relation between linear velocity and angular velocity is given as
here
v = linear speed
R = radius
= angular speed
now plug in all data in the equation
so rotating speed is 60.9 rad/s
21. Calculate the acceleration of the bus from point D to E. Show your work.
1 answer:
21) Acceleration from D to E:
22) The acceleration of the bus from D to E is
Explanation:
21)
The acceleration of an object is equal to the rate of change of velocity of the object. Mathematically:
where
u is the initial velocity
v is the final velocity
t is the time elapsed
In this problem, we want to measure the acceleration of the bus from point D to point E. We have:
- Initial velocity at point D: u = 0
- Final velocity at point E: v = 5 m/s
- Time elapsed from D to E: t = 21 - 16 = 5 s
Therefore, the acceleration between D and E is
22) This question is the same as 21), so the result is the same.
Learn more about acceleration:
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Answer:
The maximum speed of sonic at the bottom of the hill is equal to 19.85m/s and the spring constant of the spring is equal to (497.4xmass of sonic) N/m
Energy approach has been used to sole the problem.
The points of interest for the analysis of the problem are point 1 the top of the hill and point 2 the bottom of the hill just before hitting the spring
The maximum velocity of sonic is independent of the his mass or the geometry. It is only depends on the vertical distance involved
Explanation:
The step by step solution to the problem can be found in the attachment below. The principle of energy conservation has been applied to solve the problem. This means that if energy disappears in one form it will appear in another.
As in this problem, the potential and kinetic energy at the top of the hill were converted to only kinetic energy at the bottom of the hill. This kinetic energy too got converted into elastic potential energy .
x = compression of the spring = 0.89
Answer:
w = 5832.372 Joules
Explanation:
Mass of water, m = 20 kg
The water was pulled up to a height of 35 meters, i.e. h = 35 m
It takes 14 minutes to pull up the water through the height, 35 m
speed = distance/ time = 35/14 = 2.5 m/min
The bucket's height, y = speed * time = 2.5t meters
6 kg of water drips out of the bucket throughout the 14 minutes
The rate at which the water drips drips out = (6/14) = 0.4286 kg/min
Mass of water that drips out in time, t = 0.4286t kg
The mass of water remaining = (20 - 0.4286t) kg
Change in Workdone, Δw = mgΔy
Δy = 2.5 Δt
Δw = mg * 2.5 Δt
dw = (20 - 0.4286t)g2.5 dt
integrating both sides
dw = (50g - 1.07gt)dt
where b = 0, a = 14
w = 50gt - 1.07g(t²)/2 g = 9.8 m/s²
w = 490t - 5.243t²
w = (490*14 - 5.243*14²) - (490*0 - 5.243*0²)
w = 6860 - 1027.628
w = 5832.372 Joules
Answer:
center of mass of the two masses will lie at x = 2.52 cm
center of gravity of the two masses will lie at x = 2.52 cm
So center of mass is same as center of gravity because value of gravity is constant here
Explanation:
Position of centre of mass is given as
here we have
now we have
so center of mass of the two masses will lie at x = 2.52 cm
now for center of gravity we can use
here we have
now we have
So center of mass is same as center of gravity because value of gravity is constant here