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2 years ago

The frequency of a microwave is 6Ghz. Calculate the wavelength of the wave in the air.

Physics

1 answer:

lbvjy [14]2 years ago

3 0

Answer:

$\lambda=0.05\ m$

Explanation:

Given that,

The frequency of a microwave is 6Ghz.

We know that, $1\ GHz=10^9\ Hz$

We need to find the wavelength of the wave in the air.

The relation between wavelength, frequency and speed is given by :

$c=f\lambda\\\\\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{6\times 10^9}\\\\\lambda=0.05\ m$

So, the wavelength of the wave in the air is 0.05 m.

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Is it possible to produce a continuous and oriented aramid fiber–epoxy matrix composite having longitudinalandtransverse moduli

Masja [62]

Answer:

Not possible

Explanation:

$E_{cl}$ = longitudinal modulus of elasticity = 35 Gpa

$E_{ct}$ = transverse modulus of elasticity = 5.17 Gpa

$E_m$ = Epoxy modulus of elasticity = 3.4 Gpa

$V_{\rho l}$ = Volume fraction of fibre (longitudinal)

$V_{\rho t}$ = Volume fraction of fibre (transvers)

$E_f$ = Modulus of elasticity of aramid fibers = 131 Gpa

Longitudinal modulus of elasticity is given by

$E_{cl}=E_m(1-V_{\rho l})+E_fV_{\rho l}\\\Rightarrow 35=3.4(1-V_{\rho l})+131V_{\rho l}\\\Rightarrow 35=3.4-3.4V_{\rho l}+131V_{\rho l}\\\Rightarrow V_{\rho l}=\frac{35-3.4}{131-3.4}\\\Rightarrow V_{\rho l}=0.24764$

Transverse modulus of elasticity is given by

$E_{ct}=\frac{E_mE_f}{(1-V_{\rho t})E_f+V_{\rho t}E_m}\\\Rightarrow 5.17=\frac{3.4\times 131}{(1-V_{\rho t})131+V_{\rho t}3.4}\\\Rightarrow \frac{3.4\times 131}{5.17}-131=-127.6V_{\rho t}\\\Rightarrow V_{\rho t}=\frac{\frac{3.4\times 131}{5.17}-131}{-127.6}\\\Rightarrow V_{\rho t}=0.35148$

$V_{\rho l}\neq V_{\rho t}$

Hence, it is not possible to produce a continuous and oriented aramid fiber.

5 0

2 years ago

Physical activity is the most important thing you can do to improve and maintain health-related physical fitness. Which list con

katovenus [111]

Answer:

I guess it's d because age environment heredity Nd maturation r more obvious factors if we compre with only physical fitness

8 0

2 years ago

Select answer about motion and newtons law and explain why it's correct 23POINTS NEED ASAP

zlopas [31]

The train would need the greatest amount of force due to weight! If you think of it, a baseball won't need much force to stop it, but if you have a heavy train, it will need excessive force to stop the train. The answer would be #3

I hope this answer helps!

Sorry if it doesn't make sense, as I don't know that much about physics! I am just thinking of what makes sense.

3 0

2 years ago

Find the difference between two masses measured as 123.6 grams and 115.972 grams. Express the answer to the correct number of si

xxTIMURxx [149]

Answer:

The difference is 7.6 grams.

Explanation:

In mathematics the difference of two numbers is express as the subtraction between them:

$a-b$

So to find out the difference between the two measured masses, a will be represented by 123.6 grams since is the bigger number, and b by 115.972 grams.

Therefore, it is get:

$123.6grams-115.972grams = 7.6grams$

<u>Hence, the difference is 7.6 grams. </u>

The result of 7.628 will be expressed as 7.6 to have the correct number of significant figures.

Notice how that can be express in units of kilograms too since there is 1000 gram in 1 kilogram:

$7.6grams . \frac{1Kg}{1000grams}$ ⇒ $7.6x10^{-3}Kg$

8 0

2 years ago

Three packing crates of masses, M1 = 6 kg, M2 = 2 kg and M3 = 8 kg are connected by a light string of negligible mass that passe

never [62]

Answer:

39.81 N

Explanation:

I attached an image of the free body diagrams I drew of crate #1 and #2.

Using these diagram, we can set up a system of equations for the sum of forces in the x and y direction.

∑Fₓ = maₓ

∑Fᵧ = maᵧ

Let's start with the free body diagram for crate #2. Let's set the positive direction on top and the negative direction on the bottom. We can see that the forces acting on crate #2 are in the y-direction, so let's use Newton's 2nd Law to write this equation:

∑Fᵧ = maᵧ
T₁ - m₂g = m₂aᵧ

Note that the tension and acceleration are constant throughout the system since the string has a negligible mass. Therefore, we don't really need to write the subscripts under T and a, but I am doing so just so there is no confusion.

Let's solve for T in the equation...

T₁ = m₂aᵧ + m₂g
T₁ = m₂(a + g)

We'll come back to this equation later. Now let's go to the free body diagram for crate #1.

We want to solve for the forces in the x-direction now. Let's set the leftwards direction to be positive and the rightwards direction to be negative.

∑Fₓ = maₓ

F_f - F_g sinΘ = maₓ

The normal force is equal to the x-component of the force of gravity.

(F_n · μ_k) - m₁g sinΘ = m₁aₓ

(F_g cosΘ · μ_k) - m₁g sinΘ = m₁aₓ
[m₁g cos(30) · 0.28] - [m₁g sin(30)] = m₁aₓ
[(6)(9.8)cos(30) · 0.28] - [(6)(9.8)sin(30)] = (6)aₓ
[2.539595871] - [-58.0962595] = 6aₓ
60.63585537 = 6aₓ
aₓ = 10.1059759 m/s²

Now let's go back to this equation:

T₁ = m₂(a + g)

We have 3 known variables and we can solve for the tension force.

T = 2(10.1059759 + 9.8)
T = 2(19.9059759)
T = 39.8119518 N

The tension force is the same throughout the string, therefore, the tension in the string connecting M2 and M3 is 39.81 N.

3 0

2 years ago