = v / w. The following values (v=y meters / second; wavelength = x meters) must be substituted to the equation, which leaves you y/x hertz.

6 0

3 years ago

A "570-W" electric heater is designed to operate from 120-V lines. What current does it draw? If the line voltage drops to 110 V

Viefleur [7K]

Answer:

Watts=Volt*Amps

So A=570/120=4.75amps

If voltage drops to 110V We get A=570/110=5.(18...)amps

4 0

3 years ago

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