It's just speed, distance, time
Speed = D/T
Distance = SxT
Time = D/S
Therefore Speed (or velocity) = D/T = 55/1.4 = 39 MPH rounded up - FOR JOURNEY 1
Velocity = D/T = 55/2 = 22.5 MPH - FOR JOURNEY 2
To work out his average speed (velocity) you add these together and divide by 2
Average = (39+22.5)/2
Answer = 30.75MPH or 31MPH rounded up
Hope this helps :) + If I'm wrong I'm sorry lol
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:
The kinetic energy at a displacement of half the amplitude is 37.5 J
Explanation:
Given;
total energy on the spring, E = 50 J
When the displacement is half the amplitude, the total energy in the spring is sum of the kinetic energy and elastic potential energy.
E = K + U
Where;
K is the kinetic energy
U is the elastic potential energy
K = E - U
K = E - ¹/₂KA²
When the displacement is half = ¹/₂(A) = A/₂
K = E - ¹/₂K(A/₂)²
K = E - ¹/₂K(A²/₄)
K = E - ¹₄(¹/₂KA²)
Recall, E = ¹/₂KA²
K = ¹/₂KA² - ¹₄(¹/₂KA²) (recall from simple arithmetic, 1 - ¹/₄ = ³/₄)
K = 1(¹/₂KA²) - ¹₄(¹/₂KA²) = ³/₄(¹/₂KA²)
K = ³/₄(¹/₂KA²)
But E = ¹/₂KA² = 50J
K = ³/₄ (50J)
K = 37.5 J
Therefore, the kinetic energy at a displacement of half the amplitude is 37.5 J