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2 years ago

How do we solve questions C and D? I already did A and B and I am confused on how to continue

Physics

1 answer:

aleksandrvk [35]2 years ago

8 0

(a) The work done in moving the unit charge from point C to A is 7.62 x 10⁻³ J.

(b) The work done in moving the unit charge from point D to B is 7.62 x 10⁻³ J.

<h3>Work done in moving the charge from C to A</h3>

W = Fd

W = Kq²/d

from 0 origin to C, d = √(5² + 5²) = 7.07 m
from 0 origin to A, d = 5 m

W(C to A) = W(0 to C) + W(0 to A)

$W(C \ to \ A) = - \frac{Kq^2}{7.07} + \frac{Kq^2}{5} \\\\ W(C \ to \ A) = 0.0586 \ Kq^2\\\\W(C \ to \ A) = 0.0586 \times 9 \times 10^9 \times (3.8\times 10^{-6})^2\\\\W(C \ to \ A) = 7.62 \times 10^{-3} \ J$

<h3>Work done in moving the charge from D to B</h3>

from 0 origin to D, d = √(5² + 5²) = 7.07 m
from 0 origin to B, d = 5 m

W(D to B) = W(0 to D) + W(0 to B)

W(D to B) = 7.62 x 10⁻³ J

Learn more about work done here: brainly.com/question/25573309

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Which of the following are decomposition reactions?

zhenek [66]

The 2 decomposition reactions are:

B. $K_{2}$ $CO_{3}$ (s) --> $K_{2} O$ (s) + $CO_{2}$ (g)

C. $2SO_{3}$ (g) --> $2SO_{2}$ (g) + $O_{2}$ (g)

Explanation:

Step 1:

A decomposition reaction is one in which a single compound breaks down into two or more products.

Step 2:

Let us examine each reaction if it satisfies the above rule.

A. 2 LiOH + $CO_{2}$ ---> $Li_{2} CO_{3}$ + $H_{2} O$

In this reaction we have 2 reactants giving 2 products. Hence this is not a decomposition reaction.

B. $K_{2} CO_{3}$ ---> $K_{2} O$ + $CO_{2}$

In this reaction a single reactant potassium carbonate gives rise to 2 products Potassium Oxide and carbon-di-oxide. Hence this is a decomposition reaction

C. $2SO_{3}$ ---> $2SO_{2}$ + $O_{2}$

In this reaction a single reactant sulfur tri-oxide gives rise to 2 products sulfur di-oxide and oxygen. Hence this is a decomposition reaction.

D. $2Na$ + $Cl_{2}$ ---> $2NaCl_{}$

In this reaction 2 reactants sodium and chlorine gives rise to a single product sodium chloride. Hence this is not a decomposition reaction.

Step 3:

Answer:

The 2 decomposition reactions are:

B. $K_{2} CO_{3}$ (s) ---> $K_{2}O$ (s) + $CO_{2}$ (g)

C. $2SO_{3}$ ---> $2SO_{2}$ + $O_{2}$

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3 years ago

98. In Fig. 24-71, a metal sphere

yarga [219]

Answer:

(a) The potential difference between the spheres is 750 kVA

(b) The charge on the smaller sphere is 6. $\overline 6$ μC

Explanation:

(a) The given parameters are;

The charge on the inner sphere, q = 5.00 μC

The radius of the inner sphere, r = 3.00 cm = 0.03 m

The charge on the larger sphere, Q = 15.0 μμC

The radius of the larger sphere, R = 6.00 cm = 0.06 m

The potential difference between two concentric spheres is given according to the following equation;

$V_r - V_R = k \times q \times \left ( \dfrac{1}{r} - \dfrac{1}{R} \right)$

Where;

R = The radius of the larger sphere = 0.06 m

r = The radius of the inner sphere = 0.03 m

q = The charge of the inner sphere = 5.00 × 10⁻⁶ C

Q = The charge of the outer sphere = 15.00 × 10⁻⁶ C

k = 9 × 10⁹ N·m²/C²

Therefore, by plugging in the value of the variables, we have;

$V_r - V_R = 9 \times 10^9 \times 5.00 \times 10^{-6} \times \left ( \dfrac{1}{0.03} - \dfrac{1}{0.06} \right) = 750,000$

The potential difference between the spheres, $V_r - V_R$ = 750,000 N·m/C = 750 kVA

(b) When the spheres are connected with a wire, the charge, 'q', on the smaller sphere will be added to the charge, 'Q', on the larger sphere which as follows;

$Q_f$ = Q + q = (5 + 15) × 10⁻⁶ C = 20 × 10⁻⁶ C

$Q_f$ = 20 × 10⁻⁶ C

From which we have;

Q₁/Q₂ = R/r

Where;

Q₁ = The new charge on the on the larger sphere

Q₂ = The new charge on the on the smaller sphere

$Q_f$ = 20 × 10⁻⁶ C = Q₁ + Q₂

∴ Q₁ = 20 × 10⁻⁶ C - Q₂ = 20 μC - Q₂

∴ (20 μC - Q₂)/Q₂ = 0.06/(0.03) = 2

20 μC - Q₂ = 2·Q₂

20 μC = 3·Q₂

Q₂ = 20 μC/3

The charge on the smaller sphere, Q₂ = 20 μC/3 = 6. $\overline 6$ μC

The charge on the smaller sphere, Q₁ = 40 μC/3 = 13. $\overline 3$ μC.

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3 years ago

A small bouncy ball with a momentum of 8 kg∙m/s to the left approaches head-on a large door at rest. The ball bounces straight b

vlada-n [284]

Answer:

-14 kg m/s

Explanation:

Taking the direction "to the left" as positive direction, the initial momentum of the ball is

p1 = +8 kg m/s

while the final momentum is

p2 = -6 kg m/s

so the change in momentum is

$\Delta p = p_2 - p_1 = -6kg m/s - (8 kg m/s) = -14 kg m/s$

According to the impulse theorem, the impulse exerted on the ball is equal to the change in momentum of the ball, so:

$I_1 = -14 kg m/s$ (which means 14 kg m/s to the right)

While the impulse that the ball exerted on the ball is equal and opposite in direction, so:

$I_2 = + 14 kg m/s$ (which means towards the left)

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4 years ago

The law of conservation of matter states that during a chemical reaction, the amount of matter

mixas84 [53]

Answer:

The law of conservation of matter says that in chemical reactions, the total mass of the products must equal the total mass of the reactants.

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3 years ago

Car 1 of mass 1,715kg is sitting at an intersection. Car 2 of mass 2,780kg read ends Car 1, knocking it forward at 8 m/s. If Car

SVEN [57.7K]

Answer:8 miles

Explanation:

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3 years ago