What is the force in kN of work done is 1.2 ms moves through 120m

RKI Instruments borrowed $4,800,000 from a private equity firm for expansion of its facility for manufacturing carbon monoxide m

Tju [1.3M]

Answer:

n

Explanation:

6 0

3 years ago

A brass alloy is known to have a yield strength of 275 MPa (40,000 psi), a tensile strength of 380 MPa (55,000 psi), and an elas

zlopas [31]

Answer:

The magnitude of the load can be computed because it is mandatory in order to produce the change in length ( elongation )

Explanation:

Yield strength = 275 Mpa

Tensile strength = 380 Mpa

elastic modulus = 103 GPa

The magnitude of the load can be computed because it is mandatory in order to produce the change in length ( elongation ) .

Given that the yield strength, elastic modulus and strain that is experienced by the test spectrum are given

strain = yield strength / elastic modulus

= 0.0027

3 0

3 years ago

________is defined to be the probability that a failed component or system will be restored to a repaired specified condition wi

julsineya [31]

Answer:

Repairable component

Explanation:

Repairable component is defined to be the probability that a failed component or system will be restored to a repaired specified condition within a period of time when maintenance is performed in accordance with prescribed procedures.

6 0

3 years ago

Read 2 more answers

A hydraulic jump is induced in an 80 ft wide channel. The water depths on either side of the jump are 1 ft and 10 ft. Please cal

Sphinxa [80]

Answer:

a) The velocity is 42.0833 ft/s

b) The flow rate is 3366.664 ft³/s

c) The Froude number is 0.2345

d) The flow energy dissipated (expressed as percentage of the energy prior to the jump) is 18.225 ft

e) The critical depth is 3.8030 ft

Explanation:

Given data:

80 ft wide channel, L

1 ft and 10 ft water depths, d₁ and d₂

Questions: a) Velocity of the faster moving flow, v = ?

b) The flow rate (discharge), q = ?

c) The Froude number, F = ?

d) The flow energy dissipated, E = ?

e) The critical depth, dc = ?

a) For the velocity:

$\frac{d_{2} }{d_{1} } =\frac{1}{2} (\sqrt{1+8F^{2} } -1)$

$10*2=\sqrt{1+8F^{2} } -1$

Solving for F:

F = 7.4162

$v=F\sqrt{gd_{1} }$

Here, g = gravity = 32.2 ft/s²

$v=7.4162*\sqrt{32.2*1} =42.0833ft/s$

b) The flow rate:

$q=v*L*d_{1} =42.0833*80*1=3366.664ft^{3} /s$

c) The Froude number:

$v_{2} =\frac{q}{L*d_{2} } =\frac{3366.664}{80*10} =4.2083ft/s$

$F=\frac{v_{2}}{\sqrt{gd_{2} } } =\frac{4.2083}{\sqrt{32.2*10} } =0.2345$

d) The flow energy dissipated:

$E=\frac{(d_{2}-d_{1})^{3} }{4d_{1}d_{2}} =\frac{(10-1)^{3} }{4*1*10} =18.225ft$

e) The critical depth:

$d_{c} =(\frac{(\frac{q}{L})^{2} }{g} )^{1/3} =(\frac{(\frac{3366.664}{80})^{2} }{32.2} )^{1/3} =3.8030ft$

4 0

4 years ago

A high-voltage transmission line carries 1 000 A at 700 kV for a distance of 100 miles. If the resistance per length in the wire

mamaluj [8]

Answer:

50 MW

Explanation:

We are given;

Current = 1000A

Voltage = 700 KV

Resistance per length = 0.5 Ω/mile

Length = 100 miles

The total resistance would be;

R=0.5 Ω/mile x 100 miles = 50Ω.

Now, the current is the same throughout the wire, but the voltage is only at the beginning of the line, and thus the current for the power equation is;

P = I²R

So;

P=(1000A)² x 50Ω.= 50,000,000W = 50MW

3 0

3 years ago

What is the force in kN of work done is 1.2 ms moves through 120m​

What is the force in kN of work done is 1.2 ms moves through 120m