The work done by Joe is 0 J.
<u>Explanation</u>:
When a force is applied to an object, there will be a movement because of the applied force to a certain distance. This transfer of energy when a force is applied to an object that tends to move the object is known as work done.
The energy is transferred from one state to another and the stored energy is equal to the work done.
W = F . D
where F represents the force in newton,
D represents the distance or displacement of an object.
Force = 0 N, D = 20 cm = 0.20 m
W = 0 
0.20 = 0 J.
Hence the work done by Joe is 0 J.
Answer:
α(0) = 0 rad/s²
α(5) = 15 rad/s²
Explanation:
The angular velocity of the flywheel is given as follows:
w(t) = A + B t²
where, A and B are constants.
Now, for the angular acceleration, we must take derivative of angular velocity with respect to time:
Angular Acceleration = α (t) = dw/dt
α(t) = (d/dt)(A + B t²)
α(t) = 2 B t
where,
B = 1.5
<u>AT t = 0 s</u>
α(0) = 2(1.5)(0)
<u>α(0) = 0 rad/s²</u>
<u></u>
<u>AT t = 5 s</u>
α(5) = 2(1.5)(5)
<u>α(5) = 15 rad/s²</u>
Answer:
the order of importance must be b e a f c
Explanation:
Modern theories indicate that the moon was formed by the collision of a bad plant with the Earth during its initial cooling period, due to which part of the earth's material was volatilized and as a ring of remains that eventually consolidated in Moon.
Based on the aforementioned, let's analyze the statements in order of importance
b) True. Since the moon is material evaporated from Earth, its compassion is similar
e) True. If the moon is material volatilized from the earth it must train a finite receding speed
a) True. The solar system was full of small bodies in erratic orbits that wander between and with larger bodies
f) False. The moon's rotation and translation are equal has no relation to its formation phase
c) false. The amount of vaporized material on the moon is large
Therefore, the order of importance must be
b e a f c
Answer:
travilng on a curve in the road
Explanation:
Answer:
The distance will be x = 41.7 [m]
Explanation:
We must first find the components in the x & y axes of the initial velocity.
![(v_{o})_{x} = 15*cos(20)= 14.09[m/s]\\(v_{o})_{y} = 15*sin(20)= 5.13[m/s]](https://tex.z-dn.net/?f=%28v_%7Bo%7D%29_%7Bx%7D%20%3D%2015%2Acos%2820%29%3D%2014.09%5Bm%2Fs%5D%5C%5C%28v_%7Bo%7D%29_%7By%7D%20%3D%2015%2Asin%2820%29%3D%205.13%5Bm%2Fs%5D)
The acceleration is the gravity acceleration therefore.
g = 9.81 [m/s^2]
Now we can calculate how long it takes to fall.
![y=(v_{o})_{y}*t-0.5*g*t^2\\-28 = 5.13*t-0.5*9.81*t^2\\-28=-4.905*t^2+5.13*t\\4.905*t^2-5.13*t=28\\t = 2.96[s]](https://tex.z-dn.net/?f=y%3D%28v_%7Bo%7D%29_%7By%7D%2At-0.5%2Ag%2At%5E2%5C%5C-28%20%3D%205.13%2At-0.5%2A9.81%2At%5E2%5C%5C-28%3D-4.905%2At%5E2%2B5.13%2At%5C%5C4.905%2At%5E2-5.13%2At%3D28%5C%5Ct%20%3D%202.96%5Bs%5D)
With this time we can find the horizontal distance that runs the projectile.
![x=(v_{o})_{x}*t\\x=14.09*2.96\\x=41.7[m]](https://tex.z-dn.net/?f=x%3D%28v_%7Bo%7D%29_%7Bx%7D%2At%5C%5Cx%3D14.09%2A2.96%5C%5Cx%3D41.7%5Bm%5D)
