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3 years ago

A positively charged glass rod is bought close to a suspended metal needle. What

Physics

2 answers:

Darina [25.2K]3 years ago

7 0

Answer:

attracted

Explanation:

opposite charges attract each other when the rub against each other

EastWind [94]3 years ago

6 0

Answer:

This depends because the electrostatic force obeys the principle that states that force between both of the particles does not get affected by the charges of the other thus if the needle is getting attracted it possess negative charges( the opposite charge) .And if they repel it means they have the same charges ( positive charges).

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ANEK [815]

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3 years ago

As the boat in which he is riding approaches a dock at 3.0 m/s, Jasper stands up in the boat and jumps toward the dock. Jasper a

Annette [7]

Im pretty sure it’s a because it makes more sense you know?.

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2 years ago

What is the theory of light?

Lostsunrise [7]

Answer:

Corpuscular theory of light

Explanation:

In optics, the corpuscular theory of light, arguably set forward by Descartes in 1637, states that light is made up of small discrete particles called "corpuscles" which travel in a straight line with a finite velocity and possess impetus. This was based on an alternate description of atomism of the time period.

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3 years ago

Michael Jordan, el célebre basquetbolista, ganó el torneo de clavadas de la NBA en 1988. Para lograr la hazaña saltó 1.35 metros

kozerog [31]

(a) 0.40 s

First of all, let's find the initial speed at which Jordan jumps from the ground.

The maximum height is h = 1.35 m. We can use the following equation:

$v^2-u^2=2gh$

where

v = 0 is the velocity at the maximum height

u is the initial velocity

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

Solving for u,

$u=\sqrt{-2gh}=\sqrt{-2(-9.8)(1.35)}=5.14 m/s$

The time needed to reach the maximum height can now be found by using the equation

$v=u+gt$

Solving for t,

$t=\frac{v-u}{g}=\frac{0-5.14}{-9.8}=0.52s$

Now we can find the velocity at which Jordan reaches a point 20 cm below the maximum height, so at a height of

h' = 1.35 - 0.20 = 1.15 m

Using again the equation

$v'^2-u^2=2gh'$

we find

$v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(1.15)}=1.97 m/s$

And the corresponding time is

$t'=\frac{v'-u}{g}=\frac{1.97-5.14}{-9.8}=0.32s$

So the time to go from h' to h is

$\Delta t = t-t'=0.52-0.32=0.20 s$

And since we have also to take into account the fall down (after Jordan reached the maximum height), which is symmetrical, we have to multiply this time by 2 to get the total time of permanence in the highest 20 cm of motion:

$\Delta t=2\cdot 0.20 = 0.40 s$