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3 years ago

A thunderstorm is a connection between?

Physics

2 answers:

ExtremeBDS [4]3 years ago

8 0

A thunderstorm is a loose, intermittent electrical connection between charged clouds and the ground.

Current only flows in occasional pulses, called "lightning".

dangina [55]3 years ago

7 0

Temperature and pressure from lightening

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How are electromagnetic waves produced? A) An accelerating electric point charge (electron) emits electromagnetic waves. B) A ba

shusha [124]

Hello,

Here is your answer:

The proper answer to this question will be option A "An accelerating electric point charge (electron) emits electromagnetic waves." That's because electromagnetic waves are formed by motion of electrically charged particles (electrons)!

Your answer is A!

If you need anymore help feel free to ask me!

Hope this helps!

8 0

3 years ago

Read 2 more answers

What isthe correct 1/4=20%

anzhelika [568]

Answer:

Explanation:

5 0

3 years ago

An object is projected at 25m/s from the top of a building of height 50m. At the same instant,another object is projected from t

docker41 [41]

A) The objects have the same vertical position after 2 seconds

B) The objects have same vertical position at y = 30.4 m (but they do not collide since they have different x-position)

Explanation:

The motion of the first object along the vertical direction is a uniformly accelerated motion, so we can write its position at time t using the following equation:

$y_1(t)=h+u_1 t + \frac{1}{2}gt^2$

where:

h = 50 m is the initial height

$u_1=0$ is the initial vertical velocity (the object is projected horizontally, so the vertical velocity is zero at the beginning)

$g=-9.8 m/s^2$ is the acceleration of gravity

So, its vertical position can be rewritten as

$y_1(t)=50-4.9t^2$

The position of object 2 instead can be written as

$y_2(t)=(u_2 sin \theta)t + \frac{1}{2}gt^2$

where

$u_2 sin \theta$ is the initial vertical velocity, where

$u_2 = 50 m/s$ is the initial velocity

$\theta=30^{\circ}$ is the angle of projection

Substituting, we get:

$y_2(t)=(50)(sin 30^{\circ})t+\frac{1}{2}(-9.8)t^2=25t-4.9t^2$

The two objects collide when their vertical position is the same, so:

$y_1(t)=y_2(t)\\50-4.9t^2 = 25t-4.9t^2$

And solving for t, we find:

$50=25t\\t= 2 s$

Note that this means that the two object at t = 2 s have the same vertical position: however, this is not true for the horizontal position.

In order to find the point where they collide, we have to substitute the time of the collision that we found in part A into one of the expressions of the vertical position.

Substituting into the expression of object 2, we find:

$y_2(t) = 25t-4.9t^2=25(2.0)-4.9(2.0)^2=30.4 m$

We can verify that at the same time, the vertical position of object 1 is the same:

$y_1(t)=50-4.9t^2=50-4.9(2.0)^2=30.4 m$

This means that the two objects have the same vertical position at 30.4 m.

However, in reality, the two objects do not collide. In fact, object 1 is moving in the horizontal direction with constant velocity

$v_{1x}=25 m/s$

So its horizontal position at t = 2.0 s is

$x_1(2.0)=v_{1x}t=(25)(2.0)=50 m$

While object 2 is moving in the horizontal plane with velocity

$v_{2x}=u_2 cos \theta=(50)(cos 30^{\circ})=43.3 m/s$

So its horizontal position at t = 2.0 s is

$x_2(2.0)=v_{2x}t=(43.3)(2.0)=86.6 m$

So in reality, the two objects do not collide, if they start from the same x-position.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

7 0

3 years ago

The major problem with hydrogen as an alternative source of energy is that none exists on the surface of the earth. True or fals

Brut [27]

That statement is as false as false can be.

Roughly 78% of the Earth's surface is covered by water,
and Hydrogen is a major component of water.

On the surface of the Earth, you really have to try hard
to be far away from a cheap source of Hydrogen.

8 0

3 years ago

To receive AM radio, you want an RLC circuit that can be made to resonate at any frequency between 500 and 1650 kHz. This is acc

kotykmax [81]

In RLC circuit the frequency of the circuit is also known as resonance frequency of the circuit

It is calculated by

$f = \frac{1}{2\pi}\sqrt{\frac{1}{LC}}$