Answer:
40.92 m/s
Explanation:
The computation is shown below:
Ek = 1 ÷2mv²...............................(1)
v = √(2Ek/m).......................... (2)
Here EK denotes kinetic energy
m denotes mass
v denotes velocity
Given that
m = 0.25kg and Ek = 209.3J
So,
v = √(2×209.3 ÷0.25)
= √1674.4
= 40.92 m/s
Answer:
The height is 
A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )
Explanation:
From the question we are told that
The height is 
The angle of the slope is 
According to the law of conservation of energy
The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

Where I is the moment of inertia which is mathematically represented as this for a sphere

The angular velocity
is mathematically represented as

So the equation for conservation of energy becomes
![mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2](https://tex.z-dn.net/?f=mgh_s%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5B%5Cfrac%7B2%7D%7B5%7D%20mr%5E2%20%5D%5B%5Cfrac%7Bv%7D%7Br%7D%20%5D%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
![mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]](https://tex.z-dn.net/?f=mgh_s%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2%20%5B%5Cfrac%7B2%7D%7B5%7D%20%2B1%20%5D)
![mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]](https://tex.z-dn.net/?f=mgh_s%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2%20%5B%5Cfrac%7B7%7D%7B5%7D%20%5D)
![gh_s =[\frac{7}{10} ] v^2](https://tex.z-dn.net/?f=gh_s%20%3D%5B%5Cfrac%7B7%7D%7B10%7D%20%5D%20v%5E2)

Considering a circular hoop
The moment of inertial is different for circle and it is mathematically represented as

Substituting this into the conservation equation above
![mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2](https://tex.z-dn.net/?f=mgh_c%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%28mr%5E2%29%5B%5Cfrac%7Bv%7D%7Br%7D%20%5D%20%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2)
Where
is the height where the circular hoop would be released to equal the speed of the sphere at the bottom



Recall that 


Substituting values

The distance in meters she would have moved before she begins to slow down is 11.25 m
<h3>
LINEAR MOTION</h3>
A straight line movement is known as linear motion
Given that Ann is driving down a street at 15 m/s. Suddenly a child runs into the street. It takes Ann 0.75 seconds to react and apply the brakes.
To know how many meters will she have moved before she begins to slow down, we need to first list all the given parameters.
From definition of speed,
speed = distance / time
Make distance the subject of the formula
distance = speed x time
distance = 15 x 0.75
distance = 11.25m
Therefore, the distance in meters she would have moved before she begins to slow down is 11.25 m
Learn more about Linear motion here: brainly.com/question/13665920
Answer:
B. Hold each type of fabric over a candle flame and time how long it takes for the fabric to start to burn.
Explanation: