25) The curved section of a speedway is a circular arc having a radius of 190 m. This curve is properly banked for racecars moving at 34 m/s. At what angle with the horizontal is the curved part of the speedway banked? Show work. A) 32° B) 34° C) 30% D) 28° E) 26°
25) The curved section of a speedway is a circular arc having a radius of 190 m. This curve is properly banked for racecars moving at 34 m/s. At what angle with the horizontal is the curved part of the speedway banked? Show work. A) 32° B) 34° C) 30% D) 28° E) 26°
College Physics
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Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Please answer 25 and 26
![Here is the transcribed text for educational purposes on a website, including detailed explanations of any problems:
---
### Physics Practice Problems
#### Problem 25:
The curved section of a speedway is a circular arc having a radius of 190 m. This curve is properly banked for racecars moving at 34 m/s. At what angle with the horizontal is the curved part of the speedway banked? **Show work**.
- A) 32°
- B) 34°
- C) 30°
- D) 28°
- E) 26°
**Explanation:**
To find the angle (θ) at which the curve is banked, use the following formula for banked curves without friction:
\[ \tan(\theta) = \frac{v^2}{rg} \]
where:
\( v = 34 \, \text{m/s} \) (velocity),
\( r = 190 \, \text{m} \) (radius),
\( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity).
*Calculation steps for finding the banked angle should be shown here.*
---
#### Problem 26:
At their closest approach, Venus and Earth are \( 4.20 \times 10^{10} \) m apart. The mass of Venus is \( 4.87 \times 10^{24} \) kg, the mass of Earth is \( 5.97 \times 10^{24} \) kg, and \( G = 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2 \). What is the magnitude of the gravitational force exerted by Venus on Earth at that point? **Show work**.
- A) \( 1.10 \times 10^{18} \) N
- B) \( 4.62 \times 10^{28} \) N
- C) \( 5.43 \times 10^{26} \) N
- D) \( 6.30 \times 10^{20} \) N
- E) \( 1.72 \times 10^{19} \) N
**Explanation:**
To calculate the gravitational force (F) between two masses, use Newton's law of gravitation](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbc6b83f6-6fd0-4bf2-b4c8-4ee11df5765c%2Ffbe6dd63-267b-4192-926a-504ccaf723b3%2Ft0w195_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Here is the transcribed text for educational purposes on a website, including detailed explanations of any problems:
---
### Physics Practice Problems
#### Problem 25:
The curved section of a speedway is a circular arc having a radius of 190 m. This curve is properly banked for racecars moving at 34 m/s. At what angle with the horizontal is the curved part of the speedway banked? **Show work**.
- A) 32°
- B) 34°
- C) 30°
- D) 28°
- E) 26°
**Explanation:**
To find the angle (θ) at which the curve is banked, use the following formula for banked curves without friction:
\[ \tan(\theta) = \frac{v^2}{rg} \]
where:
\( v = 34 \, \text{m/s} \) (velocity),
\( r = 190 \, \text{m} \) (radius),
\( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity).
*Calculation steps for finding the banked angle should be shown here.*
---
#### Problem 26:
At their closest approach, Venus and Earth are \( 4.20 \times 10^{10} \) m apart. The mass of Venus is \( 4.87 \times 10^{24} \) kg, the mass of Earth is \( 5.97 \times 10^{24} \) kg, and \( G = 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2 \). What is the magnitude of the gravitational force exerted by Venus on Earth at that point? **Show work**.
- A) \( 1.10 \times 10^{18} \) N
- B) \( 4.62 \times 10^{28} \) N
- C) \( 5.43 \times 10^{26} \) N
- D) \( 6.30 \times 10^{20} \) N
- E) \( 1.72 \times 10^{19} \) N
**Explanation:**
To calculate the gravitational force (F) between two masses, use Newton's law of gravitation
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