Answer:
First try logging out. If that doesn't work there is a really helpful video on how to uninstall and reinstall with everything still there. Hope this was helpful.
The recommended way to hold a loose hub on a fan is by riveting. Please look at the image attached so you can know where exactly is a hub. The correct answer choice for this question is:
A) Riveting
I hope this helps, Regards.
Answer:
Explanation:welding
The updated information about specific parameters of the new hardware component can be stored in:
b. Flash Memory: Flash memory is a non-volatile storage medium that can retain information even when the power is turned off. It is commonly used to store firmware, such as the system BIOS, which contains instructions for the computer's hardware. When a new hardware component is installed, its specific parameters can be stored in the flash memory, allowing the computer to recognize and configure the component correctly.
e. CMOS RAM: CMOS RAM, or complementary metal-oxide-semiconductor random-access memory, is a type of volatile memory that stores the computer's basic hardware configuration, including information about installed devices. When a new hardware component is installed, its parameters can be stored in the CMOS RAM, enabling the computer to identify and interact with the component during the startup process.
It's important to note that both flash memory and CMOS RAM can be used to store updated hardware information, but they serve different purposes. Flash memory is typically used for firmware storage, while CMOS RAM is used for storing the computer's hardware configuration. So, either option b. Flash Memory or e. CMOS RAM could be a valid answer depending on the specific context of the question.
The System BIOS stores the specific parameters of a hardware component that has been replaced in a computer. It is built-in software that helps the computer to boot up by saving the components' information.
When you replace a modular hardware component inside a computer case, the updated specific parameters of the newly installed device are generally stored in the System BIOS (Basic Input Output System). The System BIOS consist of a built-in software that instructs the computer on how to perform a certain process like booting up. The BIOS stores all the hardware components information so it would know what it has to manage.
On the other hand, Flash Memory is a type of storage technology used for saving data, Secondary storage refers to storage devices like hard drives, Startup BIOS is the part of the BIOS responsible for booting the computer, and CMOS RAM stores the BIOS settings and other system time information.
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Answer:
The three bulbs can be connected in series to the battery B1, such that when switch SW1 is turned ON all three bulb turns ON with the same brightness. We now have the three resistors R1, R2, and R3 connected in parallel to the lamps LMP1, LMP2, and LMP3 respectively when SW2 turn on. SW2 is a Three Pole Single Throw switch. Please see attachment for the electric circuit diagram.
Explanation:
QUESTION
How do you make a circuit so 1 switch will turn on/off all the lights(3 lights) and a second switch will change the lights from all being the same brightness to all being different brightness?
ANSWER
List of Component Used For The Design Of The Electric Circuit
The three bulbs can be connected in series to the battery B1, such that when switch SW1 is turned ON all three bulb turns ON with the same brightness. We now have the three resistors R1, R2, and R3 connected in parallel to the lamps LMP1, LMP2, and LMP3 respectively when SW2 turn on. SW2 is a Three Pole Single Throw switch. Please see attachment for the electric circuit diagram.
1. 18Volt Battery ( two 9 Volts batteries connected in series)
2. Three Filament lamps LMP1=LMP2=LMP3= 6V 6W.
3. Three resistors of R1=100Ω, R2=12Ω and R3=3Ω
4. SPST Switch =SW1
5. TPST Switch = SW2
Since the battery is 18V and connected in series to all three lamps, it is capable of delivering 6V per lamp. Each lamp is 6W. So the resistance of the bulb can be determined using Ohm’s law.
V=IR, P=IV = V2/R, R=V2/P
Where I= Current
V= Voltage
R= Resistance
P = Power
Lamp Resistance RL= 62/6 =6Ω
Total resistance of the lamps connected in series Rs= 6+6+6 = 18Ω
The total current through the series resistor combination IS = Battery supplied voltage divide by RS =18/18 =1A
So each lamp will dissipate power PL= I2*R= 1*6 =6W
When switch SW2 is closed, there is now a resistor connected in parallel to each of the lamp which will now reduce the total resistance combination of the lamp and resistors to a values lower than 6Ω
Lamp1 has 100Ω connected in parallel to it to give a total resistance of 5.66Ω
Lamp2 has 12Ω connected in parallel to it to give a total resistance of 4Ω
Lamp3 has 3Ω connected in parallel to it to give a total resistance of 2Ω
The new total resistance of the circuit is now R=5.66+4+2 = 11.66Ω
The new total current flowing through the circuit I=18/11.66 = 1.54A
The power dissipated by each lamp with a new series current of 1.54A can now be recalculated as follows;
Power dissipated by Lamp 1 P1= 1.542*5.66Ω = 13.42W
Power dissipated by Lamp 2 P2= 1.542*4Ω = 9.49W
Power dissipated by Lamp 1 P1= 1.542*2Ω = 4.74W
From the power dissipated by the 3 bulbs, we can see that lamp 3 is bright, lamp 2 is brighter and Lamp 1 is the brightest of all the three lamps.
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