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1 year ago

Help me pls im sturgguleing in scince.

Physics

2 answers:

aliina [53]1 year ago

8 0

A gas plasma i just took the quiz lol

Lady bird [3.3K]1 year ago

5 0

Answer:

I would say A) a gas

Explanation:

Similar to gas plasma has no defiant shape or volume. Both plasma and gas particals are far apart. But gas is electrically neutral and plasma charged. Plasma is distinct from gas because it possess unique properties. Hope that helps

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If forces acting on an object are unbalanced, which factor may result from an unbalanced force? A.The net force is negative. B.T

bezimeni [28]

Answer:

<h2>A. The net force is negative.</h2>

Explanation:

just took the test :)

4 0

2 years ago

Read 2 more answers

Learning Goal: To understand the behavior ofthe electric field at the surface of a conductor, and itsrelationship to surface cha

Ivan

Complete Question

The complete question is shown on the first uploaded image

Answer:

a it is always zero

b 0

c $\eta = -\epsilon _o E$

Explanation:ss

Here the net charge is on the outer surface of the conductor thus this means that the net charge inside the conductor is zero

Generally the charge density of a conductor is dependent on the charge per unit area which implies that the charge density is dependent on the net charge so this means that the charge density inside the conductor is zero

Generally the direction of electric field this from the positive charge to the negative charge so from the question we can deduce that the negative charge is located on the surface of the conductor

So We can mathematically define the charge density on the surface of the electric field as

∮ $E \cdot dA = \frac{-Q}{\epsilon _o}$

Where E is the electric field

$dA$ change in unit area

$-Q$ is the negative charge

$\epsilon _o$ is the permittivity of free space

$EA = \frac{-Q}{\epsilon _o }$

$\frac{Q}{A} = -\epsilon _o E$

$\eta = -\epsilon _o E$

Where $\eta$ is the charge density

8 0

2 years ago

A 1-kg rock is suspended from the tip of a horizontal meterstick at the 0-cm mark so that the meterstick barely balances like a

tigry1 [53]

Explanation:

Given that,

Mass if the rock, m = 1 kg

It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.

We need to find the mass of the meter stick. The force acting by the stone is

F = 1 × 9.8 = 9.8 N

Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.

$9.8\times 12.5=W\times (50-12.5)\\\\W=\dfrac{9.8\times 12.5}{37.5}$