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2 years ago

________ occurs when a person is holding an object that is directly hit or splashed by lightning.

Physics

2 answers:

Papessa [141]2 years ago

7 0

When someone is holding something that has been struck or splashed by lightning, contact damage occurs.

We need additional information concerning lightning and injuries in order to identify the solution.

<h3>What types of injuries are brought on by lightning?</h3>

Lightning is the name for a natural electrical discharge that occurs quickly and with a dazzling flash.

It has a tremendous amount of energy.

Lightning-related injuries can be divided into three categories: direct strikes, side splashes, and contact injuries.

When someone is struck by lightning directly, they can get direct injury.

When a current splashes from a neighboring object, it is called a side splash.

When someone touches a lightning-hit object, contact harm results.

In light of this, we can say that contact injuries happen when a person is holding an object that has been struck by lightning or splashed by it.

Learn more about the lightning and harm here:

brainly.com/question/28055828

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gulaghasi [49]2 years ago

6 0

Contact injury happens when a person is holding an object that is directly hit or splashed by lightning.

To find answer, we need to know more about the lightning and injuries.

<h3>What are the injuries caused by lightning?</h3>

The phenomenon of natural electrical discharge happens in a short time with a bright flash is called lightning.
It carries a huge amount of energy.
The injuries caused by lightning can be classified as Direct strikes, Side splash and the Contact injury.
Direct injuries happens when a person directly hit by lightning.
Side splash occurs when a current splashes from a nearby object.
Contact injury happens when a person touching an object that is hit by lightning.

Thus, we can conclude that, contact injury occurs when a person is holding an object that is directly hit or splashed by lightning.

Learn more about lightning and injuries here:

brainly.com/question/28055828

#SPJ1

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A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 18.0 cm , giving it a ch

vladimir1956 [14]

A) The electric field inside the paint layer is zero

B) The electric field just outside the paint layer is $3.2\cdot 10^7 N/C$ (radially inward)

C) The electric field at 6.00 cm from the surface is $1.2\cdot 10^7 N/C$ (radially inward)

Explanation:

We can solve the problem by applying Gauss Law, which states that the electric flux through a Gaussian surface must be equal to the charge contained in the surface divided by the vacuum permittivity:

$\int EdS = \frac{q}{\epsilon_0}$

where

E is the magnitude of the electric field

dS is the element of the surface

q is the charge contained within the surface

$\epsilon_0$ is the vacuum permittivity

By taking a sphere centered in the origin,

$\int E dS = E \cdot 4\pi r^2$

where $4\pi r^2$ is the surface of the Gaussian sphere of radius r.

In this problem, we want to find the electric field just inside the paint layer, so we take a value of r smaller than

$R=9.0 cm = 0.09 m$ (radius of the plastic sphere is half of the diameter)

Since the charge is all distributed over the plastic sphere, the charge contained within the Gaussian sphere is zero:

$q=0$

And therefore,

$E4\pi r^2 = 0\\\rightarrow E = 0$

So, the electric field inside the plastic sphere is zero.

Here we apply again Gauss Law:

$E\cdot 4 \pi r^2 = \frac{q}{\epsilon_0}$

In this case, we want to calculate the electric field just outside the paint layer: this means that we take r as the radius of the plastic sphere, so

$r=R=0.18 m$

The charge contained within the Gaussian sphere is therefore

$q=-29.0 \mu C = -29.0\cdot 10^{-6}C$

Therefore, the electric field is

$E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-29.0\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.09)^2}=-3.2\cdot 10^7 N/C$

And the negative sign indicates that the direction of the field is radially inward (because the charge that generates the field is negative). However, the text of the question says "Enter positive value if the field is directed radially inward and negative value if the field is directed radially outward", so the answer to this part is

$E=3.2\cdot 10^7 N/C$

For this part again, we apply Gauss Law:

$E\cdot 4 \pi r^2 = \frac{q}{\epsilon_0}$

In this case, we want to calculate the field at a point 6.00 cm outside the surface of the paint layer; this means that the radius of the Gaussian sphere must be

r = 9 cm + 6 cm = 15 cm = 0.15 m

While the charge contained within the sphere is again

$q=-29.0 \mu C = -29.0\cdot 10^{-6}C$

Therefore, the electric field in this case is

$E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-29.0\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.15)^2}=-1.2\cdot 10^7 N/C$