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2 years ago

I dont know this.......

Mathematics

2 answers:

Rashid [163]2 years ago

7 0

Answer: 16

Step-by-step explanation:

Butoxors [25]2 years ago

5 0

Answer:

your ans wer is above

have a nice time

with explained

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1. Two object accumulated a charge of

In-s [12.5K]

Answer:

The magnitude of the electric force between these two objects

will be: 181.274 N.

i.e. F $=181.274$ N

Step-by-step explanation:

Two object accumulated a charge of 4.5 μC and another a charge of 2.8 μC.

q₁ = 4.5 μC = 4.5 × 10⁻⁶ C

q₂ = 2.8 μC = 2.8 × 10⁻⁶ C

separated distance = d = 2.5 cm

Calculating the magnitude of the force between two charged objects using the formula:

$F = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r^2}$

$=\:\frac{4.5\times \:\:10^{-6}\:\times \:\:2.8\:\times \:\:10^{-6}}{4\:\times \:\left(3.14\right)\:\times \:\left(8.85\times \:10^{-12}\right)\times \left(2.5\times \:\:10^{-2}\right)^2}$

$=\frac{10^{-12}\times \:12.6}{10^{-12}\times \:4\times \:8.85\pi \left(10^{-2}\times \:2.5\right)^2}$ ∵ $4.5\times \:10^{-6}\times \:2.8\times \:10^{-6}=10^{-12}\times \:12.6$

$=\frac{10^{-12}\times \:12.6}{10^{-12}\times \:35.4\pi \left(10^{-2}\times \:2.5\right)^2}$ ∵ $\mathrm{Multiply\:the\:numbers:}\:4\times \:8.85=35.4$

$\mathrm{Cancel\:the\:common\:factor:}\:10^{-12}$

$=\frac{12.6}{35.4\pi \left(10^{-2}\times \:2.5\right)^2}$

$=\frac{12.6}{0.025^2\times \:35.4\pi }$ ∵ $\left(10^{-2}\times \:2.5\right)^2=0.025^2$

$=\frac{12.6}{0.022125\pi }$ ∵ $35.4\pi\times 0.025^2=0.022125\pi$

$=\frac{12.6}{0.06950 }$

F $=181.274$ N

Therefore, the magnitude of the electric force between these two objects will be: 181.274 N.

i.e. F $=181.274$ N

7 0

2 years ago

PLEASE HELP!! I do not understand this. And can you explain it for me when I use it again?

damaskus [11]

Let's start b writing down coordinates of all points:
A(0,0,0)
B(0,5,0)
C(3,5,0)
D(3,0,0)
E(3,0,4)
F(0,0,4)
G(0,5,4)
H(3,5,4)

a.) When we reflect over xz plane x and z coordinates stay same, y coordinate changes to same numerical value but opposite sign. Moving front-back is moving over x-axis, moving left-right is moving over y-axis, moving up-down is moving over z-axis.

A(0,0,0)
Reflecting
A(0,0,0)

B(0,5,0)
Reflecting
B(0,-5,0)

C(3,5,0)
Reflecting
C(3,-5,0)

D(3,0,0)
Reflecting
D(3,0,0)

b.)
A(0,0,0)
Moving
A(-2,-3,1)

B(0,-5,0)
Moving
B(-2,-8,1)

C(3,-5,0)
Moving
C(1,-8,1)

D(3,0,0)
Moving
D(1,-3,1)

7 0

2 years ago

Can you help me please?

fgiga [73]

Answer:

Step-by-step explanation:

all real numbers

3 0

3 years ago

Read 2 more answers

Given a quadrilateral with vertices ????(0, 5), ????(6, 5), ????(4, 0), and H(−2, 0):

tankabanditka [31]