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:
42.50 dB
Explanation:
Determine the minimum voltage gain
amplitude of input signal ( Vi ) = 15 mV
amplitude of output signal ( Vo) = 2 V
Vo = 2 v
therefore ; minimum gain = Vo / Vi = 2 / ( 15 * 10^-3 )
= 133.33
Minimum gain in DB = 20 log ( 133.33 )
= 42.498 ≈ 42.50 dB
Answer:
please help you are not the intended recipient
Answer:
Production Function : TFSq = f { ingredient 1 ...... ing. i }
Increasing/ Constant/ Increasing Returns to Scale :
Output change > / = / < Input change respectively
Explanation:
Production Function is the relationship between production inputs & outputs, given technology. It denotes the maximum output that can be generated with given inputs.
Tutti Frutti Smoothie [TFS] quantity = Function of {Ingredient1.....ingredient i}
Returns to Scale represents change in output when all inputs change in same proportion.
- Constant Returns to Scale [CRS] : Output Change = All inputs change
- Increasing Returns to Scale [IRS] : Output Change > All inputs change
- Negative Returns to Scale [NRS] : Output Change < All inputs change
When all inputs (ingredients) change by same proportion i.e get twice 2X :- If output of Tutti Frutti Smoothie increases by > 2X i.e 3X - IRS. If it increases equal ie 2X - CRS. If it increases lesser i.e 1.5X - CRS.
The correct question;
An object of irregular shape has a characteristic length of L = 1 m and is maintained at a uniform surface temperature of Ts = 400 K. When placed in atmospheric air at a temperature of Tinfinity = 300 K and moving with a velocity of V = 100 m/s, the average heat flux from the surface to the air is 20,000 W/m² If a second object of the same shape, but with a characteristic length of L = 5 m, is maintained at a surface temperature of Ts = 400 K and is placed in atmospheric air at Too = 300 K, what will the value of the average convection coefficient be if the air velocity is V = 20 m/s?
Answer:
h'_2 = 40 W/K.m²
Explanation:
We are given;
L1 = 1m
L2 = 5m
T_s = 400 K
T_(∞) = 300 K
V = 100 m/s
q = 20,000 W/m²
Both objects have the same shape and density and thus their reynolds number will be the same.
So,
Re_L1 = Re_L2
Thus, V1•L1/v1 = V2•L2/v2
Hence,
(h'_1•L1)/k1 = (h'_2•L2)/k2
Where h'_1 and h'_2 are convection coefficients
Since k1 = k2, thus, we now have;
h'_2 = (h'_1(L1/L2)) = [q/(T_s - T_(∞))]• (L1/L2)
Thus,
h'_2 = [20,000/(400 - 300)]•(1/5)
h'_2 = 40 W/K.m²
