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2 years ago

What are the two (2) different design elements of scratch?

Engineering

1 answer:

Roman55 [17]2 years ago

6 0

Answer: There is four elements of Scratch. The stage, the sprites, the script and the programming palette.

Explanation:

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Interpret the assembly program below: MOV R3,R0;

Reika [66]

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.

7 0

3 years ago

Which term defines the amount of mechanical work an engine can do per unit of heat energy it uses?

skad [1K]

Answer:

Explanation:

is the because that's the amount of work in making machine can do producing heat

7 0

3 years ago

A NC drill press is to perform a series of through-hole drilling operations on a 1.75 in thick aluminum plate that is a componen

jekas [21]

Answer:

26.7 min

Explanation:

First, we will find the time required to drill each hole:

N = 300 x 12/0.75 $\pi$ = 1527.7 rev/min

fr = 1527.7 (0.015) = 22.916 in/min

Formula for distance per hole: 0.5 + A + 1.75

A = 0.5 (0.75) tan (90-100 / 2) = 0.315 in
Tm = (0.5 + 0.315 + 1.75) / 22.916 = 0.112 min

Now, we will calculate the time required to draw back the drill form hole:

= 0.112 / 2 = 0.056 min

Time to move between holes = 1.5 / 15 = 0.1 min

For 100 holes, the number of moves between holes = 99

Total time required to drill 100 holes (t):

t = 100 (0.112 + 0.056) + 99 (0.1) = 26.7 min

7 0

3 years ago

Read 2 more answers

Will Give Brainliest

Tom [10]

Answer:

A hole within a hole

Explanation:

8 0

3 years ago

Air at 80 kPa and 10Â°C enters an adiabatic diffuser steadily with a velocity of 150 m/s and leaves with a low velocity at a pre

il63 [147K]

Answer:

The exit temperature is 293.74 K.

Explanation:

Given that

At inlet condition(1)

P =80 KPa

V=150 m/s

T=10 C

Exit area is 5 times the inlet area

Now

$A_2=5A_1$

If consider that density of air is not changing from inlet to exit then by using continuity equation

$A_1V_1=A_2V_2$

So $A_1\times 150=5A_1V_2$

$V_2=30$ m/s

Now from first law for open system

$h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w$

Here Q=0 and w=0

$h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2}$

When air is treating as ideal gas

$h=C_pT$

Noe by putting the values

$h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2}$

$1.005\times 283+\dfrac{150^2}{2000}=1.005\times T_2+\dfrac{30^2}{2000}$

$T_2=293.74K$

So the exit temperature is 293.74 K.

7 0

3 years ago