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3 years ago

Electricity does not commonly occur in nature except in the form of ( hydrocarbons, photosyntheis, lightning, thunder)

Physics

2 answers:

garik1379 [7]3 years ago

4 0

Lightning i would think

yaroslaw [1]3 years ago

4 0

The correct answer to the question is : Lightning

EXPLANATION:

Hydrocarbons are the chemical compounds of carbon and hydrogen with their derivatives which are neutral in nature. Hence, they constitute no electric current.

Photosynthesis is the process of production of food by plants. It does not constitute any electric current in nature.

Thunder is created due to the collision between clouds. It can exert only sound pressure.

Lightning is created due to the same process as that of thunder.But it contains heavy electric current which is dangerous for living organisms.

Hence, the correct answer is lightning.

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What is the average speed of an object travels 60 meters in 3 seconds?

Solnce55 [7]

Answer:

20 metres per second

Explanation:

60/3

7 0

3 years ago

What must be the magnitude of a uniform electric field if it is to have the same energy density as that possessed by a 0.50 T ma

Sindrei [870]

Answer:

E = 1.50 × $10^{8}$ V/m

Explanation:

given data

B = 0.50 T

solution

we know that energy density by the magnetic field is express as

$\mu _b = \frac{B}{2\mu _o}$ ...............1

and

energy density due to electric filed is

$\mu _e = \frac{\epsilon _o E^2}{2}$ ...............2

and here $\mu _b = \mu _ e$

so that

E = $\frac{B}{\sqrt{\mu _o \times \epsilon _o}}$ ...................3

put here value and we get

$E = \frac{0.50}{\sqrt{4\pi \times 10^{-7} \times 8.852 \times 10^{-12}}}$

E = 3 × $10^{8}$ × 0.50

E = 1.50 × $10^{8}$ V/m

6 0

3 years ago

An earthquake produces longitudinal P waves that travel outward at 8000 m/s and transverse S waves that move at 4500 m/s. A seis

vivado [14]

Answer:

1234285.7 m or 1234.3 km

Explanation:

Let the distance be $d$ , the time taken by P waves be $t_P$ and the time taken by the S waves be $t_S$ .

$\text{Velocity}\dfrac{\text{Distance}}{\text{Time}}$

$\text{Time}\dfrac{\text{Distance}}{\text{Velocity}}$

For the P waves,

$t_P=\dfrac{d}{8000}$

$d=8000t_P$

For the S waves,

$t_S=\dfrac{d}{4500}$

$d=4500t_S$

Equating the $d$ ,

$8000t_P=4500t_S$

Divide both sides of the equation by 500 to reduce the terms.

$16t_P=9t_S$

Since S waves arrive 2 minutes (= 120 seconds) after P waves,

$t_S-t_P=120$

$t_S=120+t_P$

Substitute this in the equation of the distance.

$16t_P=9(t_P+120)$

$16t_P=9t_P+1080$

$7t_P=1080$

$t_P=\dfrac{1080}{7}$

Substitute this in the equation for $d$ involving $t_P$ .

$d=8000t_P$

$d=8000\times\dfrac{1080}{7}$

$d=1234285.7 \text{ m }= 1234.3 \text{ km}$

4 0

3 years ago

a railroad tie weights 920 N and is 2.6 m long. How much force is required to: pick it up off the ground? lift one end and rotat

lukranit [14]

1) The minimum force needed is 920 N

2) The minimum force is 460 N

3) The minimum force is 598 N

Explanation:

We can answer this part by simply looking at the forces involved. In fact, there are two forces acting on the railroad:

Its weight, W, acting downward
The force applied to lift it, F, upward

So the net force on the railroad is

$\sum F = F - W$

where

W = 920 N is the weight of the railroad

In order to lift the railroad, the net force must be upward, so

$\sum F \geq 0$

And therefore

$F\geq W$

which means that the minimum force needed is equal to the weight of the railroad, 920 N.

In this case, we have to use the principle of equilibrium of moments.

In fact, when the railroad rotates uniformly (=constant angular speed) about its end, it means that the moment produced by the weight (acting in one direction) is equal to the moment produced by the force applied (acting in the other direction). Therefore, we can write:

$W \frac{L}{2} = F L$

where

W = 920 N is the weight

L = 2.6 m is the length of the railroad

F is the force applied

We wrote L/2 on the left of the equation because the weight acts at the center of mass of the railroad (located at the midpoint), while on the right it is L because the force F is applied at the end of the railroad.

Solving for F,

$F=\frac{W}{2}=\frac{920}{2}=460 N$

This problem is similar to the previous part, however in this case, the force applied F is applied 0.6 m from the end, pivoting around the opposite end.

This means that the distance between the point of application of the force F and the pivot is

L' = L - 0.6

where

L = 2.6 m

Therefore the equation for the equilibrium of moments becomes

$W\frac{L}{2}=F(L-0.6)$