Answer:
S = 0.5 km
velocity of motorist = 42.857 km/h
Explanation:
given data
speed = 70 km/h
accelerates uniformly = 90 km/h
time = 8 s
overtakes motorist = 42 s
solution
we know initial velocity u1 of police = 0
final velocity u2 = 90 km/h = 25 mps
we apply here equation of motion
u2 = u1 + at
so acceleration a will be
a =
a = 3.125 m/s²
so
distance will be
S1 = 0.5 × a × t²
S1 = 100 m = 0.1 km
and
S2 = u2 × t
S2 = 25 × 16
S2 = 400 m = 0.4 km
so total distance travel by police
S = S1 + S2
S = 0.1 + 0.4
S = 0.5 km
and
when motorist travel with uniform velocity
than total time = 42 s
so velocity of motorist will be
velocity of motorist = 
velocity of motorist =
velocity of motorist = 42.857 km/h
Answer:
5.6 mm
Explanation:
Given that:
A cylindrical tank is required to contain a:
Gage Pressure P = 560 kPa
Allowable normal stress
= 150 MPa = 150000 Kpa.
The inner diameter of the tank = 3 m
In a closed cylinder there exist both the circumferential stress and the longitudinal stress.
Circumferential stress 
Making thickness t the subject; we have


t = 0.0056 m
t = 5.6 mm
For longitudinal stress.



t = 0.0028 mm
t = 2.8 mm
From the above circumferential stress and longitudinal stress; the stress with the higher value will be considered ; which is circumferential stress and it's minimum value with the maximum thickness = 5.6 mm
The total number of trips that the vehicle has to make based on the given sequence of operation is 120 trips.
<em>"Your</em><em> </em><em>question is not complete, it seems to be missing the following information;"</em>
The sequence of operation is A - E - D - C - B - A - F
The given parameters;
- <em>number of pieces that will flow from the first machine A to machine F, = 2,000 pieces</em>
- <em>initial unit load specified in the first machine, L₁ = 50</em>
- <em>final unit load, L₂ = 100 </em>
- <em>the capacity of the vehicle = 1 unit load</em>
<em />
The given sequence of operation of the vehicle;
A - E - D - C - B - A - F
<em>the vehicle makes </em><em>6 trips</em><em> for </em><em>100</em><em> unit </em><em>loads</em>
The total number of trips that the vehicle has to make, in order to transport the 2000 pieces of the load given, is calculated as follows.
100 unit loads ----------------- 6 trips
2000 unit loads --------------- ?

Thus, the total number of trips that the vehicle has to make based on the given sequence of operation is 120 trips.
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Answer:
%Reduction in area = 73.41%
%Reduction in elongation = 42.20%
Explanation:
Given
Original diameter = 12.8 mm
Gauge length = 50.80mm
Diameter at the point of fracture = 6.60 mm (0.260 in.)
Fractured gauge length = 72.14 mm.
%Reduction in Area is given as:
((do/2)² - (d1/2)²)/(do/2)²
Calculating percent reduction in area
do = 12.8mm, d1 = 6.6mm
So,
%RA = ((12.8/2)² - 6.6/2)²)/(12.8/2)²
%RA = 0.734130859375
%RA = 73.41%
Calculating percent reduction in elongation
%Reduction in elongation is given as:
((do) - (d1))/(d1)
do = 72.14mm, d1 = 50.80mm
So,
%RA = ((72.24) - (50.80))/(50.80)
%RA = 0.422047244094488
%RA = 42.20%