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2 years ago

Using a flowchart, discuss how a tsunami develops?

Physics

1 answer:

Ahat [919]2 years ago

4 0

Answer:

A tsunami is a sequence of particularly long water waves that can spread over very great distances and, as such, cause water to move.

When penetrating into areas of shallow water, the sea is compressed and thus piles up on the coasts to form several high tidal waves. These carry the water with great force far over the shoreline and usually cause great damage. During the subsequent retreat, the material carried away on the flooded land, often also people and animals, is mostly washed far out into the ocean.

Tsunamis occur as a result of sudden water displacement, such as when parts of the ocean floor are raised or lowered during an undersea earthquake or when large masses of earth and rock slide into the water as well as due to violent winds.

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Electrons and protons travel from the Sun to the Earth at a typical velocity of 3.83 ✕ 105 m/s in the positive x-direction. Thou

Leona [35]

Answer:

$F=2.84*10^{-26}N$ & -y direction

$F=2.84*10^{-26}N$ & +y direction

Explanation:

From the question we are told that:

Speed of electron $V_e=3.83 * 10^5 m/s$ +x direction

Earths magnetic field $B_e=3.04 * 10^-^8$ +z direction

Generally the equation for magnetic force $F_m$ is mathematically given by

$F=q(V_e*B_e)$

where

$q=1.6*10^{-19}c\\\=i*\=z=-\=j$

$F=1.6*10^{-19}(3.83 * 10^5 m/s*3.04 * 10^-^8)$

$F=-2.84*10^{-26}N \=j$

Magnitude & Direction

$F=2.84*10^{-26}N$ -y direction

Generally the equation for magnitude and direction of the magnetic force on an electron. is mathematically given by

$\=F'=-1.6*10^{-19}(3.83 * 10^5 m/s*3.04 * 10^-^8)$

$\=F'=-2.84*10^{-26}N \=j$

Magnitude & Direction

$F=2.84*10^{-26}N$ & +y direction

5 0

3 years ago

A large balloon of mass 210 kg is filled with helium gas until its volume is 329 m3. Assume the density of air is 1.29 kg/m3 and

Nastasia [14]

(a) See figure in attachment (please note that the image should be rotated by 90 degrees clockwise)

There are only two forces acting on the balloon, if we neglect air resistance:

- The weight of the balloon, labelled with W, whose magnitude is

$W=mg$

where m is the mass of the balloon+the helium gas inside and g is the acceleration due to gravity, and whose direction is downward

- The Buoyant force, labelled with B, whose magnitude is

$B=\rho_a V g$

where $\rho_a$ is the air density, V is the volume of the balloon and g the acceleration due to gravity, and where the direction is upward

(b) 4159 N

The buoyant force is given by

$B=\rho_a V g$

where $\rho_a$ is the air density, V is the volume of the balloon and g the acceleration due to gravity.

In this case we have

$\rho_a = 1.29 kg/m^3$ is the air density

$V=329 m^3$ is the volume of the balloon

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

So the buoyant force is

$B=(1.29 kg/m^3)(329 m^3)(9.8 m/s^2)=4159 N$

(c) 1524 N

The mass of the helium gas inside the balloon is

$m_h=\rho_h V=(0.179 kg/m^3)(329 m^3)=59 kg$

where $\rho_h$ is the helium density; so we the total mass of the balloon+helium gas inside is

$m=m_h+m_b=59 kg+210 kg=269 kg$

So now we can find the weight of the balloon:

$W=mg=(269 kg)(9.8 m/s^2)=2635 N$

And so, the net force on the balloon is

$F=B-W=4159 N-2635 N=1524 N$

(d) The balloon will rise

Explanation: we said that there are only two forces acting on the balloon: the buoyant force, upward, and the weight, downward. Since the magnitude of the buoyant force is larger than the magnitude of the weigth, this means that the net force on the balloon points upward, so according to Newton's second law, the balloon will have an acceleration pointing upward, so it will rise.

(e) 155 kg

The maximum additional mass that the balloon can support in equilibrium can be found by requiring that the buoyant force is equal to the new weight of the balloon:

$W'=(m'+m)g=B$

where m' is the additional mass. Re-arranging the equation for m', we find

$m'=\frac{B}{g}-m=\frac{4159 N}{9.8 m/s^2}-269 kg=155 kg$

(f) The balloon and its load will accelerate upward.

If the mass of the load is less than the value calculated in the previous part (155 kg), the balloon will accelerate upward, because the buoyant force will still be larger than the weight of the balloon, so the net force will still be pointing upward.

(g) The decrease in air density as the altitude increases

As the balloon rises and goes higher, the density of the air in the atmosphere decreases. As a result, the buoyant force that pushes the balloon upward will decrease, according to the formula

$B=\rho_a V g$

So, at a certain altitude h, the buoyant force will be no longer greater than the weight of the balloon, therefore the net force will become zero and the balloon will no longer rise.