How does a parallel circuit differ from a series circuit? A. A parallel circuit has one path for electrons, but a series circuit has more than one path. B. A series circuit has one path for electrons, but a parallel circuit has more than one path. C. A parallel circuit includes more than one resistor, but a series circuit has one resistor. D. A parallel circuit has a power source,

Answer:

420.17 nm

Explanation:

Given:

f = 7.14 × 10¹⁴Hz

c = 3 × 10⁸ m/s

c = fλ

λ = c/f

  = 420.17 nm

Answer:The phenomena when light passes through the object (a medium) is known as refraction. Refraction is defined as the bending of light ray when it passes from one medium to another.

Explanation:

Part A: The enmeshed cars were moving at a velocity of approximately 8.66 m/s just after the collision.

Part B: Car A was traveling at a velocity of approximately 8.55 m/s just before the collision.

How to compute the above velocities

To find the speed of car A just before the collision in Part B, you can use the principle of conservation of momentum.

The total momentum of the system before the collision should equal the total momentum after the collision. You already know the total momentum after the collision from Part A, and now you want to find the velocity of car A just before the collision.

Let's denote:

- v_A as the initial velocity of car A before the collision.

- v_B as the initial velocity of car B before the collision.

In Part A, you found that the enmeshed cars were moving at a velocity of 8.66 m/s at an angle of 60 degrees south of east. You can split this velocity into its eastward and southward components. The eastward component of this velocity is:

v_east = 8.66 m/s * cos(60 degrees)

Now, you can use the conservation of momentum to set up an equation:

Total initial momentum = Total final momentum

(mass_A * v_A) + (mass_B * v_B) = (mass_A + mass_B) * 8.66 m/s (the final velocity you found in Part A)

Plug in the known values:

(1900 kg * v_A) + (1500 kg * v_B) = (1900 kg + 1500 kg) * 8.66 m/s

Now, you can solve for v_A:

(1900 kg * v_A) + (1500 kg * v_B) = 3400 kg * 8.66 m/s

1900 kg * v_A = 3400 kg * 8.66 m/s - 1500 kg * v_B

v_A = (3400 kg * 8.66 m/s - 1500 kg * v_B) / 1900 kg

Now, plug in the values from Part A to find v_A:

v_A = (3400 kg * 8.66 m/s - 1500 kg * 8.66 m/s) / 1900 kg

v_A = (29244 kg*m/s - 12990 kg*m/s) / 1900 kg

v_A = 16254 kg*m/s / 1900 kg

v_A ≈ 8.55 m/s

So, car A was going at approximately 8.55 m/s just before the collision in Part B.

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A positive charge is placed in an electric field that points west. the direction of the force on the positive particle point toward the west.

What is an electric charge?

Due to the physical characteristic of electric charge, charged material experiences a force when it is exposed to an electromagnetic field. Electric charges can be positive or negative (commonly carried by protons and electrons respectively). While like charges repel one another, opposite charges attract. If an object has no net charge, we refer to it as neutral.

The charge on one electron is -1.6 ×10⁻¹⁹ coulomb.

As given in the problem statement when a positive charge is placed in an electric field that points west. the direction of the force on the positive particle will also point toward the west because for a positively charged

particle the direction of the electric field is the same as the force.

Thus, a positive charge is introduced into a westward-pointing electric field. the force acting on the positive particle points westward

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