<h2>
Answer: Toward the center of the circle.</h2>
This situation is characteristic of the uniform circular motion , in which the movement of a body describes a circumference of a given radius with constant speed.
However, in this movement the velocity has a constant magnitude, but its direction varies continuously.
Let's say
is the velocity vector, whose direction is perpendicular to the radius
of the trajectory, therefore
the acceleration
is directed toward the center of the circumference.
Answer:
The gravitational force on the elevator = 4500N
Explanation:
The given parameters are;
The force applied by the elevator, F = 4500 N
The acceleration of the elevator = Not accelerating
From Newton's third law of motion, the action of the cable force is equal to the reaction of the gravitational force on the elevator which is the weight, W and motion of the elevator as follows;
F = W + Mass of elevator × Acceleration of elevator
∴ F = W + Mass of elevator × 0 = W
F = 4500 N = W
The net force on the elevator is F - W = 0
The gravitational force on the elevator = W = 4500N.
The magnitude of the unknown height of the projectile is determined as 16.1 m.
<h3>
Magnitude of the height</h3>
The magnitude of the height of the projectile is calculated as follows;
H = u²sin²θ/2g
H = (36.6² x (sin 29)²)/(2 x 9.8)
H = 16.1 m
Thus, the magnitude of the unknown height of the projectile is determined as 16.1 m.
Learn more about height here: brainly.com/question/1739912
#SPJ1
Answer:
141.14098 secs
Explanation:
Time taken to see the lightning flash can be gotten from:
Velocity = distance/time
Time = distance/velocity
Time = (47 * 1000)/(3 * 10^8)
Time = 0.0001567 secs
Time taken to hear the thunder can be gotten from:
Velocity = distance/time
Time = distance/velocity
Time = (47 * 1000)/(333)
Time = 141.14114 secs
The time lapse between the lightning flash and the thunder will be:
141.14114 - 0. 0001567
= 141.14098 secs
Answer:
The height is ![h_c = 42.857](https://tex.z-dn.net/?f=h_c%20%3D%2042.857)
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 ![h_s = 30 \ cm](https://tex.z-dn.net/?f=h_s%20%3D%2030%20%5C%20cm)
The angle of the slope is ![\theta = 15^o](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2015%5Eo)
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
![mgh = \frac{1}{2} I w^2 + \frac{1}{2}mv^2](https://tex.z-dn.net/?f=mgh%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20I%20w%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
Where I is the moment of inertia which is mathematically represented as this for a sphere
![I = \frac{2}{5} mr^2](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B2%7D%7B5%7D%20mr%5E2)
The angular velocity
is mathematically represented as
![w = \frac{v}{r}](https://tex.z-dn.net/?f=w%20%3D%20%5Cfrac%7Bv%7D%7Br%7D)
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)
![v^2 = \frac{10gh_s}{7}](https://tex.z-dn.net/?f=v%5E2%20%3D%20%5Cfrac%7B10gh_s%7D%7B7%7D)
Considering a circular hoop
The moment of inertial is different for circle and it is mathematically represented as
![I = mr^2](https://tex.z-dn.net/?f=I%20%3D%20mr%5E2)
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
![mgh_c = mv^2](https://tex.z-dn.net/?f=mgh_c%20%20%3D%20mv%5E2)
![gh_c = v^2](https://tex.z-dn.net/?f=gh_c%20%3D%20v%5E2)
![h_c = \frac{v^2}{g}](https://tex.z-dn.net/?f=h_c%20%3D%20%5Cfrac%7Bv%5E2%7D%7Bg%7D)
Recall that ![v^2 = \frac{10gh_s}{7}](https://tex.z-dn.net/?f=v%5E2%20%3D%20%5Cfrac%7B10gh_s%7D%7B7%7D)
![h_c= \frac{\frac{10gh_s}{7} }{g}](https://tex.z-dn.net/?f=h_c%3D%20%5Cfrac%7B%5Cfrac%7B10gh_s%7D%7B7%7D%20%7D%7Bg%7D)
![= \frac{10h_s}{7}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B10h_s%7D%7B7%7D)
Substituting values
![h_c = \frac{10(30)}{7}](https://tex.z-dn.net/?f=h_c%20%3D%20%5Cfrac%7B10%2830%29%7D%7B7%7D)