Answer:
13177.34 J
Explanation:
Work done = force × distance
work done by the engine = kinetic energy + potential energy + work done friction
kinetic energy due to the car's speed = 1/2mv² = 4468.5 J
potential energy due to the height = mgh = 993 kg × 9.8 m/s² × 0.6 m = 5838.84 J
work done by friction = 2870 J
work done by engine = 5838.84 J + 2870 J + 4468.5 J = 13177.34 J
Answer:B
Explanation:
Given
For motor A
Characteristic life(r)=4100 hr
MTTF=4650 hrs
shape factor(B )=0.8
For motor B
Characteristic life(r)=336 hr
MTTF=300 hr
Shape Factor (B)=3
Reliability for 100 hours



For B


B is better for 100 hours
(b)For 750 hours


So here B is more Reliable.
Answer:
than look inside it
Explanation:
well if you need to inspect something, looking is very important
Answer:
The thing that separates the work of technology transfer research from implementation of the products of such research is:
Technology transfer research looks at how technology may be transferred but does not actually make the transfer.
Explanation:
This suggests that technology transfer research is different from the technology transfer (implementation) itself. The first stops at making scientific investigations into technology transfer activities while the next step performs the actual transfer or implementation. In other words, technology transfer conveys the results of scientific and technological research to the marketplace and to the wider society. It is the bridge-builder between the research and the implementation.
Answer:
T_{f} = 90.07998 ° C
Explanation:
This is a calorimetry process where the heat given by the Te is absorbed by the air at room temperature (T₀ = 25ºC) with a specific heat of 1,009 J / kg ºC, we assume that the amount of Tea in the cup is V₀ = 100 ml. The bottle being thermally insulated does not intervene in the process
Qc = -Qb
M
(T₁ -
) = m
(T_{f}-T₀)
Where M is the mass of Tea that remains after taking out the cup, the density of Te is the density of water plus the solids dissolved in them, the approximate values are from 1020 to 1200 kg / m³, for this calculation we use 1100 kg / m³
ρ = m / V
V = 1000 -100 = 900 ml
V = 0.900 l (1 m3 / 1000 l) = 0.900 10⁻³ m³
V_air = 0.100 l = 0.1 10⁻³ m³
Tea Mass
M = ρ V_te
M = 1100 0.9 10⁻³
M = 0.990 kg
Air mass
m = ρ _air V_air
m = 1.225 0.1 10⁻³
m = 0.1225 10⁻³ kg
(m c_{e_air} + M c_{e_Te}) T_{f}. = M c_{e_Te} T1 - m c_{e_air} T₀
T_{f} = (M c_{e_Te} T₁ - m c_{e_air} T₀) / (m c_{e_air} + M c_{e_Te})
Let's calculate
T_{f} = (0.990 1100 90.08– 0.1225 10⁻³ 1.225 25) / (0.1225 10⁻³ 1.225 + 0.990 1100)
T_{f} = (98097.12 -3.75 10⁻³) / (0.15 10⁻³ +1089)
T_{f} = 98097.11 / 1089.0002
T_{f} = 90.07998 ° C
This temperature decrease is very small and cannot be measured