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3 years ago

Any clue on this one

Physics

1 answer:

rodikova [14]3 years ago

3 0

Answer: First option

Explanation: The higher the frequency, the higher the energy.

λν=c where λ is the wavelength, ν is the frequency and c is the speed of light. So when wavelength decreases, v increases and so does energy.

An AM radio wave has a very long wavelength. It therefore has a very low frequency and low energy.

A light wave has a very short wavelength. It therefore has a high frequency and high energy.

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A rod 7.0 m long is pivoted at a point 2.0 m from the left end. A downward force of 50 N acts at the left end, and a downward fo

kicyunya [14]

If the rod is in rotational equilibrium, then the net torques acting on it is zero:

∑ τ = 0

Let's give the system a counterclockwise orientation, so that forces that would cause the rod to rotate counterclockwise act in the positive direction. Compute the magnitudes of each torque:

• at the left end,

τ = + (50 N) (2.0 m) = 100 N•m

• at the right end,

τ = - (200 N) (5.0 m) = - 1000 N•m

• at a point a distance d to the right of the pivot point,

τ = + (300 N) d

Then

∑ τ = 100 N•m - 1000 N•m + (300 N) d = 0

⇒ (300 N) d = 1100 N•m

⇒ d ≈ 3.7 m

6 0

2 years ago

What number is between 2000 and 2500

Kitty [74]

If its just a number 2,300 is an answer

6 0

3 years ago

There are infinite black and white dots on a plane. Prove that the distance between one black dot and one white dot is one unit.

OlgaM077 [116]

Answer:this is confusing and what subject is this

Explanation:

6 0

3 years ago

Read 2 more answers

Unpolarized light is incident upon two ideal polarizing filters that do not have their transmission axes aligned. If 19% of the

Zepler [3.9K]

Answer:

51.94°

Explanation:

$I_0$ = Unpolarized light

$I_2$ = Light after passing though second filter = $0.19I_0$

Polarized light passing through first filter

$I_1=\frac{I_0}{2}$

Polarized light passing through second filter

$I_2=\frac{I_0}{2}cos^2\theta\\\Rightarrow 0.19I_0=\frac{I_0}{2}cos^2\theta\\\Rightarrow cos^2\theta=\frac{0.19I_0}{\frac{I_0}{2}}\\\Rightarrow cos\theta=\sqrt{\frac{0.19I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{\frac{0.19I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{0.19\times 2}\\\Rightarrow \theta=cos^{-1}\sqrt{0.38}\\\Rightarrow \theta=51.94^{\circ}$

The angle between the two filters is 51.94°

5 0

2 years ago

A transverse standing wave is set up on a string that is held fixed at both ends. The amplitude of the standing wave at an antin

ZanzabumX [31]

Answer:

a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

Explanation:

Given the data in the question;

as the equation of standing wave on a string is fixed at both ends

y = 2AsinKx cosωt

but k = 2π/λ and ω = 2πf

λ = 4 × 0.150 = 0.6 m

and f = v/λ = 260 / 0.6 = 433.33 Hz

ω = 2πf = 2π × 433.33 = 2722.69

given that A = 2.20 mm = 2.2×10⁻³

so $V_{max1}$ = A × ω

$V_{max1}$ = 2.2×10⁻³ × 2722.69 m/s

$V_{max1}$ = 5.9899 m/s

therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

A' = 2AsinKx

= 2.20sin( 2π/0.6 ( 0.075) rad )

= 2.20 sin( 0.7853 rad ) mm

= 2.20 × 0.706825 mm

A' = 1.555 mm = 1.555×10⁻³

$V_{max2}$ = A' × ω

$V_{max2}$ = 1.555×10⁻³ × 2722.69

$V_{max2}$ = 4.2338 m/s

Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

8 0

2 years ago