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givi [52]
3 years ago
14

Which of these statements about half-life is true? Check all that apply. Half-lives vary dramatically from isotope to isotope. A

relationship exists between isotopes and half-lives. Half-lives are constant for a particular isotope. Half-life is an exact measure of when individual atoms will decay. Temperature, pressure, and density affect the half-life of an isotope.​
Chemistry
1 answer:
juin [17]3 years ago
7 0

Answer:A,C

Explanation:

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Ymorist [56]

Answer:

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Explanation:

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8 0
3 years ago
Only oppositely charged objects can attract each other. true false
Delicious77 [7]
No as neutral object will attract and be attracted by a positive and negative charge 

hope that helps 
4 0
3 years ago
A laser produces 17.0 mW of light. In 4.00 hr , the laser emits 6.04×1020 photons. What is the wavelength of the laser?
Vlad [161]
The unit 'mW' means milliwatts. It is a unit of work. There are 1,000 milliwatts in a 1 Watt of work. In 4 hours, there are 14,400 seconds.

Work= Energy/time
17 mW * 1 W/1000 mW = Energy/(14,400 seconds)
Solving for energy,
Energy = 244.8 J
Energy/photon = 244.8 J/(6.04×10²⁰) = 4.053×10⁻¹⁹ J/photon

Using the Planck's equation:

E = hc/λ
where h = 6.626×10⁻³⁴ m²·kg/s, c = 3,00,000,000 m/s and λ is the wavelength

4.053×10⁻¹⁹ J/photon = (6.626×10⁻³⁴ m²·kg/s)(3,00,000,000 m/s)/λ
λ = 4.9×10⁻⁷ m or 49 micrometers


5 0
3 years ago
Read 2 more answers
It has been suggested that the surface melting of ice plays a role in enabling speed skaters to achieve peak performance. Carry
vesna_86 [32]

Explanation:

Relation between pressure, latent heat of fusion, and change in volume is as follows.

          \frac{dP}{dT} = \frac{L}{T \times \Delta V}

Also, \frac{L}{T} = \Delta S^{fusion}_{m}

where, \Delta V^{fusion}_{m} is the difference in specific volumes.

Hence,    \frac{dP}{dT} = \frac{\Delta S^{fusion}_{m}}{\Delta V^{fusion}_{m}}

As, \Delta S^{fusion}_{m} = \frac{L}{T} = \frac{6010}{273.15} = 22.0 J/mol K

And,   \Delta V^{fusion}_{m} = \frac{M}{d_{H_{2}O}} - \frac{M}{d_{ice}} ...... (1)

where,    d_{H_{2}O} = density of water

              d_{ice} = density of ice

             M = molar mass of water = 18.02 \times 10^{-3} kg

Therefore, using formula in equation (1) we will calculate the volume of fusion as follows.

        \Delta V^{fusion}_{m} = \frac{M}{d_{H_{2}O}} - \frac{M}{d_{ice}}

                       = \frac{18.02 \times 10^{-3}}{997} - \frac{18.02 \times 10^{-3}}{920}  

                       = -1.51 \times 10^{-6}        

Therefore, calculate the required pressure as follows.

              \frac{dP}{dT} = \frac{22}{-1.51 \times 10^{-6}}

                              = 1.45 \times 10^{7} Pa/K

or,                           = 145 bar/K

Hence, for change of 1 degree pressure the decrease is 145 bar  and for 4.7 degree change dP = 145 \times 4.7 bar

                              = 681.5 bar

Thus, we can conclude that pressure should be increased by 681.5 bar to cause 4.7 degree change in melting point.

5 0
3 years ago
If the distance between two objects is decreased to - of the original
Charra [1.4K]

Answer:

The new force will be \frac{1}{100} of the original force.

Explanation:

In the context of this problem, we're dealing with the law of gravitational attraction. The law states that the gravitational force between two object is directly proportional to the product of their masses and inversely proportional to the square of a distance between them.

That said, let's say that our equation for the initial force is:

F = G\frac{m_1m_2}{R^2}The problem states  that  the distance decrease to 1/10 of the original distance, this means:[tex]R_2 = \frac{1}{10}R

And the force at this distance would be written in terms of the same equation:

F_2 = G\frac{m_1m_2}{R_2^2}

Find the ratio between the final and the initial force:

\frac{F_2}{F} = \frac{G\frac{m_1m_2}{R_2^2}}{G\frac{m_1m_2}{R^2}}

Substitute the value for the final distance in terms of the initial distance:

\frac{F_2}{F} = \frac{G\frac{m_1m_2}{(\frac{R}{10})^2}}{G\frac{m_1m_2}{R^2}}

Simplify:

\frac{F_2}{F} = \frac{\frac{1}{100R^2}}{\frac{1}{R^2}}=\frac{1}{100}

This means the new force will be \frac{1}{100} of the original force.

8 0
3 years ago
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