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Ainat [17]
3 years ago
11

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

onitors. The company repaid the loan after 1 year with a single payment of $5,184,000. What was the interest rate on the loan?
Engineering
1 answer:
Tju [1.3M]3 years ago
6 0

Answer:

n

Explanation:

You might be interested in
Cavitation in pump assemblies can be avoided by decreasing tank pressure. a)-True b)-False
IgorLugansk [536]

Answer:

a) True

Explanation:

Cavitation in a pump occurs when the pressure of the liquid inside the pump suction is less than the vapour pressure at the suction. And also when the pump discharge pressure is extremely high.

Therefore cavitation in the pump assemblies can be avoided by decreasing the tank pressure because discharge cavitation takes place when the pressure at the pump discharge is extremely high.  High pressure at the discharge end of the pump prevents the water from flowing easily out, thereby recirculating the water within the pump which causes cavitation. So when the tank pressure is low, the pump discharge pressure will be low thus avoiding cavitation.

7 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
An industrial load with an operating voltage of 480/0° V is connected to the power system. The load absorbs 120 kW with a laggin
Leni [432]

Answer:

Q=41.33 KVAR\ \\at\\\ 480 Vrms

Explanation:

From the question we are told that:

Voltage V=480/0 \textdegree V

Power P=120kW

Initial Power factor p.f_1=0.77 lagging

Final Power factor p.f_2=0.9 lagging

Generally the equation for Reactive Power is mathematically given by

Q=P(tan \theta_2-tan \theta_1)

Since

p.f_1=0.77

cos \theta_1 =0.77

\theta_1=cos^{-1}0.77

\theta_1=39.65 \textdegree

And

p.f_2=0.9

cos \theta_2 =0.9

\theta_2=cos^{-1}0.9

\theta_2=25.84 \textdegree

Therefore

Q=P(tan 25.84 \textdegree-tan 39.65 \textdegree)

Q=120*10^3(tan 25.84 \textdegree-tan 39.65 \textdegree)

Q=-41.33VAR

Therefore

The size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging is

Q=41.33 KVAR\ \\at\\\ 480 Vrms

6 0
3 years ago
What is meant by the dead state. Discuss its significance.
Jlenok [28]

Answer and Explanation:

Dead state can be defined as the state in which both the are ambient properties and the system properties are coincident to each other.

Since the surroundings are similar and these surroundings are fixed also, there will be no kinetic energy and also the gravitational potential energy for both the systems is also equal.

Practically, in thermodynamic studies, the Gravitational potential energy of the system is negligible. This state holds significance because it leads to the termination of all the processes that are spontaneous in nature.

4 0
4 years ago
What is the first step in the Design Process *
Gennadij [26K]

Answer:

Define the problem

I hope this helps!

Explanation:

The problem must be know before you can develop and solution, generate concepts, or construct and test a prototype

8 0
3 years ago
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