One of the best ways to increase engine power and control detonation and preignition is to?

Three capillary tubes with different radius (r1=1.0mm, r2=0.1mm, r3=0.01mm) are inserted into the same cup of water. The surface

kobusy [5.1K]

Answer:

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Explanation:

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4 0

3 years ago

a buyer can purchase 70 screwdrivers ten 4-inch length twelve 6 inch length twenty 8-inch length are needed. how many heavy 24-i

jasenka [17]

Answer:

28 , 24-inch screwdrivers

Explanation:

The total number of screwdrivers that can be purchased is = 70

4 - inch length screwdrivers = 10

6- inch length screwdrivers = 12

8- inch length screwdrivers = 20

Total = 20 +12 +10 = 42

Remaining = 70-42 = 28

So, heavy 24-inch screwdrivers = 28

3 0

3 years ago

What would happen if an exposed film was accidentally placed in the fixer before being placed in the developer

Eduardwww [97]

Do you still want this answered

5 0

3 years ago

A 1020 CD steel shaft is to transmit 15 kW while rotating at 1750 rpm. Determine the minimum diameter for the shaft to provide a

vladimir2022 [97]

Answer:

diameter is 14 mm

Explanation:

given data

power = 15 kW

rotation N = 1750 rpm

factor of safety = 3

to find out

minimum diameter

solution

we will apply here power formula to find T that is

power = 2π×N×T / 60 .................1

put here value

15 × $10^{3}$ = 2π×1750×T / 60

so

T = 81.84 Nm

and

torsion = T / Z ..........2

here Z is section modulus i.e = πd³/ 16

so from equation 2

torsion = 81.84 / πd³/ 16

so torsion = 416.75 / / d³ .................3

so from shear stress theory

torsion = σy / factor of safety

so here σy = 530 for 1020 steel

so

torsion = σy / factor of safety

416.75 / d³ = 530 × $10^{6}$ / 3

so d = 0.0133 m

so diameter is 14 mm

3 0

3 years ago

1. Calculate the battery life in years when a pacemaker has the following characteristics: Battery Ampere-hours = 1.5 Pulse volt

Wittaler [7]

Answer:

battery life in year = 9 years and 48 days

Explanation:

given data

Battery Ampere-hours = 1.5

Pulse voltage = 2 V

Pulse width = 1.5 m sec

Pulse time period = 1 sec

Electrode heart resistance = 150 Ω

Current drain on the battery = 1.25 µA

to find out

battery life in years

solution

we get first here duty cycle that is express as

duty cycle = $\frac{width}{period}$ ...............1

duty cycle = 1.5 × $10^{-3}$

and applied voltage will be

applied voltage = duty energy × voltage ...........2

applied voltage = 1.5 × $10^{-3}$ × 2

applied voltage = 3 mV

so current will be

current = $\frac{applied\ voltage}{resistance}$ ................3

current = $\frac{3}{150}$

current = 20 µA

so net current will be

net current = 20 - 1.25

net current = 18.75 µA

so battery life will be

battery life = $\frac{1.5}{18.75*10^{-6}}$

battery life = 80000 hours

battery life in year = $\frac{80000}{8760}$

battery life in year = 9.13 years

battery life in year = 9 years and 48 days

4 0

3 years ago