Answer:
Vo = 4.5 [m/s]
Explanation:
In order to solve this problem, we must use the following equation of kinematics.

where:
Vf = final velocity = 12 [m/s]
Vo = initial velocity [m/s]
a = acceleration = 1.5 [m/s²]
t = time = 5 [s]
Now replacing:
![12=v_{o}+1.5*5\\v_{o}=12- (7.5)\\v_{o}= 4.5[m/s]](https://tex.z-dn.net/?f=12%3Dv_%7Bo%7D%2B1.5%2A5%5C%5Cv_%7Bo%7D%3D12-%20%287.5%29%5C%5Cv_%7Bo%7D%3D%204.5%5Bm%2Fs%5D)
Answer:
19.01 N
Explanation:
F = Force being applied to the crate = 45 N
= Angle at which the force is being applied = 
Horizontal component of force is given by

The horizontal component of the force acting on the crate is 19.01 N.
B. Transition Metals.
Hope it helps!
Answer: 12.0 m/s^2
Explanation:
Let
be the angular acceleration of the end of the rod
Taking torque about the link, we have:

Torque is also given in terms of moment of inertia of the rod and its angular acceleration i.e.

From equations (i) and (ii) we have:

The acceleration of the end of the rod farthest from the link is given by:

Answer:

Explanation:
F = Force
C = Drag coefficient equal for both aircrafts
ρ = Density of air
A = Surface area equal for both aircrafts
v = Velocity



Dividing the above two equations we get

The ratio of the drag forces is 