Answer:
a) t₁ = 4.76 s, t₂ = 85.2 s
b) v = 209 ft/s
Explanation:
Constant acceleration equations:
x = x₀ + v₀ t + ½ at²
v = at + v₀
where x is final position,
x₀ is initial position,
v₀ is initial velocity,
a is acceleration,
and t is time.
When the engine is on and the sled is accelerating:
x₀ = 0 ft
v₀ = 0 ft/s
a = 44 ft/s²
t = t₁
So:
x = 22 t₁²
v = 44 t₁
When the engine is off and the sled is coasting:
x = 18350 ft
x₀ = 22 t₁²
v₀ = 44 t₁
a = 0 ft/s²
t = t₂
So:
18350 = 22 t₁² + (44 t₁) t₂
Given that t₁ + t₂ = 90:
18350 = 22 t₁² + (44 t₁) (90 − t₁)
Now we can solve for t₁:
18350 = 22 t₁² + 3960 t₁ − 44 t₁²
18350 = 3960 t₁ − 22 t₁²
9175 = 1980 t₁ − 11 t₁²
11 t₁² − 1980 t₁ + 9175 = 0
Using quadratic formula:
t₁ = [ 1980 ± √(1980² - 4(11)(9175)) ] / 22
t₁ = 4.76, 175
Since t₁ can't be greater than 90, t₁ = 4.76 s.
Therefore, t₂ = 85.2 s.
And v = 44 t₁ = 209 ft/s.
Answer:
a). Maximum Length L=0.929m
b). T=0.83 Hz or 1.2s
c). Longer, the effortless waling T=2.1 Hz or t=0.475s
d). t=1.2s V=0.774 
t=0.475s V=1.95
Explanation:
Length legs=L=1.1m
angle=50
the step that give the person forms a triangle whose two sides are known and the angle that forms between them, then using trigonometry as the image
Divide the original triangle in two and form a right triangle so the angle is 25 and the L is hypotenuse and the opposite is the step length
a).
Length of the step
L=0.464m*2
L=0.928m
b).
period=T
c).
The period is the inverse of the time of the motion so, the T1 is faster that the T because
d).
The speed is the relation between the distance with time so:
The velocity of the ball when it strikes the ground, given the data is 21.56 m/s
<h3>Data obtained from the question</h3>
From the question given above, the following data were obtained:
- Time to reach ground from maximum height (t) = 2.2 s
- Initial velocity (u) = 0 m/s
- Acceleration due to gravity (g) = 9.8 m/s²
- Final velocity (v) =?
<h3>How to determine the velocity when the ball strikes the ground</h3>
The velocity of the ball when it strikes the ground can be obtained as illustrated below:
v = u + gt
v = 0 + (9.8 × 2.2)
v = 0 + 21.56
v = 21.56 m/s
Thus, the velocity of the ball when it strikes the ground is 21.56 m/s
Learn more about motion under gravity:
brainly.com/question/22719691
#SPJ1
Answer:
a) αA = 4.35 rad/s²
αB = 1.84 rad/s²
b) t = 3.7 rad/s²
Explanation:
Given:
wA₀ = 240 rpm = 8π rad/s
wA₁ = 8π -αA*t₁
The angle in B is:
The velocity at the contact point is equal to:
Matching both expressions:
b) The time during which the disks slip is:
a) The angular acceleration of each disk is
