Answer:
Troposphere
High-pressure areas form due to downward motion through the troposphere, the atmospheric layer where weather occurs.
Answer:
a) V_f = 25.514 m/s
b) Q =53.46 degrees CCW from + x-axis
Explanation:
Given:
- Initial speed V_i = 20.5 j m/s
- Acceleration a = 0.31 i m/s^2
- Time duration for acceleration t = 49.0 s
Find:
(a) What is the magnitude of the satellite's velocity when the thruster turns off?
(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.
Solution:
- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:
V_f = V_i + a*t
V_f = 20.5 j + 0.31 i *49
V_f = 20.5 j + 15.19 i
- The magnitude of the velocity vector is given by:
V_f = sqrt ( 20.5^2 + 15.19^2)
V_f = sqrt(650.9861)
V_f = 25.514 m/s
- The direction of the velocity vector can be computed by using x and y components of velocity found above:
tan(Q) = (V_y / V_x)
Q = arctan (20.5 / 15.19)
Q =53.46 degrees
- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.
The use of force to move an object is called work. This only applies if the object moves.
Answer:
66w
Explanation:
p=w/t
p=660/10
p=66
prolly a bad explanation but hope it helps...
Answer:
The equation for the object's displacement is 
Explanation:
Given:
m = 16 lb
δ = 3 in
The stiffness is:
The angular speed is:
The damping force is:
Where
FD = 20 lb
u = 4 ft/s = 48 in/s
Replacing:
The critical damping is equal:
Like cc>c the system is undamped
The equilibrium expression is:
