<h3>
Answer:</h3>
The centripetal acceleration is 26.38 m/s²
<h3>
Explanation:</h3>
We are given;
- Mass of rubber stopper = 13 g
- Length of the string(radius) = 0.93 m
- Time for one revolution = 1.18 seconds
We are required to calculate the centripetal acceleration.
To get the centripetal acceleration is given by the formula;
Centripetal acc = V²/r
Where, V is the velocity and r is the radius.
Since time for 1 revolution is 1.18 seconds,
Then, V = 2πr/t, taking π to be 3.142 ( 1 revolution = 2πr)
Therefore;
Velocity = (2 × 3.142 × 0.93 m) ÷ 1.18 sec
= 4.953 m/s
Thus;
Centripetal acceleration = (4.953 m/s)² ÷ 0.93 m
= 26.38 m/s²
Hence, the centripetal acceleration is 26.38 m/s²
It's absolutely TRUE...........
Answer:
A and D are true , while B and F statements are false.
Explanation:
A) True. Since the standard gibbs free energy is
ΔG = ΔG⁰ + RT*ln Q
where Q= [P1]ᵃ.../([R1]ᵇ...) , representing the ratio of the product of concentration of chemical reaction products P and the product of concentration of chemical reaction reactants R
when the system reaches equilibrium ΔG=0 and Q=Keq
0 = ΔG⁰ + RT*ln Q → ΔG⁰ = (-RT*ln Keq)
therefore the first equation also can be expressed as
ΔG = RT*ln (Q/Keq)
thus the standard gibbs free energy can be determined using Keq
B) False. ΔG⁰ represents the change of free energy under standard conditions . Nevertheless , it will give us a clue about the ΔG around the standard conditions .For example if ΔG⁰>>0 then is likely that ΔG>0 ( from the first equation) if the temperature or concentration changes are not very distant from the standard conditions
C) False. From the equation presented
ΔG⁰ = (-RT*ln Keq)
ΔG⁰>0 if Keq<1 and ΔG⁰<0 if Keq>1
for example, for a reversible reaction ΔG⁰ will be <0 for forward or reverse reaction and the ΔG⁰ will be >0 for the other one ( reverse or forward reaction)
D) True. Standard conditions refer to
T= 298 K
pH= 7
P= 1 atm
C= 1 M for all reactants
Water = 55.6 M
Lunch of a patient has 3 oz skinless chicken, 3 oz of broccoli, 1 medium apple, and 1 cup of nonfat milk
Energy content of 3 oz skinless chicken is = 110 kcal
Energy content of 3 oz broccoli = 30 kcal
Energy content of 1 medium apple = 60 kcal
Energy content of 1 cup non-fat milk = 90 kcal
So the kilocalories of energy patient obtained from lunch
= 110 kcal+ 30 kcal + 60 kcal + 90 kcal = 290 kcal
