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son4ous [18]
3 years ago
6

1.0•10^-10 standard form

Engineering
1 answer:
Drupady [299]3 years ago
3 0

Answer:

1.0 * 10^{-10} = 0.0000000001

Explanation:

Given

1.0 * 10^{-10}

Required

Convert to standard form

1.0 * 10^{-10}

From laws of indices

a^{-x} = \frac{1}{a^x}

So, 1.0 * 10^{-10} is equivalent to

1.0 * 10^{-10} = 1.0 * \frac{1}{10^{10}}

1.0 * 10^{-10} = 1.0 * \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}

1.0 * 10^{-10} = 1.0 * \frac{1}{10000000000}

1.0 * 10^{-10} = 1.0 * 0.0000000001

1.0 * 10^{-10} = 0.0000000001

Hence, the standard form of 1.0 * 10^{-10} is 0.0000000001

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Answer:

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Explanation:

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3 years ago
For a 68 wt% Zn-32 wt% Cu alloy, make schematic sketches of the microstructure that would be observed for conditions of very slo
Rina8888 [55]

Answer:

-At 1000°C, liquid phase at 68 wt% Zn

-At 760°C, liquid+γ phase, liquid at 74 wt% and 66 wt% γ phase

-At 600°C, γ phase

-At 400°C, γ phase

Explanation:

The phase diagram will be used for the combination of the copper-zinc alloy, first 68% by weight of Zn will be located on the axis corresponding to the composition. Draw a vertical line upwards and go through the temperatures given by the exercise 1000, 760, 600 and 400 ° C. Later you have to know the phases for each of the temperatures. Draw a horizontal line within each phase and look at the adjacent phases and the composition of the alloy. For each of the temperatures, the phases are in response and the drawings are attached to the image.

6 0
4 years ago
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
Given the circuit at the right in which the following values are used: R1 = 20 kΩ, R2 = 12 kΩ, C = 10 µ F, and ε = 25 V. You clo
agasfer [191]

Answer:

a.) I = 7.8 × 10^-4 A

b.) V(20) = 9.3 × 10^-43 V

Explanation:

Given that the

R1 = 20 kΩ,

R2 = 12 kΩ,

C = 10 µ F, and

ε = 25 V.

R1 and R2 are in series with each other.

Let us first find the equivalent resistance R

R = R1 + R2

R = 20 + 12 = 32 kΩ

At t = 0, V = 25v

From ohms law, V = IR

Make current I the subject of formula

I = V/R

I = 25/32 × 10^3

I = 7.8 × 10^-4 A

b.) The voltage across R1 after a long time can be achieved by using the formula

V(t) = Voe^- (t/RC)

V(t) = 25e^- t/20000 × 10×10^-6

V(t) = 25e^- t/0.2

After a very long time. Let assume t = 20s. Then

V(20) = 25e^- 20/0.2

V(20) = 25e^-100

V(20) = 25 × 3.72 × 10^-44

V(20) = 9.3 × 10^-43 V

8 0
3 years ago
It is desired to produce and aligned carbon fiber-epoxy matrix composite having a longitudinal tensile strength of 610 MPa. Calc
julia-pushkina [17]

Answer:

volume fraction of fibers is 0.4

Explanation:

Given that for the aligned carbon fiber-epoxy matrix composite:

Diameter (D) = 0.029 mm

Length (L) = 2.3 mm

Tensile strength (\sigma_{cd}) = 610 MPa

fracture strength (\sigma_f) = 5300 MPa

matrix stress (\sigma_m) =  17.3 MPa

fiber-matrix bond strength (\tau_c) = 19 MPa

The critical length is given as:

L_C=\sigma_f(\frac{D}{2\tau_c} )=5300*10^6(\frac{0.029*10^{-3}}{19*10^6} )=8.1*10^{-3}=8.1mm

Since the critical length is greater than the length, the aligned carbon fiber-epoxy matrix composite can be produced.

The longitudinal strength is given by:

\sigma_{cd}=\frac{L*\tau_c}{D} .V_f+\sigma_m(1-V_f)

making Vf the subject of the formula:

V_f=\frac{\sigma_{cd}-\sigma_m}{\frac{L*\tau_c}{D} -\sigma_m}

Vf is the volume fraction of fibers.

Therefore:

V_f=\frac{\sigma_{cd}-\sigma_m}{\frac{L*\tau_c}{D} -\sigma_m}=\frac{610*10^6-17.3*10^6}{\frac{2.3*10^{-3}*19*10^6}{0.029*10^{-3}}-17.3*10^6} } =0.4

volume fraction of fibers is 0.4

8 0
3 years ago
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