Answer:
28.81 m
Explanation:
Ff = -123
m * a = -123
(29.8+10.3) * a = -123
a = -123/40.1 = -3.07
We know,
v^2 = u^2 + 2as
0^2 = 13.3^2 + 2*(-3.07)*s
s = 176.89/6.14 = 28.81
[ If there's a problem with the solution, pleaase let me know ]
When gases, fluids, or other solids are in contact with a moving object
heat is produced due to friction.
The solution for this problem is:
A velocity of wave is given as V = λ⋅f
V - velocity of wave
f - frequency of wave
λ - wave length
So getting the wave length is:
λ = V/f
λminimum = V / fmaximum
λminimum = 342 / 4200
λminimum = 0.081 m
λmaximum = V/ fminimum
λmaximum = 342 / 28
λmaximum = 12.214 m
Answer:
Ammonia, 
Explanation:
If a Rac variant, in which the residue at position 61 was replaced with an alanine (Rac-61A), was synthesized. Also, Wild-type Rac and Rac-61A were incubated separately with VopC.
In order to obtain data to support that VopC modifies Rac at residue 61, the samples should be analyzed for the presence of Ammonia (![NH_{3}[\tex]) compound.In chemistry, the Vanadyl Phthalocyanine (VopC) comprises of a catalytic domain that activates host cells GTPase Rac irreversibly, through the deamide side chain of the residue existing at position 61.Also, we know that a deamidation reaction gives off Ammonia, [tex](NH_{3})](https://tex.z-dn.net/?f=NH_%7B3%7D%5B%5Ctex%5D%29%20compound.%3C%2Fp%3E%3Cp%3EIn%20chemistry%2C%20the%20Vanadyl%20Phthalocyanine%20%28VopC%29%20comprises%20of%20a%20catalytic%20domain%20that%20activates%20host%20cells%20GTPase%20Rac%20irreversibly%2C%20through%20the%20deamide%20side%20chain%20of%20the%20residue%20existing%20at%20position%2061.%3C%2Fp%3E%3Cp%3EAlso%2C%20we%20know%20that%20a%20deamidation%20reaction%20gives%20off%20Ammonia%2C%20%5Btex%5D%28NH_%7B3%7D%29)
and as such when ammonia is present in the sample containing Wild-type Rac but not in Rac-61A, this simply proves or provide the data to support that VopC modifies Rac at residue 61.
<em>Additionally, deamidation can be defined as the chemical conversion (hydrolysis) of an amide functional group such as glutamine, asparagine, in a polypeptide to another functional group such as glutamic acid or isoaspartic acid respectively by treating it with a strong acid (deamidate, transamidase). </em>
Answer:
c. 5.50 km
Explanation:
8 min * 1h/(60 min) = 8/60 = 2/15 h
15 sec* 1 min/60 sec = 1/4 min * 1h/(60 min) = 1/240 h
8 min 15 sec = (2/15+1/240)h
40 km/h *(2/15 +1/240)h =5.50 km
