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3 years ago

Describe the motion of an unanchored rowboat when a water wave passes

2 answers:

8 0

<span>The boat will move up and down because of the waves but it will not move forward because the wave doesn't carry the boat with it.

Hope this helps. </span>

USPshnik [31]3 years ago

6 0

It mimics the movement of the waves

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Statements about isotopes

Iteru [2.4K]

Answer:

Key Takeaways: Isotopes

Isotopes are samples of an element with different numbers of neutrons in their atoms.

The number of protons for different isotopes of an element does not change.

Not all isotopes are radioactive. Stable isotopes either never decay or else decay very slowly. ...

When an isotope decays, the starting material is the parent isotope.

Explanation:

8 0

3 years ago

I will be so thankful if u answer correctly!!

polet [3.4K]

Explanation:

The acceleration due to gravity on earth $g_{E}$ is given by

$g_{E} = G\dfrac{M_{E}}{R_{E}^{2}}$

where $G$ = universal gravitational constant

$\:\:\:\:\:\:\:\:\:\:\:\:M_{R}$ = mass of the earth

$\:\:\:\:\:\:\:\:\:\:\:\:R_{E}$ = radius of the earth

Planet Krypton has twice the mass of earth and 3 times the radius so its acceleration due to gravity $g_{K}$ is

$g_{K} = G\dfrac{M_{K}}{R_{K}^{2}}$

$\:\:\:\:\:\: = G\dfrac{(2M_{E})}{(3R_{E})^{2}}$

$\:\:\:\:\;\:= (\dfrac{2}{9})G\dfrac{M_{E}}{R_{E}^{2}}$

$g_{K} = (\dfrac{2}{9})\:g_{E}$

If we multiply both sides by Superman's mass, we get his weights on both planets:

$mg_{K} = (\dfrac{2}{9})\:(mg_{E})$

$W_{K} = (\dfrac{2}{9})\:W_{E} = (\dfrac{2}{9})(900\:N)= 200\:N$

3 0

3 years ago

Which of the following describes deposition?

alexgriva [62]

Deposition is geological process of sediments, soil and rocks that are added to land-form or land mass. Wind, ice, water, and gravity transport weathered surface material, which, at the loss of enough kinetic energy in the fluid, is deposited. Technically Building up layers of sediment.

hope this helps

7 0

3 years ago

You are driving west in your car on a clear, warm day. Up ahead, you see a dark band of ominous clouds. Soon, you drive beneath

Neporo4naja [7]

Answer:

You have just driven across the Cold front.

Explanation:

Weather Front:

In Meteorology, weather fronts are simply boundaries between two air masses of different densities. There are four weather fronts and one of them is cold front and other fronts are warm front, stationary front and occluded front.

Cold Front:

In Meteorology, cold front is a boundary of cold air mass that is advancing under the warm air mass.

In our scenario, as the driver was moving in the west and he listened on radio, the next country on north is under tornado warning. So, when he passed the cold front, he experienced the conditions described.

6 0

3 years ago

A pump and its horizontal intake pipe are located 82 m beneath the surface of a large reservoir. the speed of the water in the i

uysha [10]

<span>Ans : Bernoulli's principle states for incompressible non-viscous flow that p/Ď + gâ™h + (1/2)â™vÂ˛ = constant Evaluate the equation along a stream line from liquid surface of the reservoir (1) to the inlet of the pipe pâ‚/Ď + gâ™hâ‚ + (1/2)â™vâ‚Â˛ = pâ‚‚/Ď + gâ™hâ‚‚ + (1/2)â™vâ‚‚Â˛ => vâ‚‚ = âš[ 2â™(pâ‚-pâ‚‚)/Ď + 2â™gâ™(hâ‚-hâ‚‚) + vâ‚Â˛ ] lets make some assumptions: - the pressure at the liquid surface is equal to the atmospheric pressure pâ‚ = 1atm = 101325Pa - the velocity of the liquid at the surface (that is the speed at which the liquid level in reservoir decreases) is quite small, so it may be ignored: vâ‚ â‰ 0 So vâ‚‚ = âš[ 2â™(pâ‚-pâ‚‚)/Ď + 2â™gâ™(hâ‚-hâ‚‚) ] The height difference is fixed. So the only variable remaining is the pressure in the pipe. As higher it is as lower the velocity in the pipe is. So you get the maximum velocity for the minimum pressure. Since pressure cannot drop below zero this is pâ‚‚ = 0 Therefore vâ‚‚max = âš[ pâ‚/Ď + gâ™(hâ‚-hâ‚‚) ] = âš[ 2â™101325Pa/1000kg/mÂł + 2â™9.81m/sÂ˛â™12m ] = 20.93m/s</span>

4 0

3 years ago