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1 year ago

In what ways did electrical switches have to change to progress from the

Engineering

1 answer:

frez [133]1 year ago

7 0

Answer:

The answer is D. Electrical switches had to become faster and much smaller.

Explanation:

I just took the test on ~ap.ex and got it right :)

You might be interested in

Different Gauss quadrature formulae predict different values for the same integral a. True b. False

scoray [572]

Answer: False

Explanation:

The given statement are false as, the gauss quadrature predicted same value only when their is a minute error at one point and two point rule. It basically given the more precise answer and the answer are only changed by the decimal place. The gauss quadrature method are basically used for calculating the certain integral.

5 0

2 years ago

A turntable A is built into a stage for use in a theatrical production. It is observed during a rehearsal that a trunk B starts

lions [1.4K]

Answer:

The coefficient of static friction, μₛ, between the trunk and turntable = 0.32

Explanation:

For this motion of the trunk B,

Initial velocity, v₀ = 0

Tangential Acceleration, a = 0.28 m/s²

Time taken, t = 10s

Using equations of motion,

v = v₀ + at

v = 0 + 0.28 × 10 = 2.8 m/s

Frictional force, Fᵣ = μₛN

μₛ = coefficient of static friction,

N = Normal reaction exerted on the trunk B as a result of its weight = mg

Doing a force balance on the trunk B,

Force keeping the trunk B in circular motion must balance the frictional forces.

Force keeping the trunk B in circular motion, F = mv²/r

Fᵣ = F

μₛN = mv²/r but N = mg

μₛmg = mv²/r

μₛg = v²/r

μₛ = v²/gr

μₛ = 2.8²/(9.8 × 2.5) = 0.32

Hope this helps!!!

3 0

2 years ago

Fixed rate mortgage offer:

jarptica [38.1K]

Answer: The answer is :

C) $887

Explanation:

5 0

2 years ago

The concrete canoe team does some analysis on their design and calculates that they need a compressive strength of 860 psi. They

vlada-n [284]

Answer:

874 psi

Explanation:

Given a sample mean (x') = 900,

and a standard error (SE) = 10

At 99% confidence, Z(critical) = 2.58

That gives 99% confidence interval as,

x' ± Z(critical) x SE = 900 ± 2.58 x 10

The value of the lower limit is,

900 - 25.8 = 874.2

≈ 874 psi

8 0

2 years ago

Assume a program requires the execution of 50 x 10^6 FP instructions, 110 x 10^6 INT instructions, 80 x 10^6 Load/Store (L/S) in

Svetradugi [14.3K]

Answer:

We can not improve CPI of FP instructions when we run the program two times faster because it would be negative.

Explanation:

Processor clock rate = 2 GHz

Execution Time = ∑ $(\frac{Clock cyles}{Clock rate})$

Clock cycles can be determined using following formula

Clock cycles = ( $CPI_{FP}$ x No. FP instructions )+ ( $CPI_{INT}$ x No. INT instructions) + ( $CPI_{L/S}$ x No. L/S instructions ) + ( $CPI_{branch}$ x No. branch instructions)

Clock cycles = ( 50 x $10^{6}$ x 1) + ( 110 x $10^{6}$ x 1) + ( 80 x $10^{6}$ x 4) + ( 16 x $10^{6}$ x 2)

Clock cycles = 512 x 10⁶

So,Initial Execution time for FP instructions is,

= $\frac{512(10^{6}) }{2(10^{9}) }$

Initial execution Time = 256 x 10⁻³

For 16 processors ,

clock cycle = 512 x 10⁶

Execution Time = 256 x 10⁻³

To run the program two times faster, half the number of clock cycles

( $\frac{Clockcycles}{2}$ )= ( $CPI_{FP}$ x No. FP instructions )+ ( $CPI_{INT}$ x No. INT instructions) + ( $CPI_{L/S}$ x No. L/S instructions ) + ( $CPI_{branch}$ x No. branch instructions)

$CPI_{FP improved}$ x No. FP instructions = ( $\frac{Clockcycles}{2}$ ) -[ ( $CPI_{INT}$ x No. INT instructions) + ( $CPI_{L/S}$ x No. L/S instructions ) + ( $CPI_{branch}$ x No. branch instructions)]

$CPI_{FP improved}$ x 50 x $10^{6}$ = ( $\frac{512(10)^{6} }{2}$ ) - [ ( 110 x $10^{6}$ x 1) + ( 80 x $10^{6}$ x 4) + ( 16 x $10^{6}$ x 2)]

$CPI_{FP improved}$ x 50 x $10^{6}$ = - 206 x $10^{6}$

$CPI_{FP improved}$ = - 206 x $10^{6}$ / 50 x $10^{6}$

$CPI_{FP improved}$ = - 4.12 < 0

3 0

3 years ago