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DIA
13 days ago
11

Which of the following are inertial reference frames? A. A car driving at steady speed on a straight and level road. B. A car dr

iving at steady speed up a 10∘ incline. C. A car speeding up after leaving a stop sign. D. A car driving at steady speed around a curve.

Physics
2 answers:
Softa [3K]13 days ago
8 0
A. A car moving at a constant speed on a flat, straight road. B. A vehicle traveling at a steady speed on a 10-degree incline. An object operates within an inertial reference frame if there is no net force acting upon it. According to Newton's second law, this implies that the object's acceleration also equals zero. Assessing the scenarios yields: A. A car moving at a constant speed on a flat road qualifies as an inertial reference frame, since its velocity and direction remain unchanged; thus, acceleration is zero. B. A car moving steadily up a 10-degree incline still constitutes an inertial reference frame, for similar reasons. C. A car accelerating after departing a stop sign does not represent an inertial frame due to its change in speed. D. A car driving at a steady speed around a curve cannot be considered an inertial reference frame since its direction is changing, resulting in a change in velocity and thus acceleration. Therefore, options A and B are correct.
Keith_Richards [3.2K]13 days ago
3 0
The correct choices are B.
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Answer:

The work performed by air resistance totals -0.0782 J

Explanation:

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According to the principle of conservation of energy, the energy of a raindrop must remain constant.

At the outset, the raindrop possesses only gravitational potential energy:

PE = m · g · h

Where:

PE = potential energy.

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g = gravitational acceleration (9.8 m/s²)

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PE = 4 × 10⁻⁶ kg · 9.8 m/s² · 2000 m

PE = 0.0784 J

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initial PE = final KE + Work by air

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KE = kinetic energy.

Work by air = work done by air resistance.

The kinetic energy at ground level is computed as follows:

KE = 1/2 · m · v²

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KE = 1/2 · 4 × 10⁻⁶ kg · (10 m/s)²

KE = 2 × 10⁻⁴ J

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0.0784 J = 2 × 10⁻⁴ J + Work by air

Work by air = 0.0784 J - 2 × 10⁻⁴ J

Work by air = 0.0782 J

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