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3 years ago

Witch measuring tool would be used to determine the diameter of a crankshaft journal

1 answer:

3 0

Answer =

dial bore gauge

a “dial bore gauge” measures the inside of round holes, such as the bearing journals . can mesure up to 2” and 6” diameter holes .

when ( “ ) is next to a number it means inches fwi - but hope this helped have a good day :)

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The Canadair CL-215T amphibious aircraft is specially designed to fight fires. It is the only production aircraft that can scoop

ioda

Answer:

Determine the added thrust required during water scooping, as a function of aircraft speed, for a reasonable range of speeds.= 132.26∪

Explanation:

check attached files for explanation

6 0

3 years ago

A 50 mm 45 mm 20 mm cell phone charger has a surface temperature of Ts 33 C when plugged into an electrical wall outlet but not

romanna [79]

Answer:

C = $0.0032 per day

Explanation:

We are given;

Dimension of cell phone; 50 mm × 45 mm × 20 mm

Temperature of charger; T1 = 33°C = 306K

Emissivity; ε = 0.92

convection heat transfer coefficient; h = 4.5 W/m².K

Room air temperature; T∞ = 22°C = 295K

Wall temperature; T2 = 20°C = 293 K

Cost of electricity; C = $0.18/kW.h

Chargers are usually in the form of a cuboid, and thus, surface Area is;

A = (50 × 45) + 2(50 × 20) + 2(45 × 20)

A = 6050 mm²

A = 6.05 × 10^(-3) m²

Formula for total heat transfer rate is;

E_t = hA(T1 - T∞) + εσA((T1)⁴ - (T2)⁴)

Where σ is Stefan Boltzmann constant with a value of; σ = 5.67 × 10^(-8) W/m².K⁴

Thus;

E_t = 4.5 × 6.05 × 10^(-3) (306 - 295) + (0.92 × 6.05 × 10^(-3) × 5.67 × 10^(-8)(306^(4) - 293^(4)))

E_t = 0.7406 W = 0.7406 × 10^(-3) KW

Now, we know C = $0.18/kW.h

Thus daily cost which has 24 hours gives;

C = 0.18 × 0.7406 × 10^(-3) × 24

C = $0.0032 per day

6 0

3 years ago

For a Cu-Ni alloy containing 53 wt.% Ni and 47 wt.% Cu at 1300°C, calculate the wt.% of the alloy that is solid and wt.% of allo

Zina [86]

Answer:

Hello your question is incomplete attached below is the complete question

answer: wt.% of alloy that is solid = 61.5%

wt.% of allot that is liquid = 38.5%

Explanation:

To determine the wt.% of the alloy that is solid

= $\frac{R}{R +S } * 100$

= $\frac{53-45}{58-45} * 100$ = 61.5%

To determine the wt.% of the alloy that is liquid

= $\frac{S}{S+R} * 100\\$

= $\frac{58-53}{58-45} *100$ = 38.5%

attached below is a free hand sketch as well

7 0

3 years ago

Each of the following activities are commonly performed during the implementation of the Database Life Cycle (DBLC). Fill in the

kicyunya [14]

Yessiree I agree with yu cause yu are right

4 0

3 years ago

5. One of the major road blocks for direct DC power in residences is the ___. A) generation of DC B) lack of standardization C)

Montano1993 [528]

Answer:

D) quantity of components required for this type of system

Explanation:

Electricity can be transmitted using the alternating and direct currents. The alternating current is one in which the flow of the current diverts at certain time intervals whereas, the direct current is one in which there is a constant one-directional flow of current. The DC is used in batteries and solar panels. Residential areas and business places make use of the AC current.

One of the several reasons why the DC is not used in homes is because unlike the AC it is not easy to build and sustain. Moreso, it has more components compared to the AC. For example, its motor consists of brushes and commutators . The components, example, switches are also large compared to the AC components.

6 0

4 years ago