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3 years ago

Which type of electromagnetic wave has the longest wavelength

Physics

1 answer:

balu736 [363]3 years ago

5 0

Answer

Using the formula of wavelength

$\lambda = \dfrac{c}{f}$

From the above expression, we can see that the wavelength is inversely proportional to frequency.

For wavelength to be the longest frequency should be minimum.

Radio wave has the lowest frequency i.e. 10⁴ so, the wavelength of radio-wave will be maximum.

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If a horse is running at a speed of 35 m/s when it's rider pulls up on the reigns. After 5 seconds the horse slows down to a pac

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A ball is thrown from ground level with an initial speed of 24.5 m/s at an angle of 35.5 degrees above the horizontal. The ball

eimsori [14]

6.07 m

Explanation:

Given:

$v_0=24.5\:\text{m/s}$

$\theta_0 = 35.5°$

First, we need to find the amount of time it takes to travel a horizontal distance of 25.8 m. We know that

$x = v_{0x}t \Rightarrow t = \dfrac{x}{v_0 \cos \theta_0}$

$t = 1.29\:\text{s}$

To find the vertical height where the ball hit the wall, we use

$y = v_{0y}t - \frac{1}{2}gt^2$

$\:\:\:\:=(24.5\:\text{m/s})\sin 35.5(1.29\:\text{s}) \\ - \frac{1}{2}(9.8\:\text{m/s}^2)(1.29\:\text{s})^2$

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2 years ago

Why do astronomers use frequencies other than the visible ones when they are

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3 years ago

P7.16 A thin flat plate 55 by 110 cm is immersed in a 6-m/s stream of SAE 10 oil at 20C. Compute the total friction drag if the

timofeeve [1]

Answer:

The total friction drag for the long side of the plate is 107 N

The total friction drag for the long side of the plate is 151.4 N

Explanation:

The first question is to obtain the friction drag when the fluid i parallel to the long side of the plate

The block representation of the this problem is shown on the first uploaded image

Where the U is the initial velocity = 6 m/s

So the equation we will be working with is

$F = \frac{1}{2} \rho C_fAU^2$

Where $\rho$ is the density of SAE 10W = $870\ kg/m^3$ This is obtained from the table of density at 20° C

$C_f$ is the friction drag coefficient

This coefficient is dependent on the Reynolds number if the Reynolds number is less than $5*10^5$ then the flow is of laminar type and

$C_f$ = $\frac{1.328}{\sqrt{Re} }$

But if the Reynolds number is greater than $5*10^5$ the flow would be of Turbulent type and

$C_f = \frac{0.074}{Re_E^{0.2}}$

Where Re is the Reynolds number

To obtain the Reynolds number

$Re = \frac{\rho UL}{\mu}$

where L is the length of the long side = 110 cm = 1.1 m

and $\mu$ is the Dynamic viscosity of SAE 10W oil $= 1.04*10^{-1} kg /m.s$

This is gotten from the table of Dynamic viscosity of oil

$Re = \frac{870 *6*1.1}{1.04*10^{-1}}$

$= 55211.54$

Since 55211.54 < $5.0*10^5$

Hence

$C_f = \frac{1.328}{\sqrt{55211.54} }$

$= 0.00565$

$A$ is the area of the plate = $\frac{ (110cm)(55cm)}{10000}$ = $0.55m^2$

Since the area is immersed totally it should be multiplied by 2 i.e the bottom face and the top face are both immersed in the fluid

$F = \frac{1}{2} \rho C_f(2A)U^2$

$F =\frac{1}{2} *870 *0.00565*(2*0.55)*6^2$

$F = 107N$

Considering the short side

To obtain the Reynolds number

$Re = \frac{\rho U b}{\mu}$

Here b is the short side

$Re =\frac{870*6*0,55}{1.04*10^{-1}}$

$=27606$

Since the value obtained is not greater than $5*10^5$ then the flow is laminar

And

$C_f = \frac{1.328}{\sqrt{Re} }$

$= \frac{1.328}{\sqrt{27606} }$

$= 0.00799$

The next thing to do is to obtain the total friction drag

$F = \frac{1}{2} \rho C_f(2A)U^2$

Substituting values

$F = \frac{1}{2} * 870 * 0.00799 * 2( 0.55) * 6^2$

$= 151.4 N$

4 0

2 years ago