Answer:
1450.4 KN
Explanation:
Pressure = ρhg
where: ρ is the density of the liquid, h is the height and g the force of gravity.
Total pressure exerted by the liquids at the base = Pressure of oil + Pressure of water + Pressure of mercury
So that,
i. Pressure of oil = ρhg
(ρ = 0.8 g/cm³ = 800 kg/m³)
= 800 x 5 x 9.8
= 39200
Pressure of oil = 39200 N
ii. Pressure of water = ρhg
(ρ = 1 g/cm³ = 1000 kg/m³)
= 1000 x 8 x 9.8
= 78400
Pressure of water = 78400 N
ii. Pressure of mercury = ρhg
(ρ = 13.6 g/cm³ = 13600 kg/m³)
= 13600 x 10 x 9.8
= 1332800
Pressure of mercury = 1332800 N
So that,
Total pressure exerted by the liquids at the base = 39200 + 78400 + 1332800
= 1450400
= 1450.4 KN
Total pressure exerted by the liquids at the base is 1450.4 KN
.
Answer:
20.35m per minute
Or
1,221.43 per hour
100/5=20
200/7= 28.57
50/4= 12.5
20+28.57+12.5= 61.07
61.07/3= 20.35m per minute
Or
20.35 x 60= 1,221.43m per hour
Answer:
Req = 564 Ω
Explanation:
The equivalent resistance between R1 and R2:
1/R =1/R1 + 1/R2
1/R =1/960 + 1/640
1/R = 1/384
R = 384
Now, the equivalent resistance between R and R3:
Req = 384 + 180
Req = 564 Ω
Answer:
b. 12.5 mAs, 70 kVp
Explanation:
The given parameter are;
The initial exposure factors := 10 mAs and 70 kVp
The initial Grid Ratio, G.R.₁ = 8:1
The Grid Ratio with which the radiographer desires to increase the scatter absorption, G.R.₂ = 12:1
Given that the lead content in the 12:1 grid, is higher than the lead content in 8:1 grid and that 12:1 grid needs more mAs to compensate, and provides a higher image contrast, the amount of extra mAs is given by the Grid Conversion Factors, GCF, as follows;
The GCF for G.R. 8:1 = 4
The GCF for G.R. 12:1 = 5
Therefore, given that the mAs used by the radiographer for 8:1 Grid Ratio is 10 mAs, the mAs required for a G.R. of 12:1 in order to maintain the same exposure is given as follows;
mAs for G.R. of 12:1 = 10 mAs × 5/4 = 12.5 mAs
Therefore the new exposure factors are;
12.5 mAs, 70 kVp