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3 years ago

As temperature decreases a batteries availability power what

1 answer:

7 0

The answer is increases because when something like that decreases it’s always decreasing that probly makes no sense Imao but it’s increases

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A train travels 650 meters in 25 seconds. What is the train's velocity?

frosja888 [35]

The train is traveling 26 meters A second .

3 0

3 years ago

A cylindrical rod 100 mm long and having a diameter of 10.0 mm is to be deformed using a tensile load of 27,500 N. It must not e

mash [69]

Answer:

The steel is a candidate.

Explanation:

Given

P = 27,500 N

d₀ = 10.0 mm = 0.01 m

Δd = 7.5×10 ⁻³ mm (maximum value)

This problem asks that we assess the four alloys relative to the two criteria presented. The first criterion is that the material not experience plastic deformation when the tensile load of 27,500 N is applied; this means that the stress corresponding to this load not exceed the yield strength of the material.

Upon computing the stress

σ = P/A₀ ⇒ σ = P/(π*d₀²/4) = 27,500 N/(π*(0.01 m)²/4) = 350*10⁶ N/m²

⇒ σ = 350 MPa

Of the alloys listed, the Ti and steel alloys have yield strengths greater than 350 MPa.

Relative to the second criterion, (i.e., that Δd be less than 7.5 × 10 ⁻³ mm), it is necessary to calculate the change in diameter Δd for these four alloys.

Then we use the equation υ = - εx / εz = - (Δd/d₀)/(σ/E)

⇒ υ = - (E*Δd)/(σ*d₀)

Now, solving for ∆d from this expression,

∆d = - υ*σ*d₀/E

For the Aluminum alloy

∆d = - (0.33)*(350 MPa)*(10 mm)/(70*10³MPa) = - 0.0165 mm

0.0165 mm > 7.5×10 ⁻³ mm

Hence, the Aluminum alloy is not a candidate.

For the Brass alloy

∆d = - (0.34)*(350 MPa)*(10 mm)/(101*10³MPa) = - 0.0118 mm

0.0118 mm > 7.5×10 ⁻³ mm

Hence, the Brass alloy is not a candidate.

For the Steel alloy

∆d = - (0.3)*(350 MPa)*(10 mm)/(207*10³MPa) = - 0.005 mm

0.005 mm < 7.5×10 ⁻³ mm

Therefore, the steel is a candidate.

For the Titanium alloy

∆d = - (0.34)*(350 MPa)*(10 mm)/(107*10³MPa) = - 0.0111 mm

0.0111 mm > 7.5×10 ⁻³ mm

Hence, the Titanium alloy is not a candidate.

7 0

3 years ago

The technique of smoothing out joint compound on either side of a joint is known as which of the following

Black_prince [1.1K]

Answer:

a is the answer to your questions

3 0

3 years ago

The flow rate in the pipe system below is 0.05 m3/s. The pressure at point 1 is measured to be 260 kPa. Point 1 is 0.60 m higher

DedPeter [7]

Answer:

Explanation:

The rate of flow in the pipe system in Figure P4.5.2 is 0.05 m3/s. The pressure at point 1 is measured to be 260 kPa. All the pipes are galvanized iron with roughness value of 0.15 mm. Determine the pressure at point 2. Take the loss coefficient for the sudden contraction as 0.05 and v = 1.141 × 10−6 m2/s.

The answer to the above question is

The pressure at point 2 = 75.959 kPa

Explanation:

Bernoulli's equation with losses gives

hL = z₁ - z₃ +(P₁-P₃)/(ρ×g) + (v₁²-v₃²)/(2×g)

Between points 1 and 2, z₁ = z₃ + 0.6 m therefore

hL = 0.6 m +(P₁-P₂)/(ρ×g) + (v₁²-v₃²)/(2×g)

hL = (f₁×L₁×v₁²)/(D₁×2×g) + (f₂×L₂×v₂²)/(D₂×2×g) + (f₃×L₃×v₃²)/(D₃×2×g) + k×V₃₂/(2×g) = 0.6 +(P₁-P₂)/(ρ×g) + (v₁²-v₃²)/(2×g)

But v = Q/A

or since A = π×D²/4 we have

A₁ = 1.77×10-2 m² , A₂ = 5.73×10-2 m², A₃ = 3.8×10-2 m²

Therefore from v = Q/A we have v₁ = 2.83 m/s v₂ = 0.87 m/s and v₃ = 1.315 m/s from there we find the friction coefficient from Moody Diagram as follows

ε = $\frac{Roughness _. value}{ Diameter}$ Which gives

the friction coefficients as f₁ = 0.02, f₂ = 0.017 and f₃ =0.0175

Substituting he above values into the $h_{l}$ equation we get $h_{l}$ = 19.761 m

Combined head loss = 19.761 m

Hence 19.743 m = 0.6 m +(260 kPa-P₃)/(ρ×9.81) + (6.276)/(2×9.81)

or 260 kPa-18.82 m × 9.81 m/s²×ρ= P₃

Where ρ = density of water, we have

260000 Pa - 18.82 m×9.81 m/s²×997 kg/m³ = 75958.598 kg/m·s² = 75.959 kPa

6 0

3 years ago

One of the dispersive components of an optical instrument is a 0.750-meter focal length monochromator equipped with the same gra

SVETLANKA909090 [29]

Answer: (a) 1.11 nm/mm

Explanation:

Monochromator is a device which is used to transmits a delectable band of wavelengths of light wavelengths available at the input.

Assuming focal length monochromator equipped with a 1200-groove/mm grating

The formula for the first-order reciprocal linear dispersion is

$D^{-1} = d/n*F$ where F is the focal length and n belongs to order of the first spectra.

$D^{-1} = \frac{(1/1200) mm * 10^6 (nm/mm)/}{1 * 0.75 m * (103 mm/m) } = 1.11 nm/mm$

Make sure you have used the conversion factors correctly as it will have a major impact on the calculation of the answer.

7 0

3 years ago