Answer:
t = 2.01 s
Vf = 19.7 m/s
Explanation:
It's know through the International System that the earth's gravity is 9.8 m/s², then we have;
Data:
- Height (h) = 20 m
- Gravity (g) = 9.8 m/s²
- Time (t) = ?
- Final Velocity (Vf) = ?
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Time
Use formula:
Replace:
Everything inside the root is solved first. So, we solve the multiplication of the numerator:
It divides:
The square root is performed:
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Final Velocity
use formula:
Replace:
Multiply:
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How long does it take to reach the ground?
Takes time to reach the ground in <u>2.01 seconds.</u>
How fast does it hit the ground?
Hits the ground with a speed of <u>19.7 meters per seconds.</u>
(a) The equation for the work done in stretching the spring from x1 to x2 is ¹/₂K₂Δx².
(b) The work done, in stretching the spring from x1 to x2 is 11.25 J.
(c) The work, necessary to stretch the spring from x = 0 to x3 is 64.28 J.
<h3>
Work done in the spring</h3>
The work done in stretching the spring is calculated as follows;
W = ¹/₂kx²
W(1 to 2) = ¹/₂K₂Δx²
W(1 to 2) = ¹/₂(250)(0.65 - 0.35)²
W(1 to 2) = 11.25 J
W(0 to 3) = ¹/₂k₁x₁² + ¹/₂k₂x₂² + ¹/₂F₃x₃
W(0 to 3) = ¹/₂(660)(0.35)² + ¹/₂(250)(0.65 - 0.35)² + ¹/₂(105)(0.89 - 0.65)
W(0 to 3) = 64.28 J
Learn more about work done here: brainly.com/question/25573309
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Answer:
Explanation:
distance of shuttle from centre of the earth = radius of the orbit
= 6300 + 300 = 6600 km
= 6600 x 10³
Formula of time period of the satellite
T = 2π R /v₀ , v₀ is orbital velocity
v₀ = √gR , ( if height is small with respect to radius )
T = 2π R /√gR
= 2π√ R /√g
= 2 x 3.14 x √ 6600 x 10³ / √9.8
= 2 x 3.14 x 256.9 x 10 / 3.13
= 5154.41 s
= 5154.41 / 60 minutes
= 85.91 m
85.9 minutes.
2 ) No of sunrise per day = no of rotation per day
= 24 x 60 / 85.9
= 16.76
or 17 sunrises.
Answer:
The heavier wagon rolls 1/2 as fast as the lighter wagon.
Explanation:
When the compressed spring that joins them is released then the force acts on both wagons will be of equal magnitude but in the opposite direction. However as the mass of one wagon is twice that of other, so the acceleration will become half of the heavier wagon in comparison with lighter one.