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2 years ago

Which image exclusively represents the principle of radiation? A) IV B) III C) II D) I

Chemistry

2 answers:

arsen [322]2 years ago

4 0

The principle of radiation protection is to trigger deterministic and stochastic effect.

Explanation:

The main aim of principle of radiation is to prevent the deterministic effects of radiation and reduce the risks of stochastic effects.

There are three general principals of radiation used for dealing with ionising radiation are Justification, Dose limitation and Optimization.

The three basic radiation principles are time, distance and shielding.

The risk of exposure to radiation is measured using the conventional unit rem or SI unit (sievert).

zaharov [31]2 years ago

4 0

Answer:

III

Explanation:

The sun gives off radiation. it will take about 8 minutes for radiation from the Sun to get to Earth. Actually, the Sun does not only produce IR, visible light, and UV. Fusion in the core actually gives off high energy gamma rays.

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Min is conducting an experiment where he compares the properties of water and lemonade. The first stage of the experiment is foc

vampirchik [111]

Answer: i need help

Explanation:

6 0

2 years ago

1. The specific heat capacity of iron is 0.461 J g–1 K–1 and that of titanium is 0.544 J g–1 K–1. A sample consisting of a mixtu

mezya [45]

Answer:

The answer is 80,1 °C

Explanation:

Let´s start from the mass of the sample and the heat capacities:

First of all, we must calculate an average heat capacity. That's because we have a mixture and it is unknown the heat capacity of the whole sample.

The way we should do this calculation is as follows:

(1) $H_{average}=Mass Fraction_{first component}* H_{first component}+MassFraction_{secondcomponent}*H_{second component}$

For example, the mass fraction of Fe is simply:

(2) $MassFraction_{Fe}=\frac{10g Fe}{10g Fe + 10gTi}=0.5$

If you combine the equations (1) and (2) you have:

(3) $H_{average}=0.5*0.461+0.5*0.544=0.5025\frac{J}{g-K}$

Once calculated the average heat capacity we can solve the problem taking into account the corresponding equation:

(4) $Q=m*H_{average}*(T_2-T_1)$

Remember that:

Q: Heat gained or lost

m: Mass of the sample you want to analize

$H_{average}$ : The value obtained in equation (3)

$T_2$ : Final temperature of the sample

$T_1$ : Initial temperature of the sample

Now we must replace the problem data in equation (4)

Take into account:

Heat gained in a system have a positive value
Heat lost in a system have a negative value

In this problem the sample loses 200 J, for this reason $Q=-200J$

The mass of the whole sample is: 10g of Fe + 10g of Ti = 20g of sample
The temperatures must be in absolute units of temperature (these are: rankine or kelvin)
The initial temperature of the system is 100°C or 373K

Now we are ready to use equation (4):

(5) $-200J=20g*0.5025\frac{J}{g*K} *(T_2-373K)$

It is clear that the unknown in equation (5) is $T_2$

The next step is to calculate $T_2$ . Don't forget the signs; these are important.

Key concept: Since the system is loosing heat, the final temperature of the system ( $T_2$ ) should be lower than the initial temperature ( $T_1$ )