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bulgar [2K]
3 years ago
11

How do you remove a key cap without a kaycap remover (keyboard keycap) ​

Computers and Technology
1 answer:
Lana71 [14]3 years ago
7 0
Butter knife will probably work
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What is the name of a button on a website?
zhuklara [117]
May I know which button?
5 0
3 years ago
The address space 0x0000-0x3FF is assigned to I/O peripherals of a given HCS12 chip. How many K bytes is this? (Points : 2) 2K
UNO [17]

Answer:

D. 1K

Explanation:

The address space 0x0000-0x3FF is of 1KB(One Kilo Byte) since 0x3ff is equal to 1023. therefore 1023=1 KB.

bitwise 0x3ff zeros out the top bits of the number such that the result is always between 0 and 1023. It is essentially the same thing as modulo( num,1024) for positive  values of num.

Hence the answer is D 1K

3 0
3 years ago
How can an exception be explicitly raised in C++? How are exceptions bound to handlers in C++?
xenn [34]

Answer:

Exceptions are raised using throw statement in c++.

Explanation:

Try block is used to keep the statements which we felt that they will raise an exception. Catch block is used to catch the exception which is thrown by the try block.

#include<iostream.h>

void main(){

int x,y,z;

try{

cout<"enter 2 numbers";

cin>>x>>y;

if(y==0)

throw y;

z=x/y;

}

catch(int x){

cout<<"exception caught";

}

}

4 0
3 years ago
Determine the hexadecimal value of a mask that can be used to complement the values of thefirst and third bits of f1ag and leave
Makovka662 [10]

Answer:

Let our mask be 1010000...000. Notice that, if we apply XOR operation to any number with this mask, the first and third bits will be complemented and all other bits will be unchanged. Since for each individual bit, XOR with 0 does not change the value of bit and XOR with 1 changes the value of bit.

Explanation:

To give a hexadecimal number, we should know the number of bits in representation of number.

8 0
4 years ago
Help its timed :plzzzzzz<br> fjndf,xn
Lerok [7]

Answer:

Please find attached the diagram of the different thrust of the spaceship created with Microsoft Visio with the numbers representing the following situations

(1) Thrust is less than weight. The spaceship does not launch

(2) Thrust is just a little greater than weight. The net force gives the spacecraft acceleration up, but not enough to gain the speed needed to reach orbit

(3) Thrust is greater than weight. The net force gives the spacecraft acceleration up. It gains enough speed to "fall around" Earth, constantly moving in a nearly circular path

(4) Thrust is a lot greater than weight. The very large net force gives the spacecraft a very large acceleration up. It gains so much speed that it escapes Earth

Explanation:

1) The thrust F < Weight = Mass of spacecraft, m × The acceleration due to gravity, g

F < W = m × g, the spacecraft remains on the Earth surface

2) The thrust, F > W, The space craft has an initial upwards acceleration, given by (F - W)/m

v < v_{orbit} \ Orbital \ velocity =\sqrt{\dfrac{G \cdot M}{r} }

3) F > W and  v  =v_{orbit} = \sqrt{\dfrac{G \cdot M}{r} }

4)  F > W and  v  \geq v_{esc} = \sqrt{\dfrac{2 \cdot G \cdot M}{r} }

4 0
3 years ago
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