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3 years ago

7

Light is an electromagnetic wave and travels at a speed of 3.00 × 108 m/s. The human eye is most sensitive to yellow-green light

, which has a wavelength of 5.04 × 10-7 m. What is the frequency of this light?

1 answer:

Nimfa-mama [501]3 years ago

6 0

Answer:

$f = 5.95 \times 10^{14} Hz$

Explanation:

As we know that the frequency of the wave is given as

$f = \frac{c}{\lambda}$

here we know that

$c = 3 \times 10^8 m/s$

also we know that

$\lambda = 5.04 \times 10^{-7} m$

now we have

$f = \frac{3 \times 10^8}{5.04 \times 10^{-7}}$

$f = 5.95 \times 10^{14} Hz$

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BaLLatris [955]

Answer:

a) $E = 6.024\,USD$ , For m kilograms, it is 4184m J., 3600000 joules, b) $i = 88.200\,A$

Explanation:

a) The amount of heat needed to warm water is given by the following expression:

$Q_{needed} = m_{w}\cdot c_{w}\cdot (T_{f}-T_{i})$

Where:

$m_{w}$ - Mass of water, measured in kilograms.

$c_{w}$ - Specific heat of water, measured in $\frac{J}{kg\cdot ^{\circ}C}$ .

$T_{f}$ , $T_{i}$ - Initial and final temperatures, measured in $^{\circ}C$ .

Then,

$Q_{needed} = (1440\,kg)\cdot \left(4184\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (40^{\circ}C - 10^{\circ}C)$

$Q_{needed} = 180748800\,J$

The energy needed in kilowatt-hours is:

$Q_{needed} = 180748800\,J\times \left(\frac{1}{3600000}\,\frac{kWh}{J} \right)$

$Q_{needed} = 50.208\,kWh$

The electric energy required to heat up the water is:

$E = \frac{50.208\,kWh}{0.75}$

$E = 66.944\,kWh$

Lastly, the cost of heating a hot tub is: (USD - US dollars)

$E = (66.944\,kWh)\cdot \left(0.09\,\frac{USD}{kWh} \right)$

$E = 6.024\,USD$

The heat needed to raise the temperature a degree of a kilogram of water is 4184 J. For m kilograms, it is 4184m J. Besides, a kilowatt-hour is equal to 3600000 joules.

b) The current required for the electric heater is:

$i = \frac{Q_{needed}}{\eta \cdot \Delta V \cdot \Delta t}$

$i = \frac{180748800\,J}{0.75\cdot (220\,V)\cdot (3.45\,h)\cdot \left(3600\,\frac{s}{h} \right)}$

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Heat in the amount of 100 kJ is transferred directly from a hot reservoir at 1320 K to a cold reservoir at 600 K. Calculate the

Zinaida [17]

Answer:0.0909 kJ/K

Explanation:

Given

Temperature of hot Reservoir $T_h=1320 K$

Temperature of cold Reservoir $T_l=600 K$

Heat of 100 kJ is transferred form hot reservoir to cold reservoir

Hot Reservoir is Rejecting heat therefore $Q_1=-100 kJ$

Heat is added to Reservoir therefore $Q_2=100 kJ$

Entropy change for system

$\Delta s=\frac{Q_1}{T_1}+\frac{Q_2}{T_2}$

$\Delta s=\frac{-100}{1320}+\frac{100}{600}$

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Silver has a mass of 10.5 grams and a volume of 19.3 cm3. What is its density?

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Answer:

<h3>The answer is 0.54 g/cm³</h3>

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The density of a substance can be found by using the formula

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We have the final answer as

<h3>0.54 g/cm³</h3>

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So now we can use eq.(1) to find the gravitational acceleration of the planet:
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