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3 years ago

After a metal object turns to dust(rust) it would have ?

Physics

1 answer:

AleksandrR [38]3 years ago

3 0

<span>After a metal object turns to dust(rust) it would have no entropy. </span>

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Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated p

Marta_Voda [28]

Answer:

Acceleration, $a=9.91\times 10^{15}\ m/s^2$

Explanation:

It is given that,

Separation between the protons, $r=3.73\ nm=3.73\times 10^{-9}\ m$

Charge on protons, $q=1.6\times 10^{-19}\ C$

Mass of protons, $m=1.67\times 10^{-27}\ kg$

We need to find the acceleration of two isolated protons. It can be calculated by equating electric force between protons and force due to motion as :

$ma=\dfrac{kq^2}{r^2}$

$a=\dfrac{kq^2}{mr^2}$

$a=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{1.67\times 10^{-27}\times (3.73\times 10^{-9})^2}$

$a=9.91\times 10^{15}\ m/s^2$

So, the acceleration of two isolated protons is $9.91\times 10^{15}\ m/s^2$ . Hence, this is the required solution.

3 0

3 years ago

Light bulb 1 operates with a filament temperature of 2700 K whereas light bulb 2 has a filament temperature of 2100 K. Both fila

Lemur [1.5K]

Answer:

0.3659

Explanation:

The power (p) is given as:

P = AeσT⁴

where,

A =Area

e = transmittivity

σ = Stefan-boltzmann constant

T = Temperature

since both the bulbs radiate same power

P₁ = P₂

Where, 1 denotes the bulb 1

2 denotes the bulb 2

thus,

A₁e₁σT₁⁴ = A₂e₂σT₂⁴

Now e₁=e₂

⇒A₁T₁⁴ = A₂T₂⁴

$\frac{A_1}{A_2} =\frac{T_{2}^{4}}{T_{1}^{4}}$

substituting the values in the above question we get

$\frac{A_1}{A_2} =\frac{2100_{2}^{4}}{2700_{1}^{4}}$

$\frac{A_1}{A_2} }$ =0.3659

6 0

3 years ago

What type of pollution did the Clean Water Act succeed in limiting?

Ilya [14]

Sewage. If thats not it, then I need to see your choices. :)

4 0

3 years ago

Read 2 more answers

A student drops two metallic objects into a 120-g steel container holding 150 g of water at 25°C. One object is a 206-g cube of

spayn [35]

Answer:

Mass of the aluminium chunk = 278.51 g

Explanation:

For an isolated system as given the energy lost and gains in the system will be zero therefore sum of all transfer of energy will be zero,as the temperature will also remain same

A specific heat formula is given as

Energy Change = Mass of liquid x Specific Heat Capacity x Change in temperature

Q = m×c×ΔT

Heat gain by aluminium + heat lost by copper = 0 (1)

For Aluminium:

Q = $m\times0.897\frac{J}{g.k}\times(25-5)$