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3 years ago

Which techniques can help reduce the effects of tornadoes?

Physics

1 answer:

murzikaleks [220]3 years ago

3 0

Answer: 1.building storm cellars and wind-resistant structures.

3.using more radars and satellites in the area.

A tornado is a storm of wind. In tornado the high speed wind takes the form of a cone that destroys anything that comes in it's path. The following techniques can help in reducing the effects of tornadoes:

1. building storm cellars and wind-resistant structures: This will provide support and protection to the building against high speed wind currents during an event of tornado.

3.using more radars and satellites in the area: This will help in identification of origin of wind producing tornado in a region so as to create awareness about the coming calamity. This will allow people to relocate to safer place and can take preventive measures.

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A 6.72 particle moves through a region of space where an electric field of magnitude 1270 N/C points in the positive direction,

Romashka [77]

Answer:

$v_{y} = -104 m/s$

Explanation:

Using:

Force = electric field * charge

$F=e*q$

Force = magnitude of charge * velocity * magnetic field * sin tither

$F_{x2}= |q|*v*B*sin \alpha$

Force on particle due to electric field:

$F_{x1}= E*q = (1270N/C)*(-6.72*10^{-6} ) = -8.53*10^{-3}$

Force on particle due to magnetic field:

$F_{x2}= |q|*v*B*sin \alpha = (6.72*10^{-6} )*(1.15)*(sin90)*v = (7.728*10^{-6})*(v)$

$F_{x2}$ is in the positive x direction as $F_{x1}$ is in the negative x direction while net force is in the positive x direction.

Magnetic field is in the positive Z direction, net force is in the positive x direction.

According to right hand rule, Force acting on particle is perpendicular to the direction of magnetic field and velocity of particle. This would mean the force is along the y-axis. As this is a negatively charged particle, the direction of the velocity of the particle is reversed. Therefore velocity of particle, v, has to be in the negative y direction.

Now,

$F_{xnet}- F_{x1 } = F_{x2 }$

$(6.13*10^{-3}) - (8.53*10^{-3} ) = (7.728*10^{-6})*(v)$

$v = (F_{xnet} - F_{x1}) / (F_{x2} )$

$=((6.13*10^{-3} ) - (8.53*10^{-3})) / (7.728*10^{-6})$

$= (- 104.25) m/s$

$v_{y} = -104 m/s$

8 0

3 years ago

A spherical drop of water carrying a charge of 42 pC has a potential of 620 V at its surface (with V = 0 at infinity). (a) What

iren [92.7K]

Answer:

0.0006091222 m

Explanation:

q = Charge = 42 pC

V = Voltage = 620 V

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

Electric potential is given by (at r = R)

$V=\dfrac{q}{4\pi\epsilon R}\\\Rightarrow R=\dfrac{q}{4\pi\epsilon V}\\\Rightarrow R=\dfrac{42\times 10^{-12}}{4\pi\times 8.85\times 10^{-12}\times 620}\\\Rightarrow R=0.0006091222\ m$