The statement is true in this situation is C. The size of Ffric is the same as the size of Fapp:
From the diagram, since the body is in equilibrium, the sum of vertical forces equals zero. Also, the sum of horizontal forces equal zero.
So, ∑Fx = 0 and ∑Fy = 0
Since Fapp acts in the negative x - direction and Ffric acts in the positive x - direction,
∑Fx = -Fapp + Ffric = 0
-Fapp + Ffric = 0
Fapp = Ffric
Also, since Fgrav acts in the negative y - direction and Fnorm acts in the positive y - direction,
∑Fy = Fnorm + (-Fgrav) = 0
Fnorm - Fgrav = 0
Fnorm = Fgrav
So, we see that the size of Fapp <u>equals</u> size of Ffric and the size of Fnorm <u>equals</u> the size of Fgrav.
So, the correct option is C
The statement which is true in this situation is C. The size of Ffric is the same as the size of Fapp.
Learn more about equilibrium of forces here:
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An element I think this is the. Aswer for it
I believe the answer would be Saturated.
The grams of Mg(OH)2 produced is calculated as below
calculate the moles of HCl produced =molarity xvolume/1000
= 0.3 x 435/1000= 0.1305 moles
write the equation for reaction
Mg(OH)2 +2HCl = MgCl2 + 2H2O
by use of mole ratio between HCl :MgCl2 which is 2 :1 the moles of HCl = 0.1305 x1/2 =0.0653 moles of MgCl2
mass of MgCl2 = moles x molar mass
= 0.0653mol x95 g/mol = 6.204 grams of MgCl2
Answer:
0%
Explanation:
- The probability that both of the parents are homozygous recessive is zero.
- In our case, The gene for tallness is dominant over the gene for dwarfism
- Therefore; TT is homozygous dominant while tt is homozygous recessive and Tt is heterozygous.
- Heterozygous genotype (Tt) will exhibit a tall phenotype.
- When two parents are crossed and all the offsprings are tall then the possible genotype of the parents is either;
- Homozygous dominant for both parents or
TT x TT
2. One parent is homozygous dominant and the other is homozygous recessive.
TT x Tt
- Therefore, the probability of having both parents as homozygous recessive when all the offspring are tall is Zero.