48, As preparation for this problem, review Example 13. Sup- pose that the pilot suddenly jettisons 2800 N of fuel. If the plane is to continue moving with the same velocity under the influence of the same air resistance R, by how much does the pilot have to reduce (a) the thrust and (b) the lift?

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48 please! Sorry I forgot to include info!

**Educational Content: Physics Problem Transcription**

---

**Problem 46:** 

A supersonic transport plane (mass = \(1.70 \times 10^6 \, \text{kg}\)) is moving with a constant velocity. It exerts a genuine forward thrust of \(7.40 \times 10^5 \, \text{N}\). Determine (a) the magnitude of the resistive force exerted on the tanker by the water and (b) the magnitude of the upward buoyant force exerted on the tanker by the water.

**Problem 47:**

A 12.0 kg loudspeaker is suspended from the ceiling by two vertical wires. What is the tension in each wire?

**Problem 48:**

As preparation for this problem, review Example 13. Suppose the pilot suddenly jettisons 2300 N of fuel. If the plane is to continue moving with the same velocity under the influence of the same air resistance \( R \), by how much does the pilot have to reduce (a) the thrust and (b) the lift?

**Problem 49:**

A 1.40 kg bottle of vintage wine is lying horizontally in the rack shown in the drawing. The two surfaces on which the bottle rests are both 90 degrees apart, and the right surface makes an angle of 45 degrees with respect to the vertical.

**Diagram Explanation:**

The diagram associated with Problem 49 illustrates a wine bottle lying horizontally on a rack, which consists of two surfaces meeting at a 90-degree angle. The right surface forms a 45-degree angle with the vertical, supporting the bottle in equilibrium.
Transcribed Image Text:**Educational Content: Physics Problem Transcription** --- **Problem 46:** A supersonic transport plane (mass = \(1.70 \times 10^6 \, \text{kg}\)) is moving with a constant velocity. It exerts a genuine forward thrust of \(7.40 \times 10^5 \, \text{N}\). Determine (a) the magnitude of the resistive force exerted on the tanker by the water and (b) the magnitude of the upward buoyant force exerted on the tanker by the water. **Problem 47:** A 12.0 kg loudspeaker is suspended from the ceiling by two vertical wires. What is the tension in each wire? **Problem 48:** As preparation for this problem, review Example 13. Suppose the pilot suddenly jettisons 2300 N of fuel. If the plane is to continue moving with the same velocity under the influence of the same air resistance \( R \), by how much does the pilot have to reduce (a) the thrust and (b) the lift? **Problem 49:** A 1.40 kg bottle of vintage wine is lying horizontally in the rack shown in the drawing. The two surfaces on which the bottle rests are both 90 degrees apart, and the right surface makes an angle of 45 degrees with respect to the vertical. **Diagram Explanation:** The diagram associated with Problem 49 illustrates a wine bottle lying horizontally on a rack, which consists of two surfaces meeting at a 90-degree angle. The right surface forms a 45-degree angle with the vertical, supporting the bottle in equilibrium.
### Equilibrium at Constant Velocity

**Example 13**

A jet plane is flying with a constant speed along a straight line at an angle of 30.0° above the horizontal, as shown in Figure 4.29a. The plane has a weight \( W \) with magnitude \( W = 86,500 \, \text{N} \), and its engines provide a forward thrust \( T \) of magnitude \( T = 103,000 \, \text{N} \). In addition, the lift force \( L \) (directed perpendicular to the wings) and the force \( R \) of air resistance (directed opposite to the motion) act on the plane. Find \( L \) and \( R \).

**Reasoning**

Figure 4.29b shows the free-body diagram of the plane, including the forces \( W \), \( L \), \( T \), and \( R \). As the plane is not accelerating, it is in equilibrium, meaning the sum of the x components and the sum of the y components of these forces must be zero. The lift force \( L \) and the force \( R \) of air resistance can be derived from these equilibrium conditions. To calculate the components, a free-body diagram rotated by 30.0° from the usual horizontal is used. This is for convenience, as the weight \( W \) then doesn't align with the other axes.

**Solution**

When determining the components of the weight, consider angle \( \beta \) in Figure 4.29a as 30.0°. Part c highlights the geometry responsible for this. The equation \( \alpha + \beta = 90° \) along with \( 30° = 90° \) clarifies \( \beta = 30° \). The following table shows the components of the forces at that angle:

| Force | x Component | y Component |
|-------|-------------|-------------|
| \( W \) | \(-W \sin 30.0°\) | \(-W \cos 30.0°\) |
| \( T \) | \(+T\) | \(0\) |
| \( L \) | \(0\) | \(+L\) |
| \( R \) | \(-R\) | \(0\) |

Setting the sum of the x components of the forces to zero:

\[
\
Transcribed Image Text:### Equilibrium at Constant Velocity **Example 13** A jet plane is flying with a constant speed along a straight line at an angle of 30.0° above the horizontal, as shown in Figure 4.29a. The plane has a weight \( W \) with magnitude \( W = 86,500 \, \text{N} \), and its engines provide a forward thrust \( T \) of magnitude \( T = 103,000 \, \text{N} \). In addition, the lift force \( L \) (directed perpendicular to the wings) and the force \( R \) of air resistance (directed opposite to the motion) act on the plane. Find \( L \) and \( R \). **Reasoning** Figure 4.29b shows the free-body diagram of the plane, including the forces \( W \), \( L \), \( T \), and \( R \). As the plane is not accelerating, it is in equilibrium, meaning the sum of the x components and the sum of the y components of these forces must be zero. The lift force \( L \) and the force \( R \) of air resistance can be derived from these equilibrium conditions. To calculate the components, a free-body diagram rotated by 30.0° from the usual horizontal is used. This is for convenience, as the weight \( W \) then doesn't align with the other axes. **Solution** When determining the components of the weight, consider angle \( \beta \) in Figure 4.29a as 30.0°. Part c highlights the geometry responsible for this. The equation \( \alpha + \beta = 90° \) along with \( 30° = 90° \) clarifies \( \beta = 30° \). The following table shows the components of the forces at that angle: | Force | x Component | y Component | |-------|-------------|-------------| | \( W \) | \(-W \sin 30.0°\) | \(-W \cos 30.0°\) | | \( T \) | \(+T\) | \(0\) | | \( L \) | \(0\) | \(+L\) | | \( R \) | \(-R\) | \(0\) | Setting the sum of the x components of the forces to zero: \[ \
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