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Kobotan [32]
3 years ago
13

Introduction to gear and gear ratios: If you have two gears of the same size does the output speed increase, decrease, or remain

the same constant?
Engineering
1 answer:
creativ13 [48]3 years ago
4 0

Answer:

remain the same constant

Explanation:

if a small gear is making 5 rpm turning a big gear thats making 1 rpm the bug gear is making lots of torque but if its going the opposite way its much harder to turn the big gear 1 rpm just to make the little gear do 5 rpm and it makes way less torque if thre both the same size there ging to stay the same speed

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Ohm's law states that the current (I) in amps equals the voltage (E) in volts decided by the resistance (R) in ohm's. If you con
Hitman42 [59]

Answer:

1.14 * 10^-3  amperes

Explanation:

According to ohms law

V = IR   ---- ( 1 )

V = voltage  = 2.4 * 10^3 v

I = current  = ?

R = resistance =  2.1 * 10^6 Ω

from equation 1

I = V / R

 = ( 2.4 * 10^3 ) / (2.1 * 10^6 )

 = 1.14 * 10^-3  amperes

7 0
3 years ago
A rod is 2m long at temperature of 10oC. Find the expansion of the rod, when the temperature is raised to 80oC. If this expansio
Damm [24]

Answer:

ΔL = 1.68 mm

σ = 84 MPa

Explanation:

Thermal expansion is:

ΔL = α ΔT L

Thermal stress is:

σ = α ΔT E

Given:

α = 1.2×10⁻⁵ /°C

E = 1.0×10⁵ MPa

ΔT = 80°C − 10°C = 70°C

L = 2 m

ΔL = (1.2×10⁻⁵ /°C) (70°C) (2 m)

ΔL = 0.00168 m

ΔL = 1.68 mm

σ = (1.2×10⁻⁵ /°C) (70°C) (1.0×10⁵ MPa)

σ = 84 MPa

8 0
3 years ago
When storing used oil, it need to be kept in________ container?
lisov135 [29]

Answer: In a clean plastic or metal container with a tightly sealed lid

6 0
3 years ago
Suppose an assembly requires five components from five different vendors. To guarantee starting the assembly on time with 90 per
spayn [35]

Answer:

The service level for each component must be 97.91%

Explanation:

If we want a 90% confidence of starting on time, that means we need

P_{\mbox{starting on time}}=P_{\mbox{every component being ready on time}}=0.9\\

As the probability of each component being ready is independent from the others, that means that the probability of the 5 components being ready is equal to multiply each probability:

0.9=P_{\mbox{component 1 ready on time} } * P_{\mbox{component 2 ready on time} } *\\ P_{\mbox{component 3 ready on time} } * P_{\mbox{component 4 ready on time} } *\\P_{\mbox{component 5 ready on time} }

The probability of being ready on time is equal to the service level (in fraction), and all 5 are equal so we can say:

0.9=(\mbox{service level(in fraction)})^5\\\\\sqrt[5]{0.9} =\mbox{service level(in fraction)}=0.9791\\\mbox{In percentage}: \mbox{service level (in fraction)}*100 = 97.91\%

6 0
3 years ago
A piston/cylinder contains 1.5 kg of water at 200 kPa, 150°C. It is now heated by a process in which pressure is linearly relate
Fofino [41]

Answer:

final volume V2 = 0.71136 m³

work done in process W = -291.24 kJ

heat transfer Q = 164 kJ

Explanation:

given data

mass = 1.5 kg

pressure p1 = 200 kPa

temperature t1 = 150°C

final pressure p2 = 600 kPa

final temperature t2 = 350°C

solution

we will use here superheated water table that is

for pressure 200 kPa and 150°C temperature

v1 = 0.95964 m³/kg

u1 = 2576.87 kJ/kg

and

for pressure 600 kPa and 350°C temperature

v2 = 0.47424 m³/kg

u2 = 2881.12 kJ/kg

so v1 is express as

V1 = v1 × m    ............................1

V1 = 0.95964 × 1.5

V1 = 1.43946 m³

and

V2 = v2 × m    ............................2

V2 = 0.47424 × 1.5

final volume V2 = 0.71136 m³

and

W = P(avg) × dV      .............................3

P(avg) = \frac{p1+p2}{2}    = \frac{200+600}{2} = 400 × 10³

put here value

W = 400 × 10³ × (0.71136 - 1.43946 )

work done in process W = -291.24 kJ

and

heat transfer is

Q = m × (u2 - u1)  + W       .............................4

Q = 1.5 × (2881.12 - 2576.87)  + 292.24

heat transfer Q = 164 kJ

7 0
3 years ago
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