Explanation:
1. A sequence of instructions is stored in memory.
2. The memory address wherever the first instruction is found is copied to the instruction pointer.
3. The CPU sends the address within the instruction pointer to memory on the address bus.
4. The CPU sends a “read” signal to the control bus.
5. Memory responds by sending a copy of the state of the bits at that memory location on the
data bus, that the CPU then copies into its instruction register.
6. The instruction pointer is automatically incremented to contain the address of the next
instruction in memory.
7. The CPU executes the instruction within the instruction register.
8. Go to step 3
Steps 3, 4, and 5 are called an instruction fetch. Notice that steps 3 – 8 constitute a cycle, the instruction execution cycle. It is shown graphically below.
A DMA controller can generate memory addresses and initiate memory read or write cycles. It contains several hardware registers that can be written and read by the CPU. These include a memory address register, a byte count register, and one or more control registers.
Answer:
None of the above cause thats what i put
inhalation is the most common way for chemical to enter the body
Answer:
a) The flow has three dimensions (3 coordinates).
b) ∇V = 0 it is a incompressible flow.
c) ap = (16/3) i + (32/3) j + (16/3) k
Explanation:
Given
V = xy² i − (1/3) y³ j + xy k
a) The flow has three dimensions (3 coordinates).
b) ∇V = 0
then
∇V = ∂(xy²)/∂x + ∂(− (1/3) y³)/∂y + ∂(xy)/∂z
⇒ ∇V = y² - y² + 0 = 0 it is a incompressible flow.
c) ap = xy²*∂(V)/∂x − (1/3) y³*∂(V)/∂y + xy*∂(V)/∂z
⇒ ap = xy²*(y² i + y k) - (1/3) y³*(2xy i − y² j + x k) + xy*(0)
⇒ ap = (xy⁴ - (2/3) xy⁴) i + (1/3) y⁵ j + (xy³ - (1/3) xy³) k
⇒ ap = (1/3) xy⁴ i + (1/3) y⁵ j + (2/3) xy³ k
At point (1, 2, 3)
⇒ ap = (1/3) (1*2⁴) i + (1/3) (2)⁵ j + (2/3) (1*2³) k
⇒ ap = (16/3) i + (32/3) j + (16/3) k
