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3 years ago

Interpret the assembly program below: MOV R3,R0;

Engineering

1 answer:

Reika [66]3 years ago

7 0

Answer:

Explanation:

1. With the operands R0, R1, the program would compute AND operation and ADD operation .

2. The operands could truly be signed 2's complement encoded (i.e Yes) .

3. The overflow truly occurs when two numbers that are unsigned were added and the result is larger than the capacity of the register, in that situation, overflow would occur and it could corrupt the data.

When the result of an operation is smaller in magnitude than the smallest value represented by the data type, then arithmetic underflow will occur.

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You leave your house at 5:02 PM and run 20 yards down the street. You don't realize that you forgot your Wallet back at home and

Sidana [21]

Answer:

The average velocity is 0.203 m/s

Explanation:

Given;

initial displacement, x₁ = 20 yards = 18.288 m

final displacement, x₂ = ¹/₃ x 18.288 = 6.096 m

change in time between 5:02 PM and 5:03 PM, Δt = 3 mins - 2 mins = 1 min = 60 s

The average velocity is given by;

V = change in displacement / change in time

V = (x₂ - x₁) / Δt

V = (18.288 - 6.096) / 60

V = 0.203 m/s

Therefore, the average velocity is 0.203 m/s

7 0

3 years ago

Two loads connected in parallel draw a total of 2.4 kW at 0.8 pf lagging from a 120-V rms, 60-Hz line. One load absorbs 1.5 kW a

stealth61 [152]

Answer: a) 0.948 b) 117.5µf

Explanation:

Given the load, a total of 2.4kw and 0.8pf

V= 120V, 60 Hz

P= 2.4 kw, cos θ= 80

P= S sin θ - (p/cos θ) sin θ

= P tan θ(cos^-1 (0.8)

=2.4 tan(36.87)= 1.8KVAR

S= 2.4 + j1. 8KVA

1 load absorbs 1.5 kW at 0.707 pf lagging

P= 1.5 kW, cos θ= 0.707 and θ=45 degree

Q= Ptan θ= tan 45°

Q=P=1.5kw

S1= 1.5 +1.5j KVA

S1 + S2= S

2.4+j1.8= 1.5+1.5j + S2

S2= 0.9 + 0.3j KVA

S2= 0.949= 18.43 °

Pf= cos(18.43°) = 0.948

b.) pf to 0.9, a capacitor is needed.

Pf = 0.9

Cos θ= 0.9

θ= 25.84 °

(WC) V^2= P (tan θ1 - tan θ2)

C= 2400 ( tan (36. 87°) - tan (25.84°)) /2 πf × 120^2

f=60, π=22/7

C= 117.5µf

7 0

3 years ago

An alloy has a yield strength of 818 MPa and an elastic modulus of 104 GPa. Calculate the modulus of resilience for this alloy [

crimeas [40]

Answer:

Modulus of resilience will be $3216942.308j/m^3$

Explanation:

We have given yield strength $\sigma _y=818MPa$

Elastic modulus E = 104 GPa

We have to find the modulus

Modulus of resilience is given by

Modulus of resilience $=\frac{\sigma _y^2}{2E}$ , here $\sigma _y$ is yield strength and E is elastic modulus

Modulus of resilience $=\frac{(818\times 10^6)^2}{2\times 104\times 10^9}=3216942.308j/m^3$

5 0

3 years ago

Sun of first 1 nayural numbers

marin [14]

Answer:

the answer is

$n(n + 1) \div 2$

n(n+1)/2

4 0

2 years ago

Read 2 more answers

According to the decreasing order of toughness. list the following materials (note: the steels are assumed to have no cold work

fiasKO [112]

Answer:

1090 Steel >1040 Steel > Pure aluminium >Diamond.

Explanation:

Toughness:

Toughness can be define as the are of load -deflection diagram up to fracture point.

Modulus of toughness can be defines as the area of stress-strain diagram up to fracture point.Modulus of toughness is the property of material.

So the decreasing order of toughness can be given as follows

1090 Steel >1040 Steel > Pure aluminium >Diamond.

8 0

3 years ago