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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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What is the magnitude of the angular momentum (in kgm2/s) of a 40 g golf ball flying through the air and spinning at 4300 rpm af

sladkih [1.3K]

Answer:

$L=0.0045\ kg-m^2/s$

Explanation:

Given that,

The mass of a golf ball, m = 40 g = 0.04 kg

Its angular velocity, $\omega=4300\ rpm=450.29\ rad/s$

The radius of the sphere is 2.5 cm or 0.025 m

We need to find the magnitude of the angular momentum of the ball. It is given by the formula as follows:

$L=I\omega$

Where I is moment of inertia

For sphere, $I=\dfrac{2}{5}mr^2$

$L=\dfrac{2}{5}mr^2\omega\\\\L=\dfrac{2}{5}\times 0.04\times (0.025)^2\times 450.29\\\\L=0.0045\ kg-m^2/s$

So, the magnitude of the angular momentum of the sphere is $0.0045\ kg-m^2/s$ .

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2 years ago

State whether the following statement are true or false .

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1. It’s true
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3 years ago

A pendulum of length L=36.1 cm and mass m=168 g is released from rest when the cord makes an angle of 65.4 degrees with the vert

pychu [463]

(a) -0.211 m

At the beginning the mass is displaced such that the length of the pendulum is L = 36.1 cm and the angle with the vertical is

$\theta=65.4^{\circ}$

The projection of the length of the pendulum along the vertical direction is

$L_y = L cos \theta = (36.1 cm)(cos 65.4^{\circ})=15.0 cm$

the full length of the pendulum when the mass is at the lowest position is

L = 36.1 cm

So the y-displacement of the mass is

$\Delta y = 15.0 cm - 36.1 cm = -21.1 cm = -0.211 m$

(b) 0.347 J

The work done by gravity is equal to the decrease in gravitational potential energy of the mass, which is equal to

$\Delta U = mg \Delta y$

where we have

m = 168 g = 0.168 kg is the mass of the pendulum

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

$\Delta y = 0.211 m$ is the vertical displacement of the pendulum

So, the work done by gravity is

$W=(0.168 kg)(9.8 m/s^2)(0.211 m)=0.347 J$

And the sign is positive, since the force of gravity (downward) is in the same direction as the vertical displacement of the mass.

(c) Zero

The work done by a force is:

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and the displacement

In this situation, the tension in the string always points in a radial direction (towards the pivot of the pendulum), while the displacement of the mass is tangential (it follows a circular trajectory): this means that the tension and the displacement are always perpendicular to each other, so in the formula

$\theta=90^{\circ}, cos \theta = 0$

and so the work done is zero.

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