Answer:
5 .07 s .
Explanation:
The child will move on a circle of radius r
r = 1.5 m
Let the velocity of rotation = v
radial acceleration = v² / r
v² / r = 2.3
v² = 2.3 r = 2.3 x 1.5
= 3.45
v = 1.857 m /s
Time of revolution = 2π r / v
= 2 x 3.14 x 1.5 / 1.857
= 5 .07 s .
The merry-go round will take 5.07 s to complete one complete revolution.
Given data:
The distance of child from the center is, r = 1.5 m.
The magnitude of radial acceleration is, [tex]a = 2.3 \;\rm m/s^{2}[/tex].
Since, the child is on merry-go round, which is undergoing a rotational motion. And radial acceleration means that it is under the acceleration, whose value is given as,
[tex]a=\dfrac{v^{2}}{r}\\\\v=\sqrt{a \times r}[/tex]
Here, v is the linear velocity.
Solving as,
[tex]v=\sqrt{2.3 \times 1.5} \\\\v=1.857 \;\rm m/s[/tex]
Now, we to obtain the time taken by merry-go round to complete one revolution. Then the expression for the time taken to complete one revolution is,
[tex]T=\dfrac{2 \pi r}{v}\\\\T=\dfrac{2 \pi \times 1.5}{1.857}\\\\T = 5.07 \;\rm s[/tex]
Thus, we can conclude that the merry-go round will take 5.07 s to complete one revolution.
Learn more about the centripetal acceleration here:
https://brainly.com/question/14465119
1500 POINTS TO WHOEVER ANSWERS THIS FIRST!!!!!!!!!!
1. You drop a bowling ball 15m from the top of the science building. What is its
velocity as it impacts the ground?
2. A cheetah accelerates from 0 to 13m/s in 3 seconds. How far does it travel
during that time?
3. Your foolish friend shoots a bullet straight in the air at a muzzle velocity of
380 m/s. How high does the bullet go (neglecting air resistance)?
4. Your foolish friend shoots a bullet straight in the air at a muzzle velocity of
380 m/s. What is the bullets velocity when it falls back down and hits your
friend?
Answer: Multiply the time by the acceleration due to gravity to find the velocity when the object hits the ground. If it takes 9.9 seconds for the object to hit the ground, its velocity is (1.01 s)*(9.8 m/s^2), or 9.9 m/s.
Explanation:
asheed and Sofia are riding a merry-go-round that is spinning steadily (uniform circular motion). Sofia is twice as far from the axis as is Rasheed. Sophia's acceleration is _________ that of Rasheed. *
Answer:
The same as
Explanation:
First of all, they are in a merry go round, which is a circular one. And thus, we would be dealing with circular motions.
The angular velocity of a merry go round, in this instance, is given by
w = Δ/Δt
The formula has nothing whatsoever to do with distance, and as such, both Rasheed and Sofia would have the same angular velocity.
Transferring this further, the angular acceleration is given as
α = Δw/Δt
Remember I said the velocity has nothing to do with distance, well, so does the acceleration as we can see from the formula stated. And therefore, both Rasheed and Sofia would have the same angular acceleration
which property of potential energy distinguishes it from kinetic energy
Answer:
Shape and position
Explanation:
Hope this helps! :)
During the slowing down process, there is a time when the angular speed is 0.5 rev/s. The fan arm is 0.8 m long. What is the speed of a point on the end of the fan arm at this time
Answer:
2.51 m/s
Explanation:
Given that,
Angular speed of a fan, [tex]\omega=0.5\ rev/s = 3.14\ rad/s[/tex]
Length of the fan arm, r = 0.8 m
We need to find the speed of a point on the end of the fan arm at this time. Let v is the speed. It is given in terms of angular speed is given by :
[tex]v=r\omega\\\\v=0.8\times 3.14\\\\v=2.51\ m/s[/tex]
So, the required speed is 2.51 m/s.
What type of numbers in a measurement are always significant?
Answer:
Non-zero digits
Explanation:
umm (ꏿ﹏ꏿ;)
Answer:
There's three rules on determining how many important quantities are in a number: Non-zero digits are always predominant. Any zeros between two significant digits are always major. The final zero or dangling zeros in the decimal segment are really only important.
I hope this answered your question, have a nice day!
Need help ASAP please help
Answer:
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Explanation: