Question

In 1933, a storm occurring in the Pacific Ocean moved with speeds reaching a maximum of 126 km/h. Suppose a storm is moving north at this speed. If a gull flies east through the storm with a speed of 40.0 km/h relative to the air, what's the velocity of the gull relative to Earth? a) [Velocity value] km/h b) [Velocity value] m/s c) [Velocity value] m/s^2 d) [Velocity value] km/h^2

Answer 1

The velocity of the gull relative to Earth when flying through a storm moving north at 126 km/h, and its own speed east at 40 km/h, is approximately 132.2 km/h or 36.7 m/s.

Step-by-step explanation:

The student's question revolves around the concept of relative velocity in Physics. To find the velocity of the gull relative to Earth, we need to use vector addition since the velocities are directed at right angles to each other. The storm moves north at 126 km/h, and the gull flies east at 40 km/h relative to the air. Applying the Pythagorean theorem to these perpendicular vectors, the magnitude of the gull's velocity relative to Earth is:

`\(v = √((126^2 + 40^2)) = √((15876 + 1600)) = √((17476))\) km/h`

`\(v \approx 132.2\) km/h`

To convert this velocity value to meters per second (m/s), we use the conversion factor:
`\(1 \text{ km/h} = (1)/(3.6) \text{ m/s}\).`

`\(v = 132.2 * (1)/(3.6) = 36.72 \text{ m/s} \approx 36.7 \text{ m/s}\)`

Note that velocity does not have a 'per second squared' or 'per hour squared' unit because these units are for acceleration, not velocity. Therefore, options c) and d) are not valid for a velocity value. The correct answers are the velocity of the gull relative to Earth: a) 132.2 km/h and b) 36.7 m/s.

Answer 2

In 1933, a storm occurring in the Pacific Ocean moved with speeds reaching a maximum of 126 km/h, the velocity of the gull relative to Earth is approximately 132 km/h. The correct answer is a) [Velocity value] km/h

Step-by-step explanation:

To find the velocity of the gull relative to Earth, we need to use vector addition.

Since the storm is moving north at 126 km/h and the gull is flying east at 40.0 km/h relative to the air, we can use the Pythagorean theorem to find the magnitude of the gull's velocity relative to Earth.

The magnitude of the gull's velocity relative to the Earth is given by:

|Vgull| = √((Vstorm)² + (Vair)²)

Plugging in the values, we get:

|Vgull| = √((126 km/h)^2 + (40.0 km/h)^2) = √(15876 + 1600) = √17476

Therefore, the velocity of the gull relative to Earth is approximately 132 km/h.

The correct answer is a) [Velocity value] km/h