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1 year ago

What are the 7 types of waves?.

1 answer:

5 0

The seven types of waves are radio waves, microwaves, infrared, visible, ultraviolet rays, x-rays and gamma rays.

The electromagnetic spectrum is a broad spectrum consisting of various waves. It contains very long radio waves to very short gamma rays. However, the human eye can detect only a small portion of this spectrum called as visible light.

The other waves that are invisible to the eye have their applications in various fields.

For example, radio waves have the longest waves on the spectrum and are said to have lowest energy. An MRI machine uses these waves to see tissues and bones up close.

To know more about waves:

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You might be interested in

Calculate the average drift speed of electrons traveling through a copper wire with a crosssectional area of 80 mm2 when carryin

Vedmedyk [2.9K]

Answer:

The correct answer is $2.8*10^{-5}ms^{-1}$

Explanation:

The formula for the electron drift speed is given as follows,

$u=I/nAq$

where n is the number of of electrons per unit m³, q is the charge on an electron and A is the cross-sectional area of the copper wire and I is the current. We see that we already have A , q and I. The only thing left to calculate is the electron density n that is the number of electrons per unit volume.

Using the information provided in the question we can see that the number of moles of copper atoms in a cm³ of volume of the conductor is $8.93/63.5 molcm^{-3}$ . Converting this number to m³ using very elementary unit conversion we get $140384molm^{-3}$ . If we multiply this number by the Avagardo number which is the number of atoms per mol of any gas , we get the number of atoms per m³ which in this case is equal to the number of electron per m³ because one electron per atom of copper contribute to the current. So we get,

$n=140384*6.02*10^{23} = 8.45*10^{28}electrons.m^{-3}$

if we convert the area from mm³ to m³ we get $A=80*10^{-6}m^{2}$ .So now that we have n, we plug in all the values of A ,I ,q and n into the main equation to obtain,

$u=30/(8.45*10^{28}*80*10^{-6}*1.602*10^{-19})\\u=2.8*10^{-5}m.s^{-1}$

which is our final answer.

6 0

3 years ago

A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the fo

LuckyWell [14K]

Complete Question:

A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the following closed cycle: - Isothermal expansion to 5000 cm3. - Isochoric cooling to 400 K . - Isothermal compression to 1000 cm3. - Isochoric heating to 500 K .

a) what is the work for one cycle

b) what is the thermal efficiency

Answer:

a) Work done for 1 cycle = 402.13

b) Thermal efficiency = 20%

Explanation:

Number of moles, n = 0.300 mol

Initial Volume, V₁ = 1000 cm³

Temperature, T = 500 K

Isothermal expansion to 5000 cm³

Final volume, V₂ = 5000 cm³

R = 8.314 J/ mol.K

Work done, W = nRT ln(V₂/V₁)

W = (0.3 * 8.314 * 500) * ln(5000/1000)

W = 1247.1 * ln5

W₁ = 2007.13 J

Isochoric cooling

In an Isochoric process, volume is constant i.e. V₂ = V₁ = V

W = nRT ln(V/V)

But ln(V/V) = ln 1 = 0

Work done, W₂ = 0 Joules

Isothermal Compression to 1000 cm³

V₂ = 1000 cm³

V₁ = 5000 cm³

W = nRT ln(V₂/V₁)

W = 0.3 * 8.314 * 400 ln(1000/5000)

W₃ = -1605 J

Isochoric heating to 500 K

Since there is no change in volume, no work is done

W₄ = 0 J

a) Work done for 1 cycle

W = W₁ + W₂ + W₃ + W₄

W = 2007.13 + 0 + 0 -1605+0

W = 402.13 Joules

b) Thermal efficiency

Thermal efficiency = (Net workdone for 1 cycle)/(Heat absorbed)

Heat absorbed = Work done due to thermal expansion = 2007.13 J

Thermal efficiency = 402.13/2007.13

Thermal efficiency = 0.2

Thermal efficiency = 0.2 * 100% = 20 %

3 0

4 years ago

A car is 200 m from a stop sign and traveling toward the sign at 40.0 m/s. At this time, the driver suddenly realizes that she m

victus00 [196]

Answer:

The acceleration of the car will be $a=9600m/sec^$

Explanation:

We have given that distance from stop sign s = 200 m

Time t = 0.2 sec

We have to find the constant acceleration

Now from second equation of motion $s=ut+\frac{1}{2}at^2$

$200=40\times 0.2+\frac{1}{2}\times a\times 0.2^2$

$a=9600m/sec^$

So the acceleration of the car will be $a=9600m/sec^$

6 0

4 years ago

I will give u BRAINILIEST and 15 points

o-na [289]

The answer is D interferometry

3 0

3 years ago

Read 2 more answers

You have a horizontal grindstone (a disk) that is 86 kg, has a 0.33 m radius, is turning at 92 rpm (in the positive direction),

Rudiy27

Answer:

a) α = 0.338 rad / s² b) θ = 21.9 rev

Explanation:

a) To solve this exercise we will use Newton's second law for rotational movement, that is, torque

τ = I α

fr r = I α

Now we write the translational Newton equation in the radial direction

N- F = 0

N = F

The friction force equation is

fr = μ N

fr = μ F

The moment of inertia of a saying is

I = ½ m r²

Let's replace in the torque equation

(μ F) r = (½ m r²) α

α = 2 μ F / (m r)

α = 2 0.2 24 / (86 0.33)

α = 0.338 rad / s²

b) let's use the relationship of rotational kinematics

w² = w₀² - 2 α θ

0 = w₀² - 2 α θ

θ = w₀² / 2 α

Let's reduce the angular velocity

w₀ = 92 rpm (2π rad / 1 rev) (1 min / 60s) = 9.634 rad / s

θ = 9.634 2 / (2 0.338)

θ = 137.3 rad

Let's reduce radians to revolutions

θ = 137.3 rad (1 rev / 2π rad)

θ = 21.9 rev

7 0

3 years ago