Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Connect with a community of experts ready to help you find accurate solutions to your questions quickly and efficiently. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
To determine the acceleration of the sled sliding down a frictionless hill inclined at [tex]\(28.0^{\circ}\)[/tex], let's break down the problem step by step.
1. Identify Known Values:
- Angle of incline ([tex]\(\theta\)[/tex]): [tex]\(28.0^{\circ}\)[/tex]
- Distance slid by the sled ([tex]\(d\)[/tex]): [tex]\(10.0 \)[/tex] meters
- Gravitational acceleration ([tex]\(g\)[/tex]): [tex]\(9.81 \, \text{m/s}^2\)[/tex]
2. Convert the Angle to Radians:
Since trigonometric functions in physics often use radians, it's practical to convert the angle from degrees to radians.
[tex]\[ \theta_{\text{radians}} = \theta \times \left( \frac{\pi}{180} \right) \][/tex]
Substituting the given angle:
[tex]\[ \theta_{\text{radians}} = 28.0 \times \left( \frac{\pi}{180} \right) \approx 0.4887 \, \text{radians} \][/tex]
3. Determine the Component of Gravitational Force:
On an incline, the component of the gravitational force causing acceleration down the hill is [tex]\( g \sin(\theta_{\text{radians}}) \)[/tex].
4. Calculate the Acceleration:
We use the formula for the acceleration of the sled along the incline:
[tex]\[ a = g \sin(\theta_{\text{radians}}) \][/tex]
Plugging in the values:
[tex]\[ a = 9.81 \times \sin(0.4887) \approx 4.6055 \, \text{m/s}^2 \][/tex]
Hence, the acceleration of the sled is [tex]\( \boxed{4.6055 \, \text{m/s}^2} \)[/tex].
1. Identify Known Values:
- Angle of incline ([tex]\(\theta\)[/tex]): [tex]\(28.0^{\circ}\)[/tex]
- Distance slid by the sled ([tex]\(d\)[/tex]): [tex]\(10.0 \)[/tex] meters
- Gravitational acceleration ([tex]\(g\)[/tex]): [tex]\(9.81 \, \text{m/s}^2\)[/tex]
2. Convert the Angle to Radians:
Since trigonometric functions in physics often use radians, it's practical to convert the angle from degrees to radians.
[tex]\[ \theta_{\text{radians}} = \theta \times \left( \frac{\pi}{180} \right) \][/tex]
Substituting the given angle:
[tex]\[ \theta_{\text{radians}} = 28.0 \times \left( \frac{\pi}{180} \right) \approx 0.4887 \, \text{radians} \][/tex]
3. Determine the Component of Gravitational Force:
On an incline, the component of the gravitational force causing acceleration down the hill is [tex]\( g \sin(\theta_{\text{radians}}) \)[/tex].
4. Calculate the Acceleration:
We use the formula for the acceleration of the sled along the incline:
[tex]\[ a = g \sin(\theta_{\text{radians}}) \][/tex]
Plugging in the values:
[tex]\[ a = 9.81 \times \sin(0.4887) \approx 4.6055 \, \text{m/s}^2 \][/tex]
Hence, the acceleration of the sled is [tex]\( \boxed{4.6055 \, \text{m/s}^2} \)[/tex].
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.