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Flura [38]
3 years ago
10

A charge of -5.02 nC is uniformly distributed on a thin square sheet of nonconducting material of edge length 21.8 cm. "What is

the surface charge density of the sheet"?
Physics
1 answer:
Maru [420]3 years ago
3 0

Answer:

A charge of -5.02 nC is uniformly distributed on a thin square sheet of nonconducting material of edge length 21.8 cm. "What is the surface charge density of the sheet"?

Explanation:

Surface charge density is a measure of how much electric charge is accumulated over a surface. It can be calculated as the charge per unit area.

We will convert all parameters in SI units.

Charge = Q = -5.02nC

Q  = -5.02×10^{-9}C

As it is clear from question that Sheet is a square (All sides will be of equal length)

Area = A = (21.8×10^{-2}m) (21.8×10^{-2}m)  = 4.75×10^{-4}m²

A  = 4.75×10^{-4}m²

Surface charge density = Q/A

Surface charge density = (-5.02×10^{-9}C)/(4.75×10^{-4}m²)

Surface charge density = -1.057×10^{-5} Cm^{-2}

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Tom [10]
Well, its in the air, so the air is "upon" the ball. and when it comes down...you catch it, and throw it, and get someone out, and win the game, and just keep doing that, and boooommm you're and pro baseball player. Life is good
8 0
3 years ago
If a force is 100 N and is pointing 37 degrees north of east. (a) Draw a diagram of this force. (b) Draw the force's x and y com
sasho [114]

Answer

given,

force = 100 N

Point 37 degrees north of east

a) and b) part is shown in the diagram attached below.

c) to find the x and y component of the force

x- component of the force

F_x = F cos \theta

F_x = 100\times cos 37^0

F_x = 79.86 N

y- component of the force

F_y = F sin \theta

F_y = 100\times sin 37^0

F_y = 60.18 N

3 0
3 years ago
What will happen if a car experiences a 300 N force to the right from the engine and a separate 150 N force due to friction and
gayaneshka [121]

Force applied on the car due to engine is given as

F_1 = 300 N towards right

Also there is a force on the car towards left due to air drag

F_2 = 150 N towards left

now the net force on the car will be given as

\vec F_{net} = \vec F_1 + \vec F_2

now we can say that since the two forces are here opposite in direction so here the vector sum of two forces will be the algebraic difference of the two forces.

So we can say

F_{net} = F_1 - F_2

F_{net} = 300 - 150

F_[net} = 150 N

So here net force on the car will be 150 N towards right and hence it will accelerate due to same force.

5 0
3 years ago
What net force would be necessary to cause a block of wood with a mass of 2.5 kg to accelerate at a rate of 3.0 m/s2
charle [14.2K]

Answer:

<h2>7.5 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question

mass = 2.5 kg

acceleration = 3.0 m/s²

We have

force = 2.5 × 3.0 = 7.5

We have the final answer as

<h3>7.5 N</h3>

Hope this helps you

3 0
2 years ago
A body oscillates with simple harmonic motion along the x-axis. Its displacement varies with time according to the equation x =
abruzzese [7]
I'll tell you how I look at this, although I may be missing something important.

Position = x(t) = 0.5 sin(pt + p/3)

Speed = position' = x'(t) = 0.5 p cos(pt + p/3)

Acceleration = speed' = position ' ' = x ' '(t) = -0.5 p² sin(pt + p/3)

At (t = 1.0),

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In order to evaluate this, don't I still have to know what 'p' is ? ?

I don't think it can be evaluated with the information given in the question.
6 0
3 years ago
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