A helicopter is accelerating up into the air while it is also accelerating due north. Which statement about the forces acting on the helicopter is true? Responses:

Question

The drag on the helicopter is directed upward and to the south.

The drag on the helicopter is directed upward and to the north.

The drag on the helicopter is directed downward and to the north.

The drag on the helicopter is directed downward and to the south.

Answer 1

When a helicopter is accelerating upward, it is experiencing a combination of forces: the lift force generated by its rotors, the gravitational force acting downward, and drag that opposes its motion through the air.

In this case, if the helicopter is also accelerating to the north, the drag force will be opposite to the direction of motion. Therefore, since the helicopter is moving north, the drag will act in the opposite direction, which is to the south.

The drag force is also generally directed opposite to the direction of travel (for an upward motion, it would typically be downward), but since we are assessing the directional effects due to the northward acceleration, we specifically focus on that.

Thus, the correct statement about the forces acting on the helicopter is:

The drag on the helicopter is directed downward and to the south.

This indicates that the drag opposes the northward acceleration while the helicopter is also experiencing a downward acceleration due to gravity.

Answer 2

Two galaxies on opposite ends of the universe are moving away from the Earth. Each has a velocity of 200,000 km/s relative to the Earth. How fast would an observer in one of those galaxies see the other galaxy moving away?
Responses

200,000 km/s

between 200,000 and 300,000 km/s

between 300,000 and 400,000 km/s

400,000 km/s

Answer 3

To determine the relative velocity of one galaxy as observed from another galaxy when both are moving away from Earth at high speeds, we must consider the effects of special relativity, particularly the relativistic velocity addition formula.

The formula for adding velocities in special relativity is given by:

\[ v_{rel} = \frac{v_1 + v_2}{1 + \frac{v_1 v_2}{c^2}} \]

where:

• \(v_{rel}\) is the relative velocity of one object as observed from another,
• \(v_1\) and \(v_2\) are the velocities of the two objects relative to a third observer (in this case, Earth),
• \(c\) is the speed of light, approximately \(300,000 \text{ km/s}\).

In your scenario:

• \(v_1 = 200,000 \text{ km/s}\)
• \(v_2 = 200,000 \text{ km/s}\)

Substituting these values into the formula, we have:

\[ v_{rel} = \frac{200,000 + 200,000}{1 + \frac{200,000 \times 200,000}{(300,000)^2}} \]

Calculating the denominator:

\[ \frac{200,000 \times 200,000}{(300,000)^2} = \frac{40,000,000,000}{90,000,000,000} \approx 0.4444 \]

So,

\[ 1 + 0.4444 \approx 1.4444 \]

Now substituting back into the equation for \(v_{rel}\):

\[ v_{rel} = \frac{400,000}{1.4444} \approx 276,000 \text{ km/s} \]

Thus, the observer in one galaxy would see the other galaxy moving away at approximately between 200,000 and 300,000 km/s.

Therefore, the correct response is:

between 200,000 and 300,000 km/s.

Answer 4

A student holds a water balloon outside of an open window and lets go. The window is 10 meters above the ground, and the balloon is falling under the acceleration of gravity, which is 9.8 m/s2. There are two equations that can be used to describe its motion over time:

x=x0+v0t+1/2at^2
v=v0+at

Would the balloon hit the ground before or after 1.0 s of falling? Which equation did you use to decide, and what comparison did you make to determine that it would or would not hit the ground by then?

Use 3–5 sentences to explain your reasoning.

Answer 5

To determine whether the water balloon would hit the ground before or after 1.0 second of falling, I can use the equation for position:

\[ x = x_0 + v_0t + \frac{1}{2}at^2 \]

In this scenario, the initial position \( x_0 \) is 10 meters (the height of the window), the initial velocity \( v_0 \) is 0 m/s (since the balloon is dropped), and the acceleration \( a \) is 9.8 m/s² (due to gravity). By substituting these values into the equation for \( t = 1.0 \) second, we can find the position of the balloon at that time:

\[ x = 10 + 0 \cdot 1 + \frac{1}{2} \cdot 9.8 \cdot (1)^2 = 10 + 4.9 = 14.9 \text{ meters} \]

Since the position of the balloon (14.9 meters) is above the ground level (0 meters), the balloon would not hit the ground before 1.0 second. To find the exact time it would take to hit the ground, I would set \( x = 0 \) and solve for \( t \). This clearly indicates it would take more than 1.0 second to reach the ground.

Answer 6

Many competitive swimmers practice in loose-fitting, baggy swimsuits called drag suits. In 1–2 sentences, explain how, in terms of the forces acting on a swimmer, wearing a drag suit differs from wearing a tight-fitting racing suit, and why practicing in a drag suit can make a swimmer stronger.
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Answer 7

Wearing a drag suit increases the drag force acting on a swimmer due to its larger surface area and looser fit, creating more water resistance compared to a tight-fitting racing suit, which minimizes drag and allows for smoother movement through the water. Practicing in a drag suit conditions swimmers to exert more force against this increased resistance, thereby building strength and improving their swimming technique for when they compete in less restrictive suits.