Answer:
A _______________ is a diagram that shows the relationship of one variable to another.
Explanation:
Radial acceleration is given by
where
then
Now
Using the relation
Putting into rpm
PWC operators must keep in mind that a jet drive needs moving water through the drive nozzle for maneuverability. In other term, you must have power applied in order to uphold steering control. If the engine shuts off during operation or If you release the throttle to idle, you will lose all steering control. In either situation, the PWC will remain in the direction it was headed before the throttle was released or the engine was shut off. Action of the steering control will have no outcome. If you are approaching a dock, shore, or other vessel at a speed bigger than you can control and you release the throttle to idle or shut off the engine, you won’t have maneuvering capability and the PWC will stay its forward movement.
Answer:
F= 4788 N
Explanation:
Because the car moves with uniformly accelerated movement we apply the following formula:
vf²=v₀²+2*a*d Formula (1)
Where:
d:displacement in meters (m)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Data
d=36.9 m
v₀=14.0 m/s m/s
vf= 0
Calculating of the acceleration of the car
We replace dta in the formula (1)
vf²=v₀²+2*a*d
(0)²=(14)²+2*a*(36.9)
-(14)²= (73.8) *a
a= - (196) / (73.8)
a= - 2.66 m/s²
Newton's second law of the car in direction horizontal (x):
∑Fx = m*ax Formula (2)
∑F : algebraic sum of the forces in direction x-axis (N)
m : mass (kg)
a : acceleration (m/s²)
Data
m=1800 Fkg
a= - 2.66 m/s²
Magnitude of the horizontal net force (F) that is required to bring the car to a halt in a distance of 36.9 m :
We replace data in the formula (2)
-F= (1800 kg) * ( -2.66 m/s²
)
F= 4788 N
Answer:
Potential gradient = 1.6 v/m
E.M.F. = potential gradient × balancing length 1.2 = 1.6 × l
l=1.2\1.6
=3\4
3\4=0.75m
convert m into cm 0.75x10
0.75m=75 cm
ans=75 cm
Explanation:
