In many problems where the potential energy is considered between two objects, the potential energy is defined as zero when the
1 answer:
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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 = ∑
Clock cycles can be determined using following formula
Clock cycles = ( x No. FP instructions )+ (
x No. INT instructions) + (
x No. L/S instructions ) + (
x No. branch instructions)
Clock cycles = ( 50 x x 1) + ( 110 x
x 1) + ( 80 x
x 4) + ( 16 x
x 2)
Clock cycles = 512 x 10⁶
So,Initial Execution time for FP instructions is,
=
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
( )= (
x No. FP instructions )+ (
x No. INT instructions) + (
x No. L/S instructions ) + (
x No. branch instructions)
x No. FP instructions = (
) -[ (
x No. INT instructions) + (
x No. L/S instructions ) + (
x No. branch instructions)]
x 50 x
= (
) - [ ( 110 x
x 1) + ( 80 x
x 4) + ( 16 x
x 2)]
x 50 x
= - 206 x
= - 206 x
/ 50 x
= - 4.12 < 0
Answer:
M_AD = 1359.17 lb-in
Explanation:
Given:
- T_ef = 46 lb
Find:
- Moment of that force T_ef about the line joining points A and D.
Solution:
- Find the position of point E:
mag(BC) = sqrt ( 48^2 + 36^2) = 60 in
BE / BC = 45 / 60 = 0.75
Hence, E = < 0.75*48 , 96 , 36*0.75> = < 36 , 96 , 27 > in
- Find unit vector EF:
mag(EF) = sqrt ( (21-36)^2 + (96+14)^2 + (57-27)^2 ) = 115 in
vec(EF) = < -15 , -110 , 30 >
unit(EF) = (1/115) * < -15 , -110 , 30 >
- Tension T_EF = (46/115) * < -15 , -110 , 30 > = < -6 , -44 , 12 > lb
- Find unit vector AD:
mag(AD) = sqrt ( (48)^2 + (-12)^2 + (36)^2 ) = 12*sqrt(26) in
vec(AD) = < 48 , -12 , 36 >
unit(AD) = (1/12*sqrt(26)) * < 48 , -12 , 36 >
unit (AD) = <0.7845 , -0.19612 , 0.58835 >
Next:
M_AD = unit(AD) . ( E x T_EF)
M_AD = 1835.73 + 116.49528 - 593.0568
M_AD = 1359.17 lb-in