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4 years ago

What is the frequency of light with a wavelength of 725 nm?

Physics

1 answer:

Dennis_Churaev [7]4 years ago

3 0

Frequency of light = speed of light/wavelength = 3x10⁸/725x10⁻⁹ =4.13 x 10¹⁴ Hz

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Strontium 3890Sr has a half-life of 28.5 yr. It is chemically similar to calcium, enters the body through the food chain, and co

patriot [66]

Answer:

Thus the time taken is calculated as 387.69 years

Solution:

As per the question:

Half life of $^{3890}Sr\, t_{\frac{1}{2}}$ = 28.5 yrs

Now,

To calculate the time, t in which the 99.99% of the release in the reactor:

By using the formula:

$\frac{N}{N_{o}} = (\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}$

where

N = No. of nuclei left after time t

$N_{o}$ = No. of nuclei initially started with

$\frac{N}{N_{o}} = 1\times 10^{- 4}$

(Since, 100% - 99.99% = 0.01%)

Thus

$1\times 10^{- 4} = (\frac{1}{2})^{\frac{t}{28.5}}}$

Taking log on both the sides:

$- 4 = \frac{t}{28.5}log\frac{1}{2}$

$t = \frac{-4\times 28.5}{log\frac{1}{2}}$

t = 387.69 yrs

5 0

3 years ago

A 62 kg bungee jumper jumps from a bridge. She is tied to a bungee cord whose unstretched length is 12 m. She falls a total of 3

Andrew [12]

Answer:

k = 104.46 N/m

Explanation:

Here we can use energy conservation

so we will have

initial gravitational potential energy = final total spring potential energy

as we know that she falls a total distance of 31 m

while the unstretched length of the string is 12 m

so the extension in the string is given as

$x = L - L_o$

$x = 31 - 12 = 19 m$

so we have

$mgH = \frac{1}{2}kx^2$

$62(9.81)(31) = \frac{1}{2}k (19^2)$

$k = 104.46 N/m$

5 0

3 years ago

A typical electric refrigerator has a power rating of 500 Watts, which is the rate (J/s) at which electrical energy is supplied

Goshia [24]

Answer:

The rate of heat removed from inside the refrigerator is 300 watts.

Explanation:

By the First Law of Thermodynamics and the definition of a Refrigeration Cycle, we have the following formula to determine the rate of heat removed from inside the refrigerator ( $\dot Q_{L}$ ), in watts:

$\dot Q_{L} = \dot Q_{H}-\dot W$ (1)

Where:

$\dot Q_{H}$ - Rate of heat released to the room, in watts.

$\dot W$ - Rate of electric energy needed by the refrigerator, in watts.

If we know that $\dot Q_{H} = 800\,W$ and $\dot W = 500\,W$ , then the rate of heat removed from inside the refrigerator is:

$\dot Q_{L} = \dot Q_{H}-\dot W$

$\dot Q_{L} = 300\,W$

The rate of heat removed from inside the refrigerator is 300 watts.

3 0

3 years ago

For Part A, Sebastian decided to use the copper cylinder. How would the magnitude of his q and ∆H compare if he were to redo Par

Vitek1552 [10]

The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

The given parameters;

initial temperature of metals, = $T_m$
initial temperature of water, = $T_i$
specific heat capacity of copper, $C_p$ = 0.385 J/g.K
specific heat capacity of aluminum, $C_A$ = 0.9 J/g.K
both metals have equal mass = m

The quantity of heat transferred by each metal is calculated as follows;

Q = mcΔt

For copper metal, the quantity of heat transferred is calculated as;

$Q_p = (m_wc_w + m_pc_p)(T_m - T_i)\\\\Q_p = (T_m - T_i)(m_wc_w ) + (T_m - T_i)(m_pc_p)\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m_p(T_m - T_i)\\\\m_p = m\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m(T_m - T_i)\\\\let \ (T_m - T_i)(m_wc_w ) = Q_i, \ \ \ and \ let \ (T_m- T_i) = \Delta t\\\\Q_p = Q_i + 0.385m\Delta t$

The change in heat energy for copper metal;

$\Delta H = Q_p - Q_i\\\\\Delta H = (Q_i + 0.385m \Delta t) - Q_i\\\\\Delta H = 0.385 m \Delta t$

For aluminum metal, the quantity of heat transferred is calculated as;

$Q_A = (m_wc_w + m_Ac_A)(T_m - T_i)\\\\Q_A = (T_m -T_i)(m_wc_w) + (T_m -T_i) (m_Ac_A)\\\\let \ (T_m -T_i)(m_wc_w) = Q_i, \ and \ let (T_m - T_i) = \Delta t\\\\Q_A = Q_i \ + \ m_Ac_A\Delta t\\\\m_A = m\\\\Q_A = Q_i \ + \ 0.9m\Delta t$

The change in heat energy for aluminum metal ;

$\Delta H = Q_A - Q_i\\\\\Delta H = (Q_i + 0.9m\Delta t) - Q_i\\\\\Delta H = 0.9m\Delta t$

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

Learn more here:brainly.com/question/15345295

6 0

3 years ago

The index of refraction of a substance is 1.5. Find the sine of the angle of refraction if the sine of the angle of incidence is

Mariulka [41]

We can answer the problem by Snell's Law:

Snell's law (also known as Snell–Descartes law and the law of refraction) is a formula used to describe the relationship between the angles of incidence and refraction, when referring to light or other waves passing through a boundary between two different isotropic media, such as water, glass, or air.

7 0

3 years ago