Answer:
The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J
Explanation:
The given variables are
Work done = 550 J
Volume change = V₂ - V₁ = -0.5V₁
Thus the product of pressure and volume change = work done by gas, thus
P × -0.5V₁ = 500 J
Hence -PV₁ = 1000 J
also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂
Also to compress the gas by a factor of 11 we have
P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11 = 909.091 J of work
Arginine is a basic aminoacid, because it has two amino groups and one acid
group.
At a low pH, every ionizable group is protoned. At a little higher pH, the
acid group looses its proton. A little higher pH, one amino group looses its
proton. At a very high pH, all ionizable groups are not protoned.
Pkas
<span>
<span><span>
<span>
pka1 = 1.82
</span>
<span>
pka2 = 8.99
</span>
<span>
pka3 = 12.48
</span>
</span>
</span></span>
So 9.20 is higher tan the second pKa and lower than the third pka. This
means the acid has already lost its proton, and one of the aminos too, but the
second amino hasn’t. When an acid is not protoned, it has a negative charge.
When an amino is not protoned, it’s neutral. When an amino is protoned, it has
a positive charge. So this amnino acid has one positive charge (one of the aminos) and one negative
charge (the acid), what makes it neutral.
Given required solution
M=10kg W=? W=Fd
v=5.0m/s F=mg
t=2.40s =10*10=100N
S=VT
=5m/s*2.4s
=12m
so W=12*100
W=1200J
Answer:
they meet from point o at distance 50.46 m and time taken is 11.6 seconds
Explanation:
given data
acceleration = 0.75 m/s²
speed B = 6 m/s
time B = 20 s
to find out
when and where the vehicles passed each other
solution
we consider here distance = x , when they meet after o point
and time = t for meet point z
we find first Bus B distance for 20 s ec
distance B = velocity × time
distance B = 6 × 20
distance B = 120 m
so
B take time to meet is calculate by distance formula
distance = velocity × time
120 - x = 6 × t
x = 120 - 6t .................1
and
distance of A when they meet by distance formula
distance = ut + 1/2 × at²
here u is initial speed = 0 and t is time
x = 0 + 1/2 × 0.75 × t²
x = 0.375 × t² .............2
so from equation 1 and 2
0.375 × t² = 120 - 6t
t = 11.6
so time is 11.6 second
and
distance from point o from equation 2
x = 0.375 (11.6)²
x = 50.46
so distance from point o is 50.46 m
We know that:
F = ma,
so
F/a = constant
F₁/a₁ = F₂/a₂
F₂ = F₁ x 5.6 / 1
F₂ = 5.6F₁
If F₁ = 45 N,
F₂ = 5.6 x 45
= 252 N
