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2 years ago

Waves with higher amplitude transfer ____energy.

Physics

1 answer:

natali 33 [55]2 years ago

3 0

Answer:

- Waves with higher amplitude transfer HIGHER energy.

- Waves with higher frequency transfer HIGHER energy.

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If earth did not rotate how would air at the equator move?

Dmitry_Shevchenko [17]

Heat rises, and it is warmer at the equator, so I think warm air would rise at the equator and move towards the cooler poles.

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2 years ago

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the brightest , hottest, and most massive stars are the brilliant blue stars designated as spectral class O. if a class O star w

4vir4ik [10]

The speed is 0.956 m / s.

<u>Explanation</u>:

The kinetic energy is equal to the product of half of an object's mass, and the square of the velocity.

K.E = 1/2 $\times$ m $\times$ $v^{2}$

where K.E represents the kinetic energy,

m represents the mass,

v represents the velocity.

K.E = 1/2 $\times$ m $\times$ $v^{2}$

1.10 $\times$ 10^42 = 1/2 $\times$ 3.26 $\times$ 10^31 $\times$ $v^{2}$

$v^{2}$ = (1.10 $\times$ 10^42 $\times$ 2) / (3.26 $\times$ 10^31)

v = 0.956 m / s.

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2 years ago

A short 6061-T6 aluminum cylindrical block, having an original diameter of 20mm and a length of 75mm, is placed in a compression

Alex Ar [27]

Answer:

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Explanation:

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2 years ago

Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. The electric field at a po

gregori [183]

Answer:

E = 0 r <R₁

Explanation:

If we use Gauss's law

Ф = ∫ E. dA = $q_{int}$ / ε₀

in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.

Consequently by Gauss's law the electric field is ZERO

E = 0 r <R₁

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2 years ago

A watermelon is dropped off of a 50 ft bridge, and it explodes upon impact with the ground. How fast was it traveling in mph upo

Drupady [299]

Answer: 56.72 ft/s

Explanation:

Ok, initially we only have potential energy, that is equal to:

U =m*g*h

where g is the gravitational acceleration, m the mass and h the height.

h = 50ft and g = 32.17 ft/s^2

when the watermelon is near the ground, all the potential energy is transformed into kinetic energy, and the kinetic energy can be written as:

K = (1/2)*m*v^2

where v is the velocity.

Then we have:

K = U

m*g*h = (m/2)*v^2

we solve it for v.

v = √(2g*h) = √(2*32.17*50) ft/s = 56.72 ft/s

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2 years ago