All categories

3 years ago

Are dimensionless quantities always unitless

2 answers:

4 0

Important thing is that all unitless quantity is dimensionless quantity. .A dimensionless physical quantity may have an unit

arsen [322]3 years ago

4 0

Important thing is that all unitless quantity is dimensionless quantity. ... A dimensionless physical quantity may have an unit (e.g. Mechanical equivalent of heat) but a unitless physical quantity is always dimensionless (e.g. Coefficient of friction , refractive index

You might be interested in

What is refraction?

Damm [24]

B- light bends as it passes through an object ( a would be reflect)

4 0

3 years ago

If the current through a 20-ω resistor is 8.0 a , how much energy is dissipated by the resistor in 1.0 h ?

fiasKO [112]

1 hour = 3600 seconds.
Energy dissipated = I²Rt = 8²×20×3600 = 4608000 J

4 0

3 years ago

ome metal oxides can be decomposed to the metal and oxygen under reasonable conditions. 2 Ag2O(s) → 4 Ag(s) + O2(g) Thermodynami

yuradex [85]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

Explanation :

The given balanced chemical reaction is,

$2Ag_2O(s)\rightarrow 4Ag(s)+O_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{product}$

$\Delta H^o=[n_{Ag}\times \Delta H_f^0_{(Ag)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta H_f^0_{(Ag_2O)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[4mole\times (0kJ/mol)+1mole\times (0kJ/mol)}]-[2mole\times (-31.1kJ/mol)]$

$\Delta H^o=62.2kJ=62200J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{product}$

$\Delta S^o=[n_{Ag}\times \Delta S_f^0_{(Ag)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta S_f^0_{(Ag_2O)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[4mole\times (42.55J/K.mole)+1mole\times (205.07J/K.mole)}]-[2mole\times (121.3J/K.mole)]$

$\Delta S^o=132.67J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 298 K.

$\Delta G^o=(62200J)-(298K\times 132.67J/K)$

$\Delta G^o=22664.34J=22.66kJ$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

6 0

4 years ago

When you turn on a lightbulb in a room, the entire room appears to flood with light at the same time. Your eyes cannot perceive

Evgen [1.6K]

Explanation:

Assuming we can turn on the lightbulb from any distance with a device. We can gradually increase the distance that separates us from lightbulb, in this way, if the speed of light is finite we can see a temporary delay between the moment we turn on the lightbulb and the moment in which we observe its light.

7 0

3 years ago

A 30-g bullet is fired with a horizontal velocity of 450 m/s and becomes embedded in block B, which has a mass of 3 kg. After th

Zolol [24]

Answer:

(a) The velocity of the bullet and B after the first impact is 4.4554 m/s.

(b) The velocity of the carrier is 0.40872 m/s.

Explanation:

(a) To solve the question, we apply the principle of conservation of linear momentum as follows.

we note that the distance between B and C is 0.5 m

Then we have

Sum of initial momentum = Sum of final momentum

0.03 kg × 450 m/s = (0.03 kg + 3 kg) × v₂

Therefore v₂ = (13.5 kg·m/s)÷(3.03 kg) = 4.4554 m/s

The velocity of the bullet and B after the first impact = 4.4554 m/s

(b) The velocity of the carrier is given as follows

Therefore from the conservation of linear momentum we also have

(m₁ + m₂)×v₂ = (m₁ + m₂ + m₃)×v₃

Where:

m₃ = Mass of the carrier = 30 kg

Therefore

(3.03 kg)×(4.4554 m/s) = (3.03 kg+30 kg) × v₃

v₃ = (13.5 kg·m/s)÷ (33.03 kg) = 0.40872 m/s

The velocity of the carrier = 0.40872 m/s.

3 0

3 years ago

Are dimensionless quantities always unitless​

Are dimensionless quantities always unitless