Answer:
The planes’ acceleration from A to B is 500m/s^2
Explanation:
Given that the initial velocity u is 8000m/s
and also given the final velocity v=10,000 m/s
the time taken to move from A to B = 40 second
The acceleration is defined as the rate of change of velocity with time
we know that the expression for acceleration is given as
a=(v-u)/t
substituting our given data into the expression for a we have
a=(10000-8000)/40
a=2000/40
a=500m/s^2
The planes’ acceleration from A to B is 500m/s^2
The period T is time it takes for one complete cycle or from "trough to trough" so the reverse is trough per sec = 1/T = frequency
Answer:25000/23 N/M^2
Explanation:
P=f/A=30/(0.23×0.12)=25000/23
Answer:
a) load in Newton is 96,138 b) 129.314mm
Explanation:
Stress = force/ area (cross sectional area of the bronze)
Force(load) = 294*10^6*327*10^-6 = 96138N
b) modulus e = stress/ strain
Strain = stress/ e = (294*10^6)/ (121*10^ 9) = 2.34* 10^ -3
Strain = change in length/ original length = DL/ 129
Change in length DL = 129 * 2.34*10^ -3 = 0.31347
Maximum length = change in length + original length = 129.314mm
Answer:
As given that the car maintains a constant speed v as it traverses the hill and valley where both the valley and hill have a radius of curvature R.
(i) At point C, the normal force acting on the car is largest because the centripetal force is up. gravity is down and normal force is up. net force is up so magnitude of normal force must be greater than the car's weight.
(ii) At point A, the normal force acting on the car is smallest because the centripetal force is down. gravity is down and normal force is up. net force is up so magnitude of normal force must be less than car's weight.
(iii) At point C, the driver will feel heaviest because the driver's apparent weight is the normal force on her body.
(iv) At point A, the driver will feel the lightest.
(v)The car can go that much fast without losing contact with the road at A can be determined as follow:
Fn=0 - lose contact with road
Fg= mv²/r
mg=mv²/r
v=sqrt (gr)
