Start by facing East. Your first displacement is the vector
<em>d</em>₁ = (225 m) <em>i</em>
Turning 90º to the left makes you face North, and walking 350 m in this direction gives the second displacement,
<em>d</em>₂ = (350 m) <em>j</em>
Turning 30º to the right would have you making an angle of 60º North of East, so that walking 125 m gives the third displacement,
<em>d</em>₃ = (125 m) (cos(60º) <em>i</em> + sin(60º) <em>j</em> )
<em>d</em>₃ ≈ (62.5 m) <em>i</em> + (108.25 m) <em>j</em>
The net displacement is
<em>d</em> = <em>d</em>₁ + <em>d</em>₂ + <em>d</em>₃
<em>d</em> ≈ (287.5 m) <em>i</em> + (458.25 m) <em>j</em>
and its magnitude is
|| <em>d</em> || = √[ (287.5 m)² + (458.25 m)² ] ≈ 540.973 m ≈ 541 m
Answer:
Tension= 21,900N
Components of Normal force
Fnx= 17900N
Fny= 22700N
FN= 28900N
Explanation:
Tension in the cable is calculated by:
Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium
FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)
Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)
Ftorque= 2/3FBcostheta+ 4/3FWcostheta
Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°
Ftorque= 21900N
b) components of Normal force
Efx=FNx-FTcos(90-theta)=0 static equilibrium
Fnx=21900cos(90-55)=17900N
Fy=FNy+ FTsin(90-theta)-FB-FW=0
FNy= -FTsin(90-55)+FB+FW
FNy= -21900sin(35)+(1350+2250)×9.81=22700N
The Normal force
FN=sqrt(17900^2+22700^2)
FN= 28.900N
Answer:
<h2>
t = 2.35s</h2>
Explanation:
Using one of the equation of motion expressed as v = u+at where;
v is the final velocity = 2.0m/s
u is the initial velocity = 25m/s
a is the acceleration = -a (since the acceleration is negative)
t is the time taken
Substituting the given parameters into the formula above;
2.0 = 25-9.81t
subtract 25 from both sides
2.0 - 25 = 25-9.81t - 25
-23 = -9.81t
Divide both sides by -9.81
-23/-9.81 = -9.81t/-9.81
t = 23.0/9.81
t = 2.35s
<em>Hence it took the car 2.35s to slow to a final velocity of 2.0 m/s</em>
Answer:
t=40s,
Explanation:
If you can swim in still water at 0.5m/s, the shortest time it would take you to swim from bank to bank across a 20m wide river, if the water flows downstream at a rate of 1.5m/s, is most nearly:
from the question the swimmer will have a velocity which is equal to the sum of the speed of the water and the velocity to swi across the bank
Vt=v1+v2
the time is takes to swim across the bank will be
DY=Dv*t
DY=distance across the bank
Dv=ther velocity of the swimmer across the bank
t=20/ 0.5m/s,
t=40s, time it takes to swim across the bank
velocity is the rate of displacement
displacement is distance covered in a specific direction
