Ordinary ocean waves are created by the interaction of the ____ and ___

A localized impediment to electron flow in a circuit is a:

hichkok12 [17]

This would be the definition of a resistor. These components inhibit or “resist” the flow of a current.

Hope this helps!

8 0

4 years ago

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Use the data provided to calculate the gravitational potential energy of each cylinder mass. Round your answers to the nearest t

pantera1 [17]

Answer: 3kg: 14.7 6kg: 29.4 9kg: 44.1

Explanation: just did it on Edge

7 0

3 years ago

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A train of mass 5.6 × 10^5kg is at rest in a station.at time t=0s, a resultant force acts on the train and it starts to accelera

lana66690 [7]

Answer:

420000N

Explanation:

Given parameters:

Mass of the train = 5.6 x 10⁵kg

Acceleration = 0.75m/s²

Unknown:

Resultant force = ?

Solution:

According to newton's second law, force is the product of mass and acceleration;

Force = mass x acceleration

Resultant force that acts on the train is given below;

Force = 5.6 x 10⁵kg x 0.75m/s² = 420000N

7 0

3 years ago

A +8.75 μC point charge is glued down on a horizontal frictionless table. It is tied to a -6.50 μC point charge by a light, nonc

lisov135 [29]

(a) The tension on the wire when the two charges have opposite signs is 383.5 N.

(b) The tension on the wire if both charges were negative is 3.640.25 N.

The given parameters;

first charge, q₁ = 8.75 μC
second charge, q₂ = -6.5 μC
electric field, E = 1.85 x 10⁸ N/C
distance between the two charges, r = 2.5 cm

(a)

The attractive force between the charges is calculated as follows;

$F_1 = \frac{kq_1q_2}{r^2} \\\\F_1 = \frac{(9\times 10^9) \times (8.75\times 10^{-6})\times (-6.5\times 10^{-6})}{(0.025)^2} \\\\F_1 = -819 \ N$

The force on the negative charge due to the electric field is calculated as follows;

$F_2 = Eq_2\\\\F_2 = (1.85 \times 10^8) \times (6.5 \times 10^{-6})\\\\F_2 = 1202.5 \ N$

The tension on the wire is the resultant of the two forces and it is calculated as follows;

$T = F_2 + F_1\\\\T = 1202.5 - 819\\\\T = 383.5 \ N$

(b) when the two charges are negative

The repulsive force between the two charges is calculated as follows;

$F_1 = \frac{kq_1q_2}{r^2} \\\\F_1 = \frac{(9\times 10^9) \times (-8.75\times 10^{-6})\times (-6.5\times 10^{-6})}{(0.025)^2} \\\\F_1 = 819 \ N$

The force on the first negative charge due to the electric field is calculated as follows;

$F_2 = Eq_1\\\\F_2 = (1.85 \times 10^8)\times (8.75 \times 10^{-6})\\\\F_2 = 1618.75 \ N$

The force on the second negative charge due to the electric field is calculated as follows;

$F_3 = Eq_2\\\\F_3 = (1.85 \times 10^8) \times (6.5 \times 10^{-6})\\\\F_3 = 1202.5 \ N$

The tension on the wire is the resultant of the three forces and it is calculated as follows;

$T= F_1 + F_2 + F_3\\\\T= 819 + 1618.75 + 1202.5\\\\T = 3,640.25 \ N$

Learn more here:brainly.com/question/19565286

5 0

3 years ago

A microphone receiving a pure sound tone feeds an oscilloscope, producing a wave on its screen. If the sound intensity is origin

Temka [501]

Answer:

The new intensity = 3.38 × 10⁻⁵ W/m²

Explanation:

Intensity of sound wave:

The intensity of sound is the rate of flow of flow of energy, per unit area, perpendicular to the direction of the sound wave.

Intensity (I) ∝ A²

where I = intensity, A = Amplitude.

∴ I₁/I₂ = A₁²/A²₂............................... equation 1

From the question, the amplitude increase by 30% of the initial

∴ A₂ = A₁ + 0.3A₁ = 1.3A₁, I₁ = 2.00×10⁻⁵ W/m²

∴ (2.00×10⁻⁵)/I₂ = A₁²/(1.3A₁)²

(2.00×10⁻⁵)/I₂ = 1/1.3²

making I₂ the subject of the equation

I₂ = (2.00 × 10⁻⁵)×1.3² = 3.38 × 10⁻⁵ W/m₂

The new intensity = 3.38 × 10⁻⁵ W/m²

7 0

4 years ago

Ordinary ocean waves are created by the interaction of the ____ and ____.

Ordinary ocean waves are created by the interaction of the and .