Una carga q1 = - 45 µC esta colocada a 30 mm a la izquierda de una carga q2 = 25 µC . ¿Cuál es la fuerza resultante sobre una carga de q3 = 20 µC localizada exactamente 50 mm arriba de la carga de 25µC ?

Answers

Answer 1

Answer:

La fuerza resultante sobre q₃ es  -1.2245 × 10⁻¹⁵ i  + -0.24 × 10⁻¹⁵ j

La magnitud de la fuerza resultante sobre q₃ es aproximadamente 1.25 × 10⁻¹⁵ N

Explanation:

q₁ = -45 μC = -45 × 10⁻⁶ C

r₁₂ = 30 mm = 30 × 10⁻³ m

q₂ = 25 μC = 25 × 10⁻⁶ C

r₂₃ = 50 mm = 50 × 10⁻³ m

q₃ = 20 μC = 20 × 10⁻⁶ C

k = 9×10⁻⁹ N·m²/C²

Por lo tanto;

r₁₃ = √(50² + 30²) = 10·√(34)

F₁₂ = 9×10⁻⁹ × (-45 × 10⁻⁶)×(25 × 10⁻⁶)/(30 × 10⁻³)² = -1.125 × 10⁻¹⁴

F₁₂ = -1.125 × 10⁻¹⁴ N

F₂₃ = 9×10⁻⁹ × (20 × 10⁻⁶)×(25 × 10⁻⁶)/(50 × 10⁻³)² = 1.8 × 10⁻¹⁵ j

F₁₃ = 9×10⁻⁹ × (-45 × 10⁻⁶)×(20 × 10⁻⁶)/(10·√34 × 10⁻³)² = -2.38× 10⁻¹⁵

Los componentes de F₁₃ son;

-2,38 × 10⁻¹⁵ × cos (arctan (30/50)) = -2,04 × 10⁻¹⁵ j

-2,38 × 10⁻¹⁵ × sin (arctan (30/50)) = -1,2245 × 10⁻¹⁵ i

La fuerza resultante sobre la carga q₃, [tex]\left | \underset {F_3} \rightarrow \right |[/tex] = [tex]\underset{F_{13}}{\rightarrow}[/tex] + [tex]\underset{F_{23}}{\rightarrow}[/tex]

∴ [tex]\left | \underset {F_3} \rightarrow \right |[/tex] = 1.8 × 10⁻¹⁵ j + -1.2245 × 10⁻¹⁵ i + -2.04 × 10⁻¹⁵ j

La fuerza resultante sobre q₃ es [tex]\left | \underset {F_3} \rightarrow \right |[/tex]  = -1.2245 × 10⁻¹⁵ i  + -0.24 × 10⁻¹⁵ j

La magnitud de la fuerza resultante sobre q₃,

[tex]\left | F_3 \right |[/tex] = √((-1.2245 × 10⁻¹⁵)² + (-0.24 × 10⁻¹⁵)²) ≈ 1.25 × 10⁻¹⁵

La magnitud de la fuerza resultante sobre q₃, [tex]\left | F_3 \right |[/tex] ≈ 1.25 × 10⁻¹⁵ N.


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Answer:

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Answers

Answer:

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Answers

Answer:

Explanation:

We first need to convert the 40 km/h to m/s. Going by the fact that 40 has only 1 significant figure in it, 40 km/h = 10 m/s. The rest of the values are in their proper labels. We will use the equation:

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They have valence electrons in the same energy level.

Answers

Answer:

Elements in the same group have the same number of valence electrons.

When moving right across a period, the valence electrons of the main group elements increase by one.

When moving down a group, the valence electrons of the main group elements increase by one.

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Elements in the same period of the periodic table have valence electrons in the same energy levels.

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Why does a ship float on water when it weighs about 100tones and why a stone of less than a kg sinks

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Answer with Explanation:

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Answers

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Answer: i think c

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Answers

Answer:

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Explanation:

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Answers

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Answer:

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Answers

i believe it would be Star A since Point A is a dwarf star and dwarf stars are usually most likely to have red coloring

Answer:

Star A

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Literally the question has provided the answer itself

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To what height, h, would the pendulum bob rise after a single swing if it was being released from the height of 0.80 m

Answers

Answer:

0.80 m

Explanation:

Neglecting friction, the total mechanical energy of the pendulum is constant.

E = K + U where K = kinetic energy and U = potential energy.

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Answers

Answer:

But negative acceleration means that the rate of change of velocity is negative or velocity decreases. Example: (1) When we apply brakes in a moving car, then negative acceleration acts on it and the car stops. (2) When we throw a ball upwards, then also negative acceleration acts on it.

Explanation:

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MATHEMATICALLY A NEGATIVE ACCELERATION MEANS YOU WILL SUBTRACT FROM THE CURRENT VALUE OF THE VELOCITY.

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Answer:

D. water

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Answer:

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Answer:

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what weight is recorded by a scale when it is placed inside a lift which is in free fall? Enplain.​

Answers

There is no pressure of your feet on the scales, and no pressure of the floor on the scales, so the scales will read zero I hope this is right

Answer:

Explanation:

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If a lever lifts a load four times the effort applied and effort distance is 5 times the load distance, calculate its efficiency​

Answers

Answer:

If effort distance was 4 times, efficiency would be 100%.

Since it takes 5 times for effort distance, efficiency drops to output/input

output is 1*F

input is (1/4*F)*5

so: F/1/5*F/4 = 4F/5F = .8 or 80%

The efficiency of the lever is 80%.

To calculate the efficiency of the lever, we can use the formula for mechanical advantage and efficiency.

Mechanical Advantage (MA) is the ratio of the load (L) to the effort (E) in a lever system:

MA = Load / Effort

Given that the load is four times the effort applied:

Load = 4 * Effort

Also, the effort distance (dEffort) is five times the load distance (dLoad):

dEffort = 5 * dLoad

Now, we can write the formula for efficiency (η) of a lever system:

Efficiency (η) = (Mechanical Advantage / Ideal Mechanical Advantage) * 100%

The Ideal Mechanical Advantage (IMA) is the ratio of the effort distance to the load distance:

IMA = dEffort / dLoad

Substitute the given values into the IMA equation:

IMA = (5 * dLoad) / dLoad

IMA = 5

Now, we can calculate the Mechanical Advantage (MA) using the relationship between the load and effort:

MA = Load / Effort

MA = (4 * Effort) / Effort

MA = 4

Finally, we can calculate the efficiency (η):

Efficiency (η) = (Mechanical Advantage / Ideal Mechanical Advantage) * 100%

η = (4 / 5) * 100%

η = 0.8 * 100%

η = 80%

The efficiency of the lever is 80%.

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Which of the following is a vector quantity? i. Force ii. Velocity iii. Acceleration iv. All of these 5771​

Answers

All of these

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Definition

A vector quantity the physical quantity that has both direction as well as magnitude.

A scientist studies how air blowing on plants affects their growth. He uses fans to create different amounts of wind and measures the growth of the plants. What would make this experiment more repeatable?

Answers

Answer:D.Keeping track of the exact amount of wind on each plant

A 2.0kg object is dropped from a height of 30m.
After it drops for 2.0 seconds, what is its kinetic
energy and what is its potential energy?
(Assume no air resistance.)

Answers

Answer:

1) The kinetic energy of the object after it drops for 2.0 seconds is approximately 384.9 Joules

2) The potential energy of the object after it drops for 2.0 seconds is approximately 204 J

Explanation:

1) The given mass of the object, m = 2.0 kg

The height from which the object is dropped, h = 30 m

The kinetic energy of the object after it drops for 2.0 seconds = Required

Kinetic energy, K.E. = (1/2)·m·v²

Where;

v = The velocity of the object

The kinematic equation for finding the velocity of the object is presented as follows;

v = u + g·t

Where;

u = The initial velocity of the object = 0

g = The acceleration due to gravity of the object ≈ 9.81 m/s²

t = The time of motion of the object = 2.0 seconds

∴ The velocity after 2 seconds, v ≈ 0 + 9.81 m/s² × 2 s = 19.62 m/s

The kinetic energy, K.E. after 2 seconds as the object drops is given as follows;

[tex]K.E._{(after \ two \ seconds)}[/tex] = (1/2) × 2.0 kg × (19.62 m/s)² = 384.9444 J ≈ 384.9 J

2) The total energy, M.E. of the object at the top, h = 30 m, u = 0, is given as follows;

The total mechanical energy, M.E. = P.E. + K.E.

M.E. = m·g·h + (1/2)·m·u²

∴ M.E. = 2.0 kg × 9.81 m/s² × 30 m + 0 = 588.6 J

M.E. = 588.6 J

Given that the total mechanical energy, M.E., is constant, we have;

At 2.0 seconds, M.E. = 588.6 J , K.E. ≈ 384.9 J, P.E. = M.E. - K.E.

∴ P.E. = 588.9 J - 384.9 J ≈ 204 J

The potential energy after it drops 2.0 seconds, P.E. ≈ 204 J

An object is thrown with velocity 7.1 m s-1 vertically upwards on the Moon. The
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A 3.5 s
B 4.4 s
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Answers

Answer:D

Explanation:

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