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4 years ago

An object is electrically charged if the amounts of positive and negative charge it contains are not _______.

Physics

2 answers:

Naddik [55]4 years ago

4 0

Equal bc it’s neutral

zlopas [31]4 years ago

3 0

Answer:oops

Explanation:

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Answer:

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3 years ago

2. A ladder 16cm long weighs 420N and it's centre of gravity is 7m from one end.It is carried horizontally by two boys,each hold

ipn [44]

Answer:

well the answers are 180 and 240 obviously

Explanation:

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8 0

2 years ago

Read 2 more answers

A teacher told a learner to react benzene (CH) with chlorine (Cl₂) to

kolezko [41]

The minimum quantity of benzene, C₆H₆ needed for the reaction is 106.67 g

<h3>How to determine the theoretical yield of chlorobenzene, C₆H₅Cl</h3>

From the question given, the following data were obtained

Actual yield = 100 g
Percentage yield = 65%
Theoretical yield =?

Percentage yield = (Actual / Theoretical) × 100

65% = 100 / Theoretical

0.65 = Actual / Theoretical

Cross multiply

0.65 × Theoretical = 100

Divide both sides by 0.65

Theoretical = 100 / 0.65

Theoretical yield = 153.85 g

<h3>How to determine the mass of benzene, C₆H₆ needed</h3>

Balanced equation

C₆H₆ + Cl₂ → C₆H₅Cl + HCl

Molar mass of C₆H₆ = 78 g/mol

Mass of C₆H₆ from the balanced equation = 1 × 78 = 78 g

Molar mass of C₆H₅Cl = 112.5 g

Mass of C₆H₅Cl from the balanced equation = 1 × 112.5 = 112.5 g

SUMMARY

From the balanced equation above,

112.5 g of C₆H₅Cl were obtained from 78 g of C₆H₆

Therefore,

153.85 g of C₆H₅Cl will be produced from = (153.85 × 78) / 112.5 = 106.67 g of C₆H₆

Thus, the minimum amount of benzene, C₆H₆ needed for the reaction is 106.67 g

Learn more about stoichiometry:

brainly.com/question/16735180

#SPJ1

3 0

2 years ago

A positively charged particle 1 is at the origin of a Cartesian coordinate system, and there are no other charged objects nearby

8090 [49]

Answer:

P=(2 nm, 8mn)

Explanation:

Given :

Position of positively charged particle at origin, $O=(0\ nm,0\ nm)$

Position of desired magnetic field, $D\equiv(1\ nm,8\ nm)$

Magnitude of desired magnetic field, $E=0\ N.C^{-1}$

Let q be the positive charge magnitude placed at origin.

<u>We know the distance between the two Cartesian points is given as:</u>

$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

<u>For the electric field effect to be zero at point D we need equal and opposite field at the point.</u>

$\frac{1}{4\pi.\epsilon_0} \times \frac{q}{r^2 } =\frac{1}{4\pi.\epsilon_0} \times \frac{q}{r^2 }$

$\therefore (1-0)^2+(8-0)^2=r^2$

$r^2=65\ nm$

$r=\sqrt{65}$

as we know that the electric field lines emerge radially outward of a positive charge so the second charge will be at equally opposite side of the given point.

assuming that the second charge is placed at (x,y) nano-meters.

Therefore,

$x=2\times 1=2\ nm$