Answer:
Explanation:
The radius of the smaller bubble, r1 will decrease and that of the bigger bubble, r2 will increase.
The pressure that is present in the smaller bubble usually is greater than the pressure that exists inside that of the bigger bubble. This then makes air to flow from r1 to r2 thereby making the radius of the smaller bubble r1, to decrease while keeping that of the bigger bubble r2 higher.
Answer:
"What is the best advice a parent can give a child?"
Explanation:
The other answers have one clear response. Ex: Orange popsicles melt faster than grape popsicles. That would be a fact.
But parental advice can vary, depending on your opinion. I may say that all parents must teach their children not to talk to strangers, while someone else may say that parents should advice their kids to treat everyone fairly. Nothing can be proven as the only appropriate response.
Hope this helps!
(t) = 2t = 1.22 sec. I believe ...
Answer:
A. The number of valence electrons increases by 1.
Explanation:
As you move across any period on the periodic table, the number of valence electrons increases by a value of 1.
- The periodic table of elements contains an arrangement of element by their atomic numbers.
- From left to right, number of valence electrons increases.
- Down a group, the valence electrons are the same.
- Across a period, the number of valence electrons increases.
Answer:
a) T ’= 0.999 s
, b) t = 3596.4 s
Explanation:
The angular velocity of a simple pendulum is
w = √g / L
The angular velocity, frequency and period are related
w = 2π f = 2π / T
2π / T = √ g / L
T = 2π √ L / g
L = T² g / 4π²
L = 1² 9.8 / 4π²
L = 0.248 m
To know the effect of the temperature change let's use the thermal expansion ratios
ΔL = α L ΔT
ΔL = 24 10⁻⁶ 0.248 (-4 - 20)
ΔL = 142.8 10⁻⁶ m
Lf - L = -142. 8 10⁻⁶
Lf = 142.8 10⁻⁶ + 0.248
Lf = 0.2479 m
Let's calculate new period
T ’= 2π √ L / g
T ’= 2π √ (0.2479 / 9.8)
T ’= 0.999 s
We can see that the value of the period is reduced so that the clock is delayed
b) change of time in 1 hour
When the clock is at 20 ° C in one hour it performs 3600 oscillations, for the new period the time of this number of oscillations is
t = 3600 0.999
t = 3596.4 s
Therefore the clock is delayed almost 4 s
