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3 years ago

How can people help stop erosion on beach

Physics

1 answer:

seraphim [82]3 years ago

7 0

Answer:

<h3>since erosion is unavoidable the problem becomes discovering ways to prevent it. present beach erosion prevention methods include sand bags,vegetation,seawalls,sand dunes,and sand fences.</h3>

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Which would be the best way to represent the concentration of a 1.75 M K2CrO4 solution

AlekseyPX

1.75 moles per litre K2CrO4

7 0

4 years ago

Read 2 more answers

How many liters are in 0.034 moles of chlorine gas?

Anuta_ua [19.1K]

The volume of the gas is $7.6\cdot 10^{-4}m^3$

Explanation:

At standard temperature and pressure (stp), the volume occupied by 1 mole of an ideal gas is always equal to

$V=22.4 L$

In this problem, we have an amount of moles of gas of

n = 0.034 mol

Since 1 mol of gas occupies a volume of 22.4 L, we can set up the following ratio:

$\frac{1 mol}{22.4 L}=\frac{0.034 mol}{V}$

where V is the volume occupied by 0.034 mol of gas. Solving for V, we find

$V=\frac{0.034 mol}{1 mol}(22.4 L)=0.76 L$

Or, in cubic metres,

$V=0.76 L = 7.6\cdot 10^{-4}m^3$

Learn more about ideal gases:

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#LearnwithBrainly

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3 years ago

An object of mass m is dropped from height h above a planet of mass M and radius R. Find an expression for the objectâ€™s speed

Oksana_A [137]

Answer:

Explanation:

Given

mass of object is m

Mass of planet is M

radius of planet is R

Total Energy associated with mass m at a height h above planet is Gravitational Potential Energy which is given by

$E_1=-\frac{GMm}{R+h}$

When it falls on earth with some velocity v

$E_2$ =Kinetic Energy+Potential Energy

$=\frac{1}{2}mv^2+\frac{-GMm}{R}$

As Energy is conserved therefore

$=E_2$

$\frac{-GMm}{R+h}=\frac{1}{2}mv^2+\frac{-GMm}{R}$

$\frac{1}{2}mv^2=\frac{GMmh}{R(R+h)}$

$v=\sqrt{\frac{2GMh}{R(R+h)}}$

6 0

3 years ago

Under electrostatic conditions, the electric field just outside the surface of any charged conductor:

lesantik [10]

Answer:

D. is always perpendicular to the surface of the conductor

Explanation:

1) Answer is (D) option. Electric field just outside surface of charged conductor is normal to conductor at that point.

It can be explained on the basis of the fact that, Electric field inside conductor under static condition is zero. As a result potential difference between any two points with in conductor is zero. So whole of conductor is equipotential body.

Equipotential surface and Electric field lines always cut at 90 degrees to each other. Conductor being equipotential body, Electric field lines starting or terminating at conductor must be normal to surface. Hence electric field just outside conductor is perpendicular or normal to surface.

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3 years ago

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frozen [14]

A. Since it does not say the ball is moving your answer is A.

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3 years ago