False as there are many risk in driving
Answer:
μb = 0.096
μc = 0.073
Explanation:
member AB:
-800( 4/3 ) + Nb (2) = 0
Nb (2) = 3200/3
Nb = 533.3N
Post BC:
summation of force along the y axis=0
Nc + Nb + 150(3/5 ) -50(9.81)=0
Nc + 533.3 + 150(3/5 ) -50(9.81)=0
Nc = 933.83 N
Also (-4/5)(150)(3) + Fb(0.7)= 0
Fb = (4/5)(150)(3)/0.7 = 51.429 N
Likewise alog the x axis,
4/5(150) - Fc -Fb = 0
4/5(150) - Fc -51.429 = 0
Fc = 4/5(150) -51.429 =68.571 N
μb = Fb/Nb = 51.429/533.3 = 0.096
μc = Fc/Nc = 68.571 / 933.83 = 0.073
Answer:
7 bits
Explanation:
Given
Instruction Set = 110 operation
Memory unit = 32 bits per word.
We get the required bits by using the following formula
2^n = 110
But 110 is not a factor of 2.
So, we pick the nearest decimal number greater than 110 that is a power of 2.
The number is 128
2^n = 110 becomes
2^n = 128
2^n = 2^7 ---- 2 cancels out
So,
n = 7
Hence, the required number of bits needed for the opcode is 7 bits
Please add more details because I don’t know if you are using a book or passage, therefore I cannot help you unless you add more detail