Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Discover reliable solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
To determine the final velocity of the sled just before it reaches the bottom of the hill, we will break down the problem into several steps:
1. Given Information:
- The inclined angle of the hill, [tex]\(\theta = 38.0^{\circ}\)[/tex]
- The time taken to reach the bottom of the hill, [tex]\( t = 4.24 \, \text{s} \)[/tex]
- The acceleration due to gravity, [tex]\( g = 9.81 \, \text{m/s}^2 \)[/tex]
2. Convert the angle from degrees to radians:
To use trigonometric functions, we need the angle in radians.
[tex]\[ \theta \text{ (in radians)} = 0.6632251157578453 \][/tex]
3. Calculate the acceleration along the incline:
The effective acceleration [tex]\( a \)[/tex] down the incline can be calculated using the component of gravitational acceleration along the slope, which is given by [tex]\( g \sin(\theta) \)[/tex].
[tex]\[ a = 6.039639072944708 \, \text{m/s}^2 \][/tex]
4. Determine the initial velocity:
Since the sled starts from rest, the initial velocity [tex]\( u \)[/tex] is:
[tex]\[ u = 0 \, \text{m/s} \][/tex]
5. Use the kinematic equation to find the final velocity:
We use the equation [tex]\( v = u + at \)[/tex] where:
- [tex]\( v \)[/tex] is the final velocity.
- [tex]\( u \)[/tex] is the initial velocity, which is [tex]\( 0 \, \text{m/s} \)[/tex].
- [tex]\( a \)[/tex] is the acceleration along the incline.
- [tex]\( t \)[/tex] is the time taken.
Plugging in the known values:
[tex]\[ v = 0 + 6.039639072944708 \, \text{m/s}^2 \times 4.24 \, \text{s} \][/tex]
[tex]\[ v = 25.608069669285566 \, \text{m/s} \][/tex]
So, the final velocity of the sled just before it reaches the bottom of the hill is:
[tex]\[ v_f = 25.608069669285566 \, \text{m/s} \][/tex]
1. Given Information:
- The inclined angle of the hill, [tex]\(\theta = 38.0^{\circ}\)[/tex]
- The time taken to reach the bottom of the hill, [tex]\( t = 4.24 \, \text{s} \)[/tex]
- The acceleration due to gravity, [tex]\( g = 9.81 \, \text{m/s}^2 \)[/tex]
2. Convert the angle from degrees to radians:
To use trigonometric functions, we need the angle in radians.
[tex]\[ \theta \text{ (in radians)} = 0.6632251157578453 \][/tex]
3. Calculate the acceleration along the incline:
The effective acceleration [tex]\( a \)[/tex] down the incline can be calculated using the component of gravitational acceleration along the slope, which is given by [tex]\( g \sin(\theta) \)[/tex].
[tex]\[ a = 6.039639072944708 \, \text{m/s}^2 \][/tex]
4. Determine the initial velocity:
Since the sled starts from rest, the initial velocity [tex]\( u \)[/tex] is:
[tex]\[ u = 0 \, \text{m/s} \][/tex]
5. Use the kinematic equation to find the final velocity:
We use the equation [tex]\( v = u + at \)[/tex] where:
- [tex]\( v \)[/tex] is the final velocity.
- [tex]\( u \)[/tex] is the initial velocity, which is [tex]\( 0 \, \text{m/s} \)[/tex].
- [tex]\( a \)[/tex] is the acceleration along the incline.
- [tex]\( t \)[/tex] is the time taken.
Plugging in the known values:
[tex]\[ v = 0 + 6.039639072944708 \, \text{m/s}^2 \times 4.24 \, \text{s} \][/tex]
[tex]\[ v = 25.608069669285566 \, \text{m/s} \][/tex]
So, the final velocity of the sled just before it reaches the bottom of the hill is:
[tex]\[ v_f = 25.608069669285566 \, \text{m/s} \][/tex]
We hope this was helpful. Please come back whenever you need more information or answers to your queries. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.