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3 years ago

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Why do you sometimes get shocked by static electricity? Explain how it happens and give an example.

1 answer:

laiz [17]3 years ago

5 0

Answer:

That tiny shock you feel is a result of the quick movement of these electrons. You can think of a shock as a river of millions of electrons flying through the air. ... Static electricity happens more often during the colder seasons because the air is drier, and it's easier to build up electrons on the skin's surface.

Explanation:

Don't copy excatly but here it is.

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What is the acceleration of a 4,000 kg car pushed with a<br> force of 12,000 N?

jek_recluse [69]

Answer:

3 m/s

Explanation:

A= F/m

12,000/ 4000 = 3

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2 years ago

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How know that rocks formed from the seafloor sediments deposited in the Amadeus basin were softer than the arkose sandstone

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The rocks formed from the seafloor sediments deposited in the Amadeus basin were softer than the Arkose sandstone because the Amadeus basin were made up of marine and non-marine sedimentary rocks which are softer compared to quarts which make up mostly the Arkose sandstone.

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3 years ago

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A student is watching waves come in from the ocean. He noticed that the first wave he saw (Wave A) had twice the amplitude of th

Alika [10]

Answer:

Wave A

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2 years ago

A 23.5 g piece of aluminum metal is initially at 100.0°C. It is dropped into a coffee cup-calorimeter containing 130.0 g of wate

vivado [14]

Answer: The molar heat capacity of aluminum is $25.3J/mol^0C$

Explanation:

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of water = 130.0 g

$m_2$ = mass of aluminiunm = 23.5 g

$T_{final}$ = final temperature = $26.0^oC=(273+26)K=299K$

$T_1$ = temperature of water = $23^oC=(273+23)K=296K$

$T_2$ = temperature of aluminium = $100^oC=273+100=373K$

$c_1$ = specific heat of water= $4.184J/g^0C$

$c_2$ = specific heat of aluminium= ?

Now put all the given values in equation (1), we get

$130.0\times 4.184\times (299-296)=-[23.5\times c_2\times (299-373)]$

$c_2=0.938J/g^0C$

Molar mass of Aluminium = 27 g/mol

Thus molar heat capacity = $0.938J/g^0C\times 27g/mol=25.3J/mol^0C$

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3 years ago

A black, totally absorbing piece of cardboard of area a = 2.1 cm2 intercepts light with an intensity of 8.8 w/m2 from a camera s

PolarNik [594]

Answer:

The value of radiation pressure is $2.933 \times 10^{-8}$ Pa

Explanation:

Given:

Intensity $I = 8.8$ $\frac{W}{m^{2} }$

Area of piece $A = 2.1 \times 10^{-4}$ $m^{2}$

From the formula of radiation pressure in terms of intensity,

$P = \frac{I}{c}$

Where $P =$ radiation pressure, $c =$ speed of light

We know value of speed of light,

$c = 3 \times 10^{8}$ $\frac{m}{s}$

Put all values in above equation,

$P = \frac{8.8}{3 \times 10^{8} }$

$P = 2.933 \times 10^{-8}$ Pa

Therefore, the value of radiation pressure is $2.933 \times 10^{-8}$ Pa

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3 years ago