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3 years ago

Disconnecting a circuit while in operation can create a voltage blank

Engineering

1 answer:

zlopas [31]3 years ago

5 0

Answer:

what is the question

Explanation:

confused

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According to the amortization table, Demarco and Tanya will pay a total of in interest over the life of their loan.

Ymorist [56]

Answer:

(Interest rate/number of payments)*$170000= interest for the first month.

Interest amounts for all the months of repayment plus $170000=Total loan cost

Explanation:

Interest is the amount you pay for taking a loan from a bank on top of the original amount borrowed.

Factors affecting how much interest is paid are; the principal amount, the loan terms, repayment schedule, the repayment amount and the rate of interest.

The interest paid=(rate of interest/number of payments to make)*principal amount borrowed.

You divide the interest with number of payments done in a year where monthly are divided by 12.Multiplying it by loan balance in the first month which is your principal amount gives the interest rate to pay for that month.

You new loan balance will be= Principal -(repayment-interest)

Do this for the period the loan should take.

Add all the interest amount to original borrowed amount to get total cost of the loan after the period of time.

8 0

3 years ago

Read 2 more answers

1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil

kakasveta [241]

Answer:

a) $n = 1000\,users$ , b) $\Delta t_{min} = \frac{1}{30}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$ , c) $n = 10000000\,users$ , $\Delta t_{min} = \frac{1}{3000}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

Explanation:

a) The total number of users that can be accomodated in the system is:

$n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )$

$n = 1000\,users$

b) The length of the side of each cell is:

$l = \sqrt{1\,km^{2}}$

$l = 1\,km$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{30}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$

c) The total number of users that can be accomodated in the system is:

$n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )$

$n = 10000000\,users$

The length of each side of the cell is:

$l = \sqrt{100\,m^{2}}$

$l = 10\,m$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{3000}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

8 0

3 years ago

Charging method .Constant current method

mina [271]

Answer:

There are three common methods of charging a battery; constant voltage, constant current and a combination of constant voltage/constant current with or without a smart charging circuit.

Constant voltage allows the full current of the charger to flow into the battery until the power supply reaches its pre-set voltage. The current will then taper down to a minimum value once that voltage level is reached. The battery can be left connected to the charger until ready for use and will remain at that “float voltage”, trickle charging to compensate for normal battery self-discharge.

Constant current is a simple form of charging batteries, with the current level set at approximately 10% of the maximum battery rating. Charge times are relatively long with the disadvantage that the battery may overheat if it is over-charged, leading to premature battery replacement. This method is suitable for Ni-MH type of batteries. The battery must be disconnected, or a timer function used once charged.

Constant voltage / constant current (CVCC) is a combination of the above two methods. The charger limits the amount of current to a pre-set level until the battery reaches a pre-set voltage level. The current then reduces as the battery becomes fully charged. The lead acid battery uses the constant current constant voltage (CC/CV) charge method. A regulated current raises the terminal voltage until the upper charge voltage limit is reached, at which point the current drops due to saturation.

4 0

3 years ago

Write a Nios II assembly program that reads binary data from the Slider Switches, SW11-0, on the DE2-115 Simulator, and display

Citrus2011 [14]

Algorithm of the Nios II assembly program.

Attain data for simulation from the SW11-0, on the DE2-115 Simulator
The data will be read from the switches in loop.
The decimal output is displayed using the seven-segment displays and done using the loop.
The program is ended by the user operating the SW1 switch

and

The decimal equivalent on the seven-segment displays HEX3-0 is

DE2-115
DE2-115_SW11
DE2-115_HEX3
DE2-115_HEX4
DE2-115_HEX5
DE2-115_HEX6
DE2-115_HEX7

<h3>The Algorithm and decimal equivalent on the seven-segment displays HEX3-0</h3>

Generally, the program will be written using a cpulator simulator in order to attain best result.

We are to

Attain data for simulation from the SW11-0, on the DE2-115 Simulator
The data will be read from the switches in loop.
The decimal output is displayed using the seven-segment displays and done using the loop.
The program is ended by the user operating the SW1 switch

This will be the Algorithm of the Nios II assembly program .

Hence, the decimal equivalent on the seven-segment displays HEX3-0 is

DE2-115
DE2-115_SW11
DE2-115_HEX3
DE2-115_HEX4
DE2-115_HEX5
DE2-115_HEX6
DE2-115_HEX7

For more information on Algorithm

brainly.com/question/11623795

6 0

2 years ago

At 45° latitude, the gravitational acceleration as a function of elevation z above sea level is given by g = a − bz , where a =

Ahat [919]

Answer:

8861.75 m approximately 8862 m

Explanation:

We need to remember Newton's 2nd Law which says that the force experienced by an object is proportional to his acceleration and that the constant of proportionality between those two vectors correspond to the mass of the object.

$F=ma$ for the weight of an object (which is a force) we have that the acceleration experienced by that object is equal to the gravitational acceleration, obtaining that $W = mg$

For simplicity we work with $g =9.807$ $\frac{m}{s^{2}}$ despiting the effect of the height above sea level. In this problem, we've been asked by the height above sea level that makes the weight of an object 0.30% more lighter.

In accord with the formula $g = a-bz$ the "normal" or "standard" weight of an object is given by $W = mg = ma$ when $z = 0$ , so we need to find the value of $z$ that makes $W = m(a-bz) = 0.997ma$ meaning that the original weight decrease by a 0.30%, so now we operate...

$m(a-bz) = 0.997ma$ now we group like terms on the same sides $ma(1-0.997) = mbz$ we cancel equal tems on both sides and obtain that $z = \frac{a}{b} (0.003) = \frac{9.807 \frac{m}{s^{2} } }{3.32*10^{-6} s^{-2} } (0.003) = 8861.75 m$

7 0

3 years ago