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3 years ago

With which field of science is Albert Einstein associated? biology chemistry medecine physics

Physics

2 answers:

vlabodo [156]3 years ago

7 0

Answer

Albert Einstein associated with physics field of science .

Albert Einstein is a physicist, his best known theory are

(1 )Theory of relativity and the equation i.e E = MC²

(2) Photo electric effect

For his discovery of the law of the photoelectric effect Albert einstein was awarded by Nobel Prize in Physics for his work in theoretical physics,

sveta [45]3 years ago

3 0

He was a theoretical physicist

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Gaseous helium is in thermal equilibrium with liquid helium at 6.4 K. The mass of a helium atom is 6.65 × 10−27 kg and Boltzmann

chubhunter [2.5K]

Answer:

162.78 m/s is the most probable speed of a helium atom.

Explanation:

The most probable speed:

$v_{mp}=\sqrt{\frac{2K_bT}{m}}$

$K_b$ = Boltzmann’s constant = $1.38066\times 10^{-23} J/K$

T = temperature of the gas

m = mass of the gas particle.

Given, m = $6.65\times 10^{-27} kg$

T = 6.4 K

Substituting all the given values :

$v_{mp}=\sqrt{\frac{2\times 1.38066\times 10^{-23} J/K\times 6.4 K}{6.65\times 10^{-27} kg}}$

$v_{mp}=162.78 m/s$

162.78 m/s is the most probable speed of a helium atom.

4 0

3 years ago

A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0484 kg·m2 about a line through its center of mass, rolls witho

d1i1m1o1n [39]

Answer:

Part a)

$KE_r = 8 J$

Part b)

v = 3.64 m/s

Part c)

$KE_f = 12.7 J$

Part d)

$v = 2.9 m/s$

Explanation:

As we know that moment of inertia of hollow sphere is given as

$I = \frac{2}{3}mR^2$

here we know that

$I = 0.0484 kg m^2$

R = 0.200 m

now we have

$0.0484 = \frac{2}{3}m(0.200)^2$

$m = 1.815 kg$

now we know that total Kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2$

$20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2$

$20 = 1.5125 v^2$

$v = 3.64 m/s$

Part a)

Now initial rotational kinetic energy is given as

$KE_r = \frac{1}{2}I(\frac{v}{R})^2$

$KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2$

$KE_r = 8 J$

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

$mgh = (KE_i) - KE_f$

$1.815(9.8)(0.900sin27.1) = 20- KE_f$

$7.30 = 20 - KE_f$

$KE_f = 12.7 J$

Part d)

Now we know that

$\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7$

$\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7$

$1.5125 v^2 = 12.7$

$v = 2.9 m/s$

8 0

3 years ago

the length of iron rod at 100 C is 300.36 cm and at 159 C is 300.54 cm.Calculate its length at 0 c and coefficient of linear exp

Ugo [173]

Answer:

The length at 0 °C is 300.05 cm

Coefficient of linear expansion of iron is 1.02×10¯⁵ C¯¹

Explanation:

From the question given above, the following data were obtained:

Length (L₁) at 100 °C = 300.36 cm

Temperature 1 (θ₁) = 100 °C

Length (L₂) at 159 °C = 300.54 cm

Temperature 2 (θ₂) = 159 °C

Length (L₀) at 0 °C =?

Coefficient of linear expansion (α) =?

L₁ = L₀ (1 + θ₁α)

300.36 = L₀ (1 + 100α) ....(1)

L₂ = L₀ (1 + θ₂α)

300.54 = L₀ (1 + 159α) ..... (2)

Divide equation (2) by (1)

300.54 / 300.36 = L₀ (1 + 159α) / L₀ (1 + 100α)

1.0006 = (1 + 159α) / (1 + 100α)

Cross multiply

1.0006 (1 + 100α) = (1 + 159α)

1.0006 + 100.06α = 1 + 159α

Collect like terms

1.0006 – 1 = 159α – 100.06α

0.0006 = 58.94α

Divide both side by 58.94

α = 0.0006 / 58.94

α = 1.02×10¯⁵ C¯¹

Substitute the value of α into anything of the equation to obtain L₀. Here we shall use equation (2).

300.54 = L₀ (1 + 159α)

α = 1.02×10¯⁵ C¯¹

300.54 = L₀ (1 + 159 ×1.02×10¯⁵)

300.54 = L₀ (1 + 0.0016218)

300.54 = L₀ (1.0016218)

Divide both side by 1.0016

L₀ = 300.54 / 1.0016

L₀ = 300.05 cm

Summary:

The length at 0 °C is 300.05 cm

Coefficient of linear expansion of iron is 1.02×10¯⁵ C¯¹

6 0

3 years ago

A steam engine absorbs 4 x 105 J and expels 3.5 x 105 J in each cycle. What is its efficiency?

Wewaii [24]

Input heat, Qin = 4 x 10⁵ J
Output heat, Qout = 3.5 x 10⁵ J

From the first Law of thermodynamics, obtain useful work performed as
W = Qin - Qout
= 0.5 x 10⁵ J

By definition, the efficiency is
η = W/Qin
= 100*(0.5 x 10⁵/4 x 10⁵)
= 12.5%

Answer: The efficiency is 12.5%

3 0

3 years ago

Two objects must be in contact for them to exert a force on each other. <br> True or False

nevsk [136]

Answer:

False

Explanation:

This proposition is false because by example the sun exerts a force over the earth and them are not in contact

7 0

3 years ago