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4 years ago

Using at least 3 to 4 complete content related sentences describe the first and second postulate of special relativity.

2 answers:

8 0

Nothing can travel faster than the speed of light. As such, perceptions of objects and time change as they approach light speed, but the laws of physics remain consistent regardless of speed. Objects will appear shortened and time will appear to slow down around an observer approaching near light speeds, but all quantities still exist as they did before and all causality is preserved, even if observers in different points or traveling at different speeds will report different things.

andreev551 [17]4 years ago

6 0

Answer:

Inertial frames of reference are at rest or move at constant speed unlike non-inertial frames of reference which are accelerating frames of reference.

According to first postulate of the special theory of relativity, all the laws of physics hold true in all the inertial frames of reference. According to the second postulate, the speed of light (c) is constant everywhere and is independent of the relative motion of the source of light.

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Which of the following shows resistors in a parallel circuit

Alexeev081 [22]

The bottom drawing do

7 0

3 years ago

Read 2 more answers

A particular balloon can be stretched to a maximum surface area of 1257 cm2. The balloon is filled with 3.1 L of helium gas at a

chubhunter [2.5K]

Answer:

The ballon will brust at

Pmax = 518 Torr ≈ 0.687 Atm

Explanation:

Hello!

To solve this problem we are going to use the ideal gass law

PV = nRT

Where n (number of moles) and R are constants (in the present case)

Therefore, we can relate to thermodynamic states with their respective pressure, volume and temperature.

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$ --- (*)

Our initial state is:

P1 = 754 torr

V1 = 3.1 L

T1 = 294 K

If we consider the final state at which the ballon will explode, then:

P2 = Pmax

V2 = Vmax

T2 = 273 K

We also know that the maximum surface area is: 1257 cm^2

If we consider a spherical ballon, we can obtain the maximum radius:

$R_{max} = \sqrt{\frac{A_{max}}{4 \pi}}$

Rmax = 10.001 cm

Therefore, the max volume will be:

$V_{max} = \frac{4}{3} \pi R_{max}^3$

Vmax = 4 190.05 cm^3 = 4.19 L

Now, from (*)

$P_{max} = P_1 \frac{V_1T_2}{V_2T_1}$

Therefore:

Pmax= P1 * (0.687)

That is:

Pmax = 518 Torr

6 0

3 years ago

Some hypothetical alloy is composed of 12.5 wt% of metal A and 87.5 wt% of metal B. If the densities of metals A and B are 4.27

densk [106]

Answer:

The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

Explanation:

Given that,

Weight of metal A = 12.5%

Weight of metal B = 87.5%

Length of unit cell = 0.395 nm

Density of A = 4.27 g/cm³

Density of B= 6.35 g/cm³

Weight of A = 61.4 g/mol

Weight of B = 125.7 g/mol

We need to calculate the density of the alloy

Using formula of density

$\rho=n\times\dfrac{m}{V_{c}\times N_{A}}$

$n=\dfrac{\rho\timesV_{c}\times N}{m}$ ....(I)

Where, n = number of atoms per unit cells

m = Mass of the alloy

V=Volume of the unit cell

N = Avogadro number

We calculate the density of alloy

$\rho=\dfrac{1}{\dfrac{12.5}{4.27}+\dfrac{87.5}{6.35}}\times100$

$\rho=5.98$

We calculate the mass of the alloy

$m=\dfrac{1}{\dfrac{12.5}{61.4}+\dfrac{87.5}{125.7}}\times100$

$m=111.15$

Put the value into the equation (I)

$n=\dfrac{5.9855\times(0.395\times10^{-9}\times10^{2})^3\times6.023\times10^{23}}{111.15}$

$n=1.99\approx 2\ atoms/cell$

Hence, The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

5 0

4 years ago

Can an object be in motion and not in motion at the same time? Explain.

kolezko [41]

It could rotate while not advancing distance

5 0

3 years ago

An object is dropped from a bridge. A second object is thrown downwards 1.0 s later. They both reach the water 20 m below at the

Lerok [7]

Answer:

Explanation:
Kinematics equation for first Object:

but:
The initial velocity is zero

it reach the water at in instant, t1, y(t)=0:

Kinematics equation for the second Object:
The initial velocity is zero

but:

it reach the water at in instant, t2, y(t)=0. If the second object is thrown 1s later, t2=t1-1=1.02s

The velocity is negative, because the object is thrown downwards

7 0

3 years ago