Answer:
2132hp ed enregia
e
Explanation:
dawdsawdsawdsawdsawdsawdsawdaas
Answer:
The smaller gear will rotate faster.
Explanation:
If a larger gear is driven by a smaller gear, the large gear will rotate slower than the smaller gear but will have a greater moment. For example, a low gear on a bike or car. If a smaller gear is driven by a larger gear, the smaller gear will rotate quicker than the larger gear but will have a smaller moment.
I hope this helps! :)
Answer:
The time necessary to purge 95% of the NaOH is 0.38 h
Explanation:
Given:
vfpure water(i) = 3 m³/h
vNaOH = 4 m³
xNaOH = 0.2
vfpure water(f) = 2 m³/h
pwater = 1000 kg/m³
pNaOH = 1220 kg/m³
The mass flow rate of the water is = 3 * 1000 = 3000 kg/h
The mass of NaOH in the solution is = 0.2 * 4 * 1220 = 976 kg
When the 95% of the NaOH is purged, thus the NaOH in outlet is = 0.95 * 976 = 927.2 kg
The volume of NaOH in outlet after time is = 927.2/1220 = 0.76 m³
The time required to purge the 95% of the NaOH is = 0.76/2 = 0.38 h
Answer:
45.3 MN
Explanation:
The forging force at the end of the stroke is given by
F = Y.π.r².[1 + (2μr/3h)]
The final height, h is given as h = 100/2
h = 50 mm
Next, we find the final radius by applying the volume constancy law
volumes before deformation = volumes after deformation
π * 75² * 2 * 100 = π * r² * 2 * 50
75² * 2 = r²
r² = 11250
r = √11250
r = 106 mm
E = In(100/50)
E = 0.69
From the graph flow, we find that Y = 1000 MPa, and thus, we apply the formula
F = Y.π.r².[1 + (2μr/3h)]
F = 1000 * 3.142 * 0.106² * [1 + (2 * 0.2 * 0.106/ 3 * 0.05)]
F = 35.3 * [1 + 0.2826]
F = 35.3 * 1.2826
F = 45.3 MN
Answer:
Q = -68.859 kJ
Explanation:
given details
mass 
initial pressure P_1 = 104 kPa
Temperature T_1 = 25 Degree C = 25+ 273 K = 298 K
final pressure P_2 = 1068 kPa
Temperature T_2 = 311 Degree C = 311+ 273 K = 584 K
we know that
molecular mass of 
R = 8.314/44 = 0.189 kJ/kg K
c_v = 0.657 kJ/kgK
from ideal gas equation
PV =mRT






WORK DONE

w = 586*(0.1033 -0.514)
W =256.76 kJ
INTERNAL ENERGY IS



HEAT TRANSFER

= 187.902 +(-256.46)
Q = -68.859 kJ