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3 years ago

The wavelength and energy of a wave are inversely proportional. Based on this, which wave listed would have the LOWEST energy.

2 answers:

5 0

D because it has the lowest wavelength, meaning the wave is the longest. Its crest are more spaced out creating lower energy. The smaller distance between the crests equals the higher the sound. Creating higher energy.

I am 95% sure this is right. Very sorry if it isn't.
Hope it helps. ;3

serg [7]3 years ago

4 0

Answer:

D) wave with wavelength of 600 km

Explanation:

If the wavelength and energy of a wave are inversely proportional, then a high wavelength corresponds to a low energy, and low wavelength corresponds to a high energy. So, the lowest energy correspond the highest wavelength. From the options that is option D. Remember that 600 km > 600 m > 600 cm > 600 mm

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An uncrewed mission to the nearest star, Proxima Centauri, is launched from the Earth's surface as a projectile with an initial

Anna [14]

Answer:

42.96 km/s

Explanation:

From the conservation of Energy

$(PE+KE)_i=(PE+KE)_f\\\Rightarrow -\frac{GmM}{R}+\frac{1}{2}mv_i^2=0+\frac{1}{2}mv_f^2$

Mass gets cancelled

$-\frac{GM}{R}+\frac{1}{2}v_i^2=0+\frac{1}{2}v_f^2\\\Rightarrow -2\frac{GM}{R}+v_i^2=v_f^2\\\Rightarrow -v_e^2+v_i^2=v_f^2\\\Rightarrow v_f=\sqrt{v_i^2-v_e^2}$

$v_e=\sqrt{\frac{2Gm}{R}}$ = Escape velocity of Earth = 11.2 km/s

$v_i$ = Velocity of projectile = 44.4 km/s

$v_f=\sqrt{44.4^2-11.2^2}\\\Rightarrow v_f=42.96\ km/s$

The velocity of the spacecraft when it is more than halfway to the star is 42.96 km/s

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3 years ago

What is the focal length of a lens

goldenfox [79]

Answer:

The focal length of a lens is refers to the distance from the center of the lens to the principal foci.

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3 years ago

In the diagram, a force of 20 newtons is applied to a block. The block is in dynamic equilibrium. What is the magnitude and dire

sladkih [1.3K]

Answer:B 20 newtons opposite to the direction of the applied force

Explanation:

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2 years ago

Read 2 more answers

A 500 kg wrecking ball is knocking down a wall. When it is pulled back to its highest point, it is at a height of 6.2 m. When it

svetlana [45]

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3 years ago

A batter hits a foul ball. The 0.140-kg baseball that was approaching him at 40.0 m/s leaves the bat at 30.0 m/s in a direction

lara31 [8.8K]

<h3>Answer</h3><h3>7 Ns</h3><h3>Explanation</h3>

Given in the question,

mass of foul ball = 0.140 kg

initial speed with which ball was hit with the bat = 30 m/s

final speed = 40 m/s

According to the scenario the whole scene is making a right angle triangle

So, to the solve the question we will use pythagorus theorem

<h3>Hypotenuse² = base² + height²</h3>

Here,

Hypotenuse= Magnitude of impulse

Base = 1st change of momentum

height = 2nd change of momentum

1st impulse (1st change of momentum)

p = m(1)v(1) = (0.14 kg)(40.0 m/s) = 5.6 kg m / s = 5.6 N s

2nd impulse (2nd change of momentum)

p = m(2)v(2) = (0.14 kg)(30.0 m/s) = 4.2 kg m / s = 4.2 N s

Magnitude of impulse (hypotenuse of triangle)

impulse² = (5.6)² + (4.2)²

impulse² = 31.36 + 17.64

impulse² = 49

impulse² = √49

impulse = 7.0 N s

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3 years ago