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3 years ago

9

There is a dispute between the multiple parties storing financial transaction data on a blockchain over the validity of a transa

ction which happened over a year ago.
Which information would help the disputing parties verify if the data on the blockchain has been tampered with?

1 answer:

spayn [35]3 years ago

5 0

Answer:

dgjkkkkkkkfdcvv

Explanation:

hjklllgfddsssssyjjjkkkk

p.s sorry

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A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 82.4 M P a m. If the plate is expos

bekas [8.4K]

Answer:

Answer for the question is given in the attachment.

Explanation:

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3 years ago

Read 2 more answers

Charra [1.4K]

Answer: Projectile

Explanation:
A projectile is any object that once projected or dropped continues in motion by its own inertia and is influenced only by the downward force of gravity.

8 0

3 years ago

1. What is the linear distance traveled in one revolution of a 9 in diameter wheel?

zmey [24]

Explanation:

1. Circumference is pi times diameter.

C = πd

C = 9π inches

C ≈ 28.3 inches

2. Mechanical advantage is ratio of output force to input force.

MA = Fout / Fin

MA = 155 lb / 15 lb

MA ≈ 10.3

3. Mechanical advantage is ratio of input distance over output distance.

MA = din / dout

MA = 10.5 in / 0.85 in

MA ≈ 12.4

4. Mechanical advantage of pulley system is the number of strands.

MA = 8

5. Mechanical advantage is ratio of output force to input force.

MA = Fout / Fin

8 = Fout / 200 lb

Fout = 1600 lb

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4 years ago

Oil with a density of 850 kg/m3 and kinematic viscosity of 0.00062 m2 /s is being discharged by a 5 mm diameter, 40 m long horiz

ra1l [238]

Answer:

Flow rate is $1.82\times 10^{-8} m^{3}/s$

Explanation:

Given information

Density of oil, $\rho_{oil}= 850 Kg/m^{3}$

kinematic viscosity, $v= 0.00062 m^{2} /s$

Diameter of pipe, D= 5 mm= 0.005 m

Length of pipe, L=40 m

Height of liquid, h= 3 m

Volume flow rate for horizontal pipe will be given by

$\bar v=\frac {\triangle P\pi D^{4}}{128\mu L}$ where $\mu$ is dynamic viscosity and $\triangle P$ is pressure drop

At the bottom of the tank, pressure is given by

$P_{bottom}=\rho_{oil} gh=850 Kg/m^{3}\times 9.81 m/s^{2}\times 3 m= 25015.5 N/m^{2}$

Since at the top pressure is zero, therefore change in pressure a difference between the pressure at the bottom and the top. It implies that change in pressure is still $25015.5 N/m^{2}$

Dynamic viscosity, $\mu=\rho_{oil}v= 850 Kg/m^{3}\times 0.00062 m^{2}/s=0.527 Kg/m.s$

Now the volume flow rate will be

$\bar v=\frac {25015.5 N/m^{2}\times \pi \times 0.005^{4}}{128\times 0.527 Kg/m.s \times 40}=1.82037\times 10^{-8} m^{3}/s\approx 1.82\times 10^{-8} m^{3}/s$

Proof of flow being laminar

The velocity of flow is given by

$V_{flow}=\frac {\bar v}{A}=\frac {1.82\times 10^{-8} m^{3}/s}{0.25\times \pi\times 0.005^{2}}=0.000927104 m/s$

Reynolds number, $Re=\frac {\rho_{oil} v_{flow} D}{\mu}=\frac {850 Kg/m^{3}\times 0.000927104 m/s\times 0.005}{0.527 kg/m.s}=0.007476648$

Since the Reynolds number is less than 2300, the flow is laminar and the assumption is correct.

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4 years ago

10.0 kmol of a 40.0 mol% methanol and 60.0 mol% water mixture is processed in a normal batch distillation system with a still po

Serggg [28]

Answer:

[a]. 0.49.

[2]. 0.536

[c]. 4.15 kmol; 5.84 kmol.

Explanation:

Without mincing words let's dive straight into the solution to the question above.

[a].

The initial external reflux ratio, LD that must be used = [(0.85 - 0.57)/ 0.85 - 0]/ [ 1 - (0.85 - 0.57)/ 0.85 - 0].

The initial external reflux ratio, LD that must be used = 0.329/ 1- 0.329 = 0.49.

[b].

The final external reflux ratio that must be used = [ 0.85 - 0.13/ 0.85 - 0]/ [ 1 - 0.85 - 0.13/ 0.85 - 0].

Hence, the final external reflux ratio that must be used =0.847/ 1 - 0.847 = 5.536.

[c].

The amount of distillate product that is withdrawn:

4 = 0.85 H(t) + 0.8 - 0.08.

H(t) = 4.15 kmol, and the value of Wfinal = 5.84 kmol.

3 0

3 years ago