Answer: required tensile stress is 0.889 MPa
Explanation:
Given that;
tensile load is oriented along the [1 1 1] direction
shear stress is 0.242 MPa along [1 0 1] in the (1 1 -1) plane
first we determine
λ which is Angle between [1 1 1] and [1 0 1]
so
cosλ = [ 1(1) + 1(0) + 1(1) ] / [ √(1² + 1² + 1²) √(1² + 0² + 1²)]
= 2 / √3√2 = 2/√6
Next, we determine ∅ which is angle between [1 1 1] and [1 1 -1]
so,
cos∅ = [ 1(1) + 1(1) + 1(-1) ] / [ √(1² + 1² + 1²) √(1² + 1² + (-1)²)]
cos∅ = [ 2-1] / [√3√3 ]
cos∅ = 1/3
Now, we know that;
σ = T_stress/cosλcosθ
so we substitute
σ = 0.242 / ( 2/√6 × 1/3 )
σ = 0.242 / 0.2721
σ = 0.889 MPa
Therefore, required tensile stress is 0.889 MPa
Answer:
0.25 J/K
Explanation:
Given data in given question
heat (Q) = 100 J
temperature (T) = 400 K
to find out
the change in entropy of the given system
Solution
we use the entropy change equation here i.e
ΔS = ΔQ / T ...................a
Now we put the value of heat (Q) and Temperature (T) in equation a
ΔS is the entropy change, Q is heat and T is the temperature,
so that
ΔS = 100/400 J/K
ΔS = 0.25 J/K
Answer:
the order higher is 3p79g5t88=yv5379
Answer: Option D) 298 g/mol is the correct answer
Explanation:
Given that;
Mass of sample m = 13.7 g
pressure P = 2.01 atm
Volume V = 0.750 L
Temperature T = 399 K
Now taking a look at the ideal gas equation
PV = nRT
we solve for n
n = PV/RT
now we substitute
n = (2.01 atm x 0.750 L) / (0.0821 L-atm/mol-K x 399 K
)
= 1.5075 / 32.7579
= 0.04601 mol
we know that
molar mass of the compound = mass / moles
so
Molar Mass = 13.7 g / 0.04601 mol
= 297.7 g/mol ≈ 298 g/mol
Therefore Option D) 298 g/mol is the correct answer
Answer:
The correct/closest option is b
Explanation:
Restriction enzymes are enzymes (endonucleases) that cut short DNA strands at specific sites. Hence, each restriction enzyme has it's own specific site (between two bases) it cuts at. There are two types of end that can be produced by this cut; the blunt end and the sticky end.
A restriction enzyme recognizes (palindromic sequence) and cut in it's own specific end.
For example, if a restriction enzyme cuts between a guanine (G) and an adenine (A), and it cuts a palindromic double stranded DNA in the manner below, it produces a sticky end.
G║AATTC
CTTAA║G
And if a restriction enzyme cuts between guanine (G) and cytosine (C) in the manner below, it produces a blunt end.
GGG║CCC
CCC║GGG
Hence, from the question, restriction enzymes (although chosen by the scientist based on desired sequence to be cut) recognize the sticky or blunt ends itself.
