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3 years ago

Does a brick have thermal energy? Also does paper haven't thermal energy?

Physics

1 answer:

Hunter-Best [27]3 years ago

7 0

Anything that's not at absolute zero temperature has thermal energy, and can warm something that's colder.

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An example of an exothermic reaction is:

elena55 [62]

Answer:

Explanation:

Exothermic refers to a RELEASE of energy through a reaction(usually chemical). When salt dissociates, polar water molecules pull apart ions, releasing energy.

6 0

4 years ago

Unpolarized light passes through a combination of two ideal polarizers. The transmission axes of the first polarizer and the sec

Yuliya22 [10]

Answer:

62.5 %

Explanation:

Let the initial intensity of unpolarized light is Io.

After first polariser the intensity of light becomes I'.

So, $I' = \frac{I_{0}}{2}$

Now it passes through another polariser. The angle between the first polariser and the second polariser is given by Ф. The intensity is I''.

According to the law of Malus

$I'' = I' Cos^{2}\phi$

Here, Ф = 30 degree

$I'' = \frac{I_{0}}{2} Cos^{2}30=0.375I_{0}$

The percentage change in the intensity is given by

$\frac{I_{0}-I''}{I_{0}}\times 100=\frac{I_{0}-0.375I_{0}}{I_{0}}\times100$

= 62.5 %

7 0

3 years ago

How do you find mass, volume and density? Any formulas?

Andrei [34K]

Density=mass/volume. if you know any 2 of these variables, you can figure out the missing one

5 0

3 years ago

Atmospheric pressure is reported in a variety of units depending on local meteorological preferences. In many European countries

Taya2010 [7]

Answer:

$h_w=10415.7664\ mm$ of water column.

Explanation:

Given:

density of mercury, $S_m=13.534$

height of the mercury column supported by the atmosphere, $h_m=769.6\ mm$

As we know that the equivalent pressure in terms of liquid column is given as:

$P=\rho.g.h$

so,

$S_m\times 1000\times g.h_m=\rho_w.g.h_w$

where:

$g=$ gravity

$h_w=$ height of water column

$\rho_w=$ density of water

$13534\times 9.8\times 769.6=1000\times 9.8\times h_w$

$h_w=10415.7664\ mm$ of water column.

5 0

3 years ago

This exercise uses the radioactive decay model. After 3 days a sample of radon-222 has decayed to 58% of its original amount. (a

Vadim26 [7]

Answer: a) 3.85 days

b) 10.54 days

Explanation:-

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = ?

t = time taken for decomposition = 3 days

a = let initial amount of the reactant = 100 g

a - x = amount left after decay process = $\frac{58}{100}\times 100=58g$

First we have to calculate the rate constant, we use the formula :

Now put all the given values in above equation, we get

$k=\frac{2.303}{3}\log\frac{100}{58}$

$k=0.18days^{-1}$

a) Half-life of radon-222:

$t_{\frac{1}{2}}=\frac{0.693}{k}$

$t_{\frac{1}{2}}=\frac{0.693}{0.18}=3.85days$

Thus half-life of radon-222 is 3.85 days.

b) Time taken for the sample to decay to 15% of its original amount:

where,

k = rate constant = $0.18days^{-1}$

t = time taken for decomposition = ?

a = let initial amount of the reactant = 100 g

a - x = amount left after decay process = $\frac{15}{100}\times 100=15g$

$t=\frac{2.303}{0.18}\log\frac{100}{15}$

$t=10.54days$

Thus it will take 10.54 days for the sample to decay to 15% of its original amount.

3 0

4 years ago