Answer:
a)
v = 14.1028 m/s
∅ = 83.0765° north of east
b)
the required distance is 40.98 m
Explanation:
Given that;
velocity of the river u = 1.70 m/s
velocity of boat v = 14.0 m/s
Now to get the velocity of the boat relative to shore;
( north of east), we say
a² + b² = c²
(1.70)² + (14.0)² = c²
2.89 + 196 = c²
198.89 = c²
c = √198.89
c = 14.1028 m/s
tan∅ = v/u = 14 / 1.7 = 8.23529
∅ = tan⁻¹ ( 8.23529 ) = 83.0765° north of east
Therefore, the velocity of the boat relative to shore is;
v = 14.1028 m/s
∅ = 83.0765° north of east
b)
width of river = 340 m,
ow far downstream has the boat moved by the time it reaches the north shore in meters = ?
we say;
340sin( 90° - 83.0765°)
⇒ 340sin( 6.9235°)
= 40.98 m
Therefore, the required distance is 40.98 m
J can get answer on this way:
Ek=m*V*V/2= (24kg*2m/s*2m/s)/2=48 Ј
Some examples of stable system are:
1) functions of sine
2) DC
3) signum
4) unit step
5) cosine.
Happy Studying! ^^
As it is given that the air bag deploy in time
total distance moved by the front face of the bag
Now we will use kinematics to find the acceleration
now as we know that
so we have
so the acceleration is 400g for the front surface of balloon
Answer:
300 Pascal
Explanation:
Given
weight or force (F) = 6000 N
area (A) = 20 m²
pressure (p) = ?
we know
the force acting normally per unit area is pressure. So
P = F / A
= 6000 / 20
= 300 Pascal
Hope it will help :)
