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Hunter-Best [27]
3 years ago
15

Technician A says that after replacing a power steering hose, the system should be flushed, refilled, and bled. Technician B say

s to replace any O-rings
when installing a new hose. Who is correct?
A. Technician A only
B. Technician B only
C. Both Technician
A and B
D. Neither Technician
A nor B
Engineering
1 answer:
dedylja [7]3 years ago
3 0
The. Answer will be D
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Two parallel Rivers (A and B) are separated by confined and unconfined aquifer estimate the RATE of seepage of river A to River
gizmo_the_mogwai [7]

Answer:

Explanation:

A confined aquifer is one that's impermeable and does not allow seepage's  of water through it.

An unconfined aquifer is a body of water that has permeable membranes which allows passage of water through it.

There will be little or no transmission of water trough these bodies of water.

3 0
3 years ago
For powder compaction using a single-action punch, derive an expression for the distribution of axial pressure within a die of r
Zielflug [23.3K]

Answer:

a) 2∪p/lb (l+b)dH

b) po exp( 4∪x/l)

Explanation:

please check the attachment for proper explanation and proper sign notations thanks.

3 0
3 years ago
Explain briefly the power analogy between the fluid “water” and the Word of God.
melomori [17]

Throughout the Word of God, water is a symbol, a "type," of the spirit of God, and is ... to represent holy spirit, that is, the divine nature and power of God. ... The average person can go about 60 days without food before he ... record illustrating the analogous relationship between water and the holy spirit.

3 0
3 years ago
If the maximum allowable shear stress is 70 MPa, find the shaft diameter needed to transmit 40 kW when the shaft speed is 250 rp
victus00 [196]

Answer:

The diameter is 50mm

Explanation:

The answer is in two stages. At first the torque (or twisting moment) acting on the shaft and needed to transmit the power needs to be calculated. Then the diameter of the shaft can be obtained using another equation that involves the torque obtained above.

T=(P×60)/(2×pi×N)

T is the Torque

P is the the power to be transmitted by the shaft; 40kW or 40×10³W

pi=3.142

N is the speed of the shaft; 250rpm

T=(40×10³×60)/(2×3.142×250)

T=1527.689Nm

Diameter of a shaft can be obtained from the formula

T=(pi × SS ×d³)/16

Where

SS is the allowable shear stress; 70MPa or 70×10⁶Pa

d is the diameter of the shaft

Making d the subject of the formula

d= cubroot[(T×16)/(pi×SS)]

d=cubroot[(1527.689×16)/(3.142×70×10⁶)]

d=0.04808m or 48.1mm approx 50mm

7 0
3 years ago
Compute the atomic density (the number of atoms per cm3 rather than mass density g/cm3) for a perfect crystal of silicon at room
Ann [662]

Answer:

5E22 atoms/cm³

Explanation:

We need to find the number of moles of silicon per cm³

number of moles per cm³ = density/atomic weight = 2.33 g/cm ÷ 28.09 g/mol = 0.083 mol/cm³.

Since there are 6.022 × 10²³ atoms/mol, then the number of atoms of silicon per cm³ = number of atoms per mol × number of moles per cm³

= 6.022 × 10²³ atoms/mol × 0.083 mol/cm³

= 0.4995 × 10²³ atoms/cm³

= 4.995 × 10²² atoms/cm³

≅ 5 × 10²² atoms/cm³

= 5E22 atoms/cm³

5 0
3 years ago
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