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stiks02 [169]
3 years ago
6

If you wish to observe features that are around the size of atoms, say 5.5 × 10^-10 m, with electromagnetic radiation, the radia

tion must have a wavelength of about the size of the atom itself.
Required:

a. What is its frequency?

b. What type of electromagnetic radiation might this be?
Physics
1 answer:
Tanya [424]3 years ago
3 0

Answer:

a) 5.5×10^17 Hz

b) visible light

Explanation:

Since the wavelength of the electromagnetic radiation must be about the size of the about itself, this implies that;

λ= 5.5 × 10^-10 m

Since;

c= λ f and c= 3×10^8 ms-1

f= c/λ

f= 3×10^8/5.5 × 10^-10

f= 5.5×10^17 Hz

The electromagnetic wave is visible light

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A periodic wave with wavelength 2m has a speed of 4m/s. What is the waves frequency?.
Gekata [30.6K]

The wave frequency is 2 Hz.

What is wave frequency ?

The number of waves that pass through a fixed point in a given amount of time is referred to as the wave frequency. The hertz is the SI unit for wave frequency (Hz).

f = v / w

where,

f = frequency\\v = speed \\w = wavelength

Given,

v = 4 m/s, w = 2 m

f = 4/2 \\f = 2 Hz\\

The waves frequency is 2 Hz.

To know more about wave frequency,check out:

brainly.com/question/15830195

#SPJ4

5 0
1 year ago
A gymnast is swinging on a high bar. The distance between his waist and the bar is 0.905 m, as the drawing shows. At the top of
Citrus2011 [14]

Answer:

5.959 m/s

Explanation:

m = Mass of gymnast

u = Initial velocity

v = Final velocity

h_i = Initial height

h_f = Final height

From conservation of Energy

\frac{1}{2}mv^2+mgh_f=\frac{1}{2}mu^2+mgh_i\\\Rightarrow\frac{1}{2}mv^2+mg0=\frac{1}{2}m0^2+mgh_i\\\Rightarrow \frac{1}{2}mv^2=mgh_i\\\Rightarrow v=\sqrt{2gh_i}

h_i=2r

v=\sqrt{4gr}\\\Rightarrow v=\sqrt{4\times 9.81\times 0.905}\\\Rightarrow v=5.959\ m/s

Velocity of gymnast at bottom of swing is 5.959 m/s

5 0
3 years ago
Please help ASAP!!
inessss [21]

Answer:

at t=46/22, x=24 699/1210 ≈ 24.56m

Explanation:

The general equation for location is:

x(t) = x₀ + v₀·t + 1/2 a·t²

Where:

x(t) is the location at time t. Let's say this is the height above the base of the cliff.

x₀ is the starting position. At the base of the cliff we'll take x₀=0 and at the top x₀=46.0

v₀ is the initial velocity. For the ball it is 0, for the stone it is 22.0.

a is the standard gravity. In this example it is pointed downwards at -9.8 m/s².

Now that we have this formula, we have to write it two times, once for the ball and once for the stone, and then figure out for which t they are equal, which is the point of collision.

Ball: x(t) = 46.0 + 0 - 1/2*9.8 t²

Stone: x(t) = 0 + 22·t - 1/2*9.8 t²

Since both objects are subject to the same gravity, the 1/2 a·t² term cancels out on both side, and what we're left with is actually quite a simple equation:

46 = 22·t

so t = 46/22 ≈ 2.09

Put this t back into either original (i.e., with the quadratic term) equation and get:

x(46/22) = 46 - 1/2 * 9.806 * (46/22)² ≈ 24.56 m

4 0
3 years ago
When the temperature of the air is 50°C, the velocity of a sound wave traveling through the air is approximately?
noname [10]

The answer is:

C. 361 m/s

The explanation:

To calculate  the speed of sound at a given temperature (50°C) we are going to use this formula:

v = 331 + 0.6T

when V is the velocity

and T is the temperature = 50°C

by substitution:

v = 331 + 0.6(50)

v = 361 m/s

So, The correct answer is C.

because of the variation of the motion of the molecules of air with change of temperature so, the velocity (V) of the  sound in the air is change with temperature.

5 0
3 years ago
Read 2 more answers
A small grinding wheel has a moment of inertia of 4.0*10-5kgm2. What net torque must be applied to the wheel for its angular acc
kvv77 [185]

Hi there!

We can use the rotational equivalent of Newton's Second Law:

\huge\boxed{\Sigma \tau = I \alpha}

Στ = Net Torque (Nm)

I = Moment of inertia (kgm²)

α = Angular acceleration (rad/sec²)

We can plug in the given values to solve.

\Sigma \tau = (4 * 10^{-5})(150) = \boxed{0.006 Nm}

4 0
2 years ago
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