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4 years ago

The orientation of which of the following does not influence the phases of the moon?

Physics

2 answers:

Semmy [17]4 years ago

7 0

Answer:

So the correct answer would be the stars do NOT have an influence on the Moon phases.

klasskru [66]4 years ago

6 0

<span>So we want to know what is not influencing the phases of the Moon. What influences Moon phases are celestial bodies that are close enough to have enough gravitational force to have an influence. Those celestial bodies would be the Earth and the Sun, since moon itself cant influence itself and all other stars are too far to exert gravitational force. So the correct answer would be the Moon and the stars do NOT have an influence on the Moon phases. </span>

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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

A dog is walking at 2m/s and then begins to run at a speed of 6m/s. What is his acceleration if his total travel time is 2 secon

fiasKO [112]

The formula for velocity vf = vi + at

First list your given information

2m/s Is your initial velocity (vi)
6m/s is you final velocity (vf)
2 seconds is your time (t)

Since you want the a for acceleration get a by itself

a = (vf-vi)/t

So a= (6-2)/2

a= 4/2

a=2

Now units

the units for acceleration are m/s $x^{2}$

2m/s $x^{2}$

7 0

3 years ago

A dentist’s drill starts from rest. After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min. Find the

mylen [45]

Answer:

Angular acceleration, is $708.07\ rad/s^2$

Explanation:

Given that,

Initial speed of the drill, $\omega_i=0$

After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, $\omega_f=28940\ rev/min=3030.58\ rad/s$

We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

$\alpha =\dfrac{\omega_f}{t}\\\\\alpha =\dfrac{3030.58\ rad/s}{4.28\ s}\\\\\alpha =708.07\ rad/s^2$

So, the drill's angular acceleration is $708.07\ rad/s^2$ .

4 0

3 years ago

Three moles of a monatomic ideal gas are heated at a constant volume of 1.20 m3. The amount of heat added is 5.22x10^3 J.(a) Wha

k0ka [10]

Answer:

A) 140 k

b ) 5.22 *10^3 J

c) 2910 Pa

Explanation:

Volume of Monatomic ideal gas = 1.20 m^3

heat added ( Q ) = 5.22*10^3 J

number of moles (n) = 3

A ) calculate the change in temp of the gas

since the volume of gas is constant no work is said to be done

heat capacity of an Ideal monoatomic gas ( Q ) = n.(3/2).RΔT

make ΔT subject of the equation

ΔT = Q / n.(3/2).R

= (5.22*10^3 ) / 3( 3/2 ) * (8.3144 J/mol.k )

= 140 K

B) Calculate the change in its internal energy

ΔU = Q this is because no work is done

therefore the change in internal energy = 5.22 * 10^3 J

C ) calculate the change in pressure

applying ideal gas equation

P = nRT/V

therefore ; Δ P = ( n*R*ΔT/V )

= ( 3 * 8.3144 * 140 ) / 1.20

= 2910 Pa

3 0

3 years ago

The human body has an average density of 979 kg/m3 , what fraction of a person is submerged when floating gently in fresh water?

FromTheMoon [43]

A person is submerged of about 97.9%.

The average density of the human body is given as 979 kg/ m³.

<h3>Define Law of floatation.</h3>

Law of floatation can be defined as the volume of the liquid displaced when a body floats on the liquid surface is equal to the body submerged in the water.

As body has the stable equilibrium state, the buoyancy of the fluid will be equal to the weight.

Weight of the body floating = Weight of the body immersed in fluid

Law of floatation = Density of the floating object / density of fluid

As fluid is the freshwater here, the density of fluid will be 1000 kg/ m³.

= (979 kg/ m³) / ( 1000 kg/ m³)

= 97.9 %

A person is submerged when floating gently in fresh water about 97.9%.

Learn more about law of floatation,

brainly.com/question/17032479

#SPJ4

4 0

1 year ago