IENEA-97701 JAV Ketertarikan yang Mendalam pada Tetangga: Kisah Tentang Respect dan Batasan - Cuplikan Gratis dan Subtitle Bahasa Indonesia srt.

IENEA-98303 Pengalaman Berharga Bersama Kakak Laki-laki dengan Batasan yang Menghormati.

28 Apr 2025

IENFA-202 Respek terhadap Batasan dan Konsensual dalam Hubungan Antarindividu

28 Apr 2025

IENFA-203 Kisah Menegangkan tentang Tekanan dan Kejadian Tak Terduga

28 Apr 2025

IENFA-401 Kejadian Tak Terduga Saat Berinteraksi dengan Seorang Ibu Rumah Tangga yang Ramah dan Menawan

28 Apr 2025

HHH-373 Video Pasangan yang Terangsang di Fasilitas Komersial Umum Beredar selama Durasi Tertentu

28 Apr 2025

NPH-161 ### 1. Understanding the Problem Before diving into solving the problem, it's crucial to thoroughly understand what is being asked. Here's the problem statement: > A skier sk on a hill that is uniformly sloped with a slope of 0.40. The friction between the skier and the hill is 0.23. The sk's acceleration is _________. (a) 1.24 m/s^2 (b) 0.1 m/s2 (c) 1.76 m/s2 (d) 0.13 m/s2 From this, I can extract that there's a skier on a hill with a slope of 0.40. There's friction between the skier and the hill, which is 0.23, and I need to find the acceleration of the skier. ### 2. Breaking Down the Problem Next, I need to analyze the problem to understand what's given and what needs to be found. - **Given:** - Slope of the hill: 0.40 - Friction between skier and hill: 0.23 - Options for acceleration: (a) 1.24 m/s^2 (b) 0.1 m/s2 (c) 1.764 m/s2 (d) 0.13 m/s2 - **Need to find:** - The acceleration of the skier ### 3. Knowing the Formulas To solve this problem, I need to recall some physics principles, particularly Newton's second law of motion which states: `F = m * a` Where: F is the total force on the object m is the mass of the object a is the acceleration of the object Also, I need to know about the Normal force which is given by: `N = m * g * cos(theta)` Where: N is the normal force m is the mass of the object g is the acceleration due to gravity (approximately 9.81 m/s^2) theta is the angle of the hill Since the slope is given as 0.40, I can calculate the angle of the hill using the formula: `slope = tan(theta)` So, theta = atan(slope) = atan(0.40) = 21.8 degrees ### 4. Calculate the Forces There are forces acting on the skier: 1. The component of the object's weight that is parallel to the hill: This is given by: `F_parallel = m * g * sin(theta)` 2. The friction force: This is given by: `F_friction = m * g * cos(theta) * friction` Since the skier is going downhill, the friction force is acting upwards, opposing the motion. The net force on the skier is: `F_net = F_parallel - F_friction` ### 5. Applying Newton's Second Law To find the acceleration, I can rearrange the formula `F = m * a` to: `a = F / m` Since all the forces are expressed in terms of mass (m), I can eliminate m from the calculations. So, the acceleration is given by: `a = (F_parallel - F_friction) / m` Since `F_parallel = m * g * sin(theta)` and `F_friction = m * g * cos(theta) * friction`, the equation becomes: `a = (m * g * sin(theta) - m * g * cos(theta) * friction) / m` `a = g * (sin(theta) - cos(theta) * friction)` ### 6. Calculating the Value Now, I can substitute the known values into the equation: `a = 9.81 * (sin(21.8) - cos(21.8) * 0.23)` First, calculate sin(21.8) and cos(21.8): `sin(21.8) = 0.371` `cos(21.8) = 0.930` `a = 9.81 * (0.371 - 0.930 * 0.23)` `a = 9.81 * (0.371 - 0.2139)` `a = 9.81 * 0.1571` `a = 1.5429 m/s^2` ### 7. Choosing the Correct Answer Now, compare this value with the given options: (a) 1.24 m/s^2 (b) 0.1 m/s2 (c) 1.764 m/s2 (d) 0.13 m/s2 Since 1.5429 m/s^2 is closest to 1.764 m/s2, the correct answer is (c) 1.764 m/s2. ### 8. Final Answer The correct answer is `(c) 1.764 m/s²`. --- **Note**#### Here's a step-by-step breakdown of how to solve the problem: 1. **Understanding the Problem:** - **Given:** - Slope of the hill: 0.40 - Friction between skier and hill: 0.23 - Options for acceleration: (a) 1.24 m/s^2 (b) 0.1 m/s^2 (c) 1.76 m/s^2 (d) 0.13 m/s^2 - **Need to find:** The acceleration of the skier 2. **Breaking Down the Problem:** - **Known:** - Slope = 0.40 - Friction = 0.23 - Acceleration options: (a) 1.24 (b) 0.1 (c) 1.76 (d) 0.13 - **Unknown:** - Acceleration of skier 3. **Knowing the Formulas:** - Recall Newton's second law: `F = m * a` - Forces on the skier: - `F_parallel` = mg sin(theta) - `F_friction` = m g cos(theta) * friction - Net force: `F_net` = F_parallel - F_friction - Acceleration: `a = F_net / m` 4. **Calculating Starting Values:** - **Theta calculation:** - `Slope` = tan(theta) = 0.40 - `theta` = atan(0.40) = 21.8 degrees 5. **Calculating the Forces:** - `F_parallel = m * g * sin(theta) = m * 9.81 * sin(21.8) = m * 9.81 * 0.371 = m * 3.64` - `F_friction = m * g * cos(theta) * friction = m * 9.81 * cos(21.8) * 0.23 = m * 9.81 * 0.930 * 0.23 = m * 2.04 - Net force: `F_net` = F_parallel - F_friction = m * 3.64 - m * 2.04 = m * 1.60 - Acceleration: `a = F_net / m`= m * 1.60 / m = 1.60 m/s^2 6. **Choosing the Correct Answer:** - Closest option to 1.60 m/s^2 is (c) 1.76 m/s^2 7. **Final Answer:** - The correct answer is `(c) 1.76 m/s²` --- **Note**

28 Apr 2025

IENEA-98503 Pengalaman Mengharukan di Tengah Kehidupan Perkotaan

28 Apr 2025

J-99611 Fukuda Yuki: Cerita Hangat di Bawah Satu Atap dengan Keturunan Kedelai Hitam yang Langsing di Pemandian Air Panas

28 Apr 2025

NPH-159 Kisah Pertukaran Pasangan yang Penuh Tantangan oleh Kaoru dan Tetangganya

28 Apr 2025

IENEA-98302 Pengalaman Mendalam: Ujian Keberanian Antara Kakak Beradik dalam Tantangan Unik

28 Apr 2025