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Arte-miy333 [17]
3 years ago
10

At what water depth will a deep-water wave with a wavelength of 40 meters begin to "feel bottom"? 5 meters 15 meters 20 meters 1

0 meters
Physics
1 answer:
ella [17]3 years ago
8 0
The answer is 20 meters would be the water depth started to feel bottom. This is half of the wavelength. The reason behind this is the wave travels half or the partial of the wavelength underneath the surface. For that reason that is where the wave would be starting to feel the ground.
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Explain why the mechanical advantage of a single fixed probably is always one
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Ideal mechanical advantage doesn't take energy lost to friction into account. Explain why the mechanical advantage of a single fixed pulley is always 1. A single fixed pulley changes only the direction of the effort force. ... Energy transforms from the object supplying the force to the object being moved.
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Read the following paragraph from the introduction
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The statement 'energy cannot be created or destroyed' BEST supports the idea that energy remains constant during an energy transformation. It is the first law of thermodynamics.

<h3>Law of Conservation of Energy</h3>

The law of conservation of energy, also known as the first law of thermodynamics, indicates that energy can neither be created nor destroyed.

According to this law, the energy can be interchanged from one type of energy (e.g., kinetic energy) form to another (e.g., potential energy).

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2 years ago
A bar magnet is doplar. This means is has ___ poles.<br> A. Two <br> B. Zero<br> C. Three<br> D. One
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3 years ago
In all atoms of bismuth the number of electrons must equal to
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3 years ago
A launched hopper reach to 1.20 m maximum height. How much is it’s launch velocity?
garri49 [273]

The launch velocity is 4.8 m/s

Explanation:

We can solve this problem by applying the law of conservation of energy. In fact, the mechanical energy of the hopper (equal to the sum of the potential energy + the kinetic energy) is conserved. So we can write:

U_i +K_i = U_f + K_f

where:

U_i is the initial potential energy, at the bottom

K_i is the initial kinetic energy, at the bottom

U_f is the final potential energy, at the top

K_f is the final kinetic energy, at the top

We can rewrite the equation as:

mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2

where:

m is the mass of the hopper

g=9.8 m/s^2 is the acceleration of gravity

h_i = 0 is the initial height

u is the launch speed of the hopper

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v = 0 is the final speed (which is zero when the hopper reaches the maximum height)

Solving the equation for u, we find the launch speed of the hopper:

u=\sqrt{2gh_g}=\sqrt{2(9.8)(1.20)}=4.8 m/s

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