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3 years ago

A toy car is tied to a string and pulled across a table horizontally. Which is the

Physics

2 answers:

Arturiano [62]3 years ago

7 0

Answer:

Correct answer is option D

Explanation:

In this question we have given that toy car is tied to a string and pulled across the table

Therefore there must be three forces which will act on toy car, these are as follows

1. Force due to gravity

2. Normal reaction force

3. Tension in string

Force due to gravity $F_{g}$ will act in downward direction therefore to balance it there must be a normal force which will act in upward direction $F_{norm}$ since the toy car is not accelerating in the y direction.

Since the car is accelerating in horizontal direction therefore tension force will act in the horizontal direction therefore, $F_{t}$ will act to the right.

DerKrebs [107]3 years ago

4 0

Answer:

Explanation:

With $F_{g}$ acting as a downward force, there must be a normal force counteracting it, because the toy car is not accelerating in the y direction. This force is the Normal Force or $F_{norm}$ .

The Tension force must be going in the horizontal direction because the toy car is being pulled horizontally, so $F_{t}$ mist be going to the right.

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Now let’s apply Coulomb’s law and the superposition principle to calculate the force on a point charge due to the presence of ot

Mnenie [13.5K]

Answer:

F = - 1.68 10⁻⁴ N

, it is directed to the left of the x-axis

Explanation:

Coulomb's law is

F = k q₁ q₂ / r²

Where K is the Coulomb constant that value 8.99 10⁹ N m²/ C², q are the electric charges and r is the distance between them. Let's apply to our problem for each pair of charges

Let's reduce the magnitudes to the SI system

q₁ = 3.0 nc (1C / 10 9 nC) = 3.0 10⁻⁹ C

x₁ = 2.0 cm (1m / 100cm) = 2.0 10⁻² m

q₂ = -6.0 nC = -6.0 10⁻⁹ C

x₂ = 4.0 cm = 4.0 10⁻² m

q₃ = 5.0 nC = 5.0 10⁻⁹ C

x3 = 0 m

Charges q1 and q3

r = x₁ -x₃

r = 2.0 10⁻² -0

r = 2.0 10⁻² m

F₁₃ = 8.99 10⁹ 3.0 10⁻⁹ 5.0 10⁻⁹ / (2.0 10⁻²)²

F₁₃ = 33.7 10⁻⁵ N

As the charges are of the same sign, the force is repulsive, therefore it is directed to the left of the x-axis

Charges q2 and q3

r = r₂ –r₃

r = 4.0 10⁻² - 0 = 4.0 10⁻² m

F₂₃ = 8.99 10⁹ 6.0 10⁻⁹ 5.0 10⁻⁹ / (4.0 10⁻²)²

F₂₃ = 16.86 10⁻⁵ N

As the charges are of different sign, the force is attractive, therefore it is directed to the right of the x-axis

The force is a vector magnitude, so each component must be added independently, in this case all the forces are on the x-axis, let's take the right direction as positive

F = F₂₃ - F₁₃

F = 16.86 10⁻⁵ - 33.7 10⁻⁵

F = - 16.84 10⁻⁵ N

F = - 1.68 10⁻⁴ N

The negative sign means that it is directed to the left of the x-axis

5 0

4 years ago

A 20.0 μf capacitor is charged to a potential difference of 850 v. the terminals of the charged capacitor are then connected to

liberstina [14]

(a) $Q = 1.70\times 10^{-2}\;\text{C}$ ;
(b) $V_\text{final} = 5.31\times 10^{2}\;\text{V}$ ;
(c) $E_\text{final} = 4.52\;\text{J}$ ;
(d) $\Delta E = 2.82\;\text{J}$ .

All four values are in 3 sig. fig.

<h3>Explanation</h3>

(a)

$Q = C\cdot V = 20.0\times 10^{-6} \times 850\;\text{V} = 1.70\times 10^{-2}\;\text{J}$ .

(b)

Sum of the final charge on the two capacitors should be the same as the sum of the initial charge. Voltage of the two capacitors should be the same. That is:

$C_1\cdot V_\text{final} +C_2 \cdot V_\text{final} = C_1\cdot V_\text{initial}$ ;

$(C_1+C_2)\cdot V_\text{final} = C_1\cdot V_\text{initial}$ ;

$\displaystyle V_\text{final} = \frac{C_1}{C_1+C_2}\cdot V_\text{initial}\\\phantom{V_\text{final}} = \frac{20.0\;\mu\text{F}}{20.0\;\mu\text{F} + 12.0\;\mu\text{F}} \times 850\;\text{V}\\\phantom{V_\text{final}} =531\;\text{V}$ .

(c)

$\displaystyle E = \frac{1}{2}\cdot C\cdot V^{2}$ .

$\displaystyle E_\text{final} = \frac{1}{2} (C_1 + C_2) \cdot {V_\text{final}}^{2} \\\phantom{E_\text{final}} = \frac{1}{2} \times (20.0\times 10^{-6} + 12.0\times 10^{-6}) \times 531.25\\\phantom{E_\text{final}} = 4.52\;\text{J}$ .