Answer:
Explanation:
If the initial velocity is U
Then the horizontal component of the velocity is
Ux= Ucosθ
Then the range for a projectile is give as
R=Ux.t
Where t is the time of flight
The time of flight is given as
t=2USinθ/g
Therefore,
R=Ux.t
R=UCosθ.2USinθ/g
R=U^2×2SinθCosθ/g
Then, from trigonometric ratio
2SinθCosθ= Sin2θ
R=U^2Sin2θ/g
Given that θ=32° and g=9.81m/s^2
Then
R=U^2Sin2×32/9.81
R=U^2Sin64/9.81
R=0.0916U^2
Then, range is given by R=0.0916U^2
A=0.0916U^2.
T
The box is at a distance A from the point of projection. Then the range R=A
R=0.0916U^2
A=0.0916U^2
Then,
U^2=A/0.0916
U^2=10.915A
Then the initial velocity should be
U=√10.915A
U=3.3√A
<span>If there isn't any force then the normal contact force will be
N=m*g=7.5*9.81=73.58N
which is 73.58-23=50.58N less
so, there the person must pull at 23 degree upward
break down the tension in two components, vertical and horizontal.
vertical tension= 50.58=T*sin23
T=50.58/sin23=129.45N</span>
Answer:
Entropy is increasing. Entropy is decreasing.
Explanation:
The Entropy doesn't change.
A.. in case of any problems that may occur you would know what company to call
Answer:
This shows inertia because inertia is an object's resistance to change in motion. When the person (imma call them a she) who pulled the chair from under the guy did that, the chair was the one affected by the force of the girl, not the guy. The guy continued heading in the direction he was originally going, which was down.
At least, that's about how I would answer this question.
