Answer:
Without units, the results are unclear and it is hard to keep track of what each seperate measurement entails.
Answer : The specific heat of aluminum is, 
Solution : Given,
Heat absorbs = 677 J
Mass of the substance = 10 g
Final temperature = 
Initial temperature = 
Formula used :
or,
Q = heat absorbs
m = mass of the substance
c = heat capacity of aluminium
= final temperature
= initial temperature
Now put all the given values in the above formula, we get the specific heat of aluminium.

Therefore, the specific heat of aluminum is, 
Given k=100, Area if plate A = 1.00 cm2

Thickness of rutile 

Without any dielectric medium capacitance of a parallel plate capacitor is c = oAd
When a dielectric is placed between capacitors plates then capacitance c1 = KoAd
Here
, 
Now 



Correct Option: A
What purposes serve parallel plate capacitors?
The following are some uses for parallel plate capacitors:
- This kind of capacitor is utilized in rechargeable energy systems such as batteries.
- These capacitors are used in systems for dynamic digital memory.
- Such capacitors are used in pulsed laser and radar circuits.
To learn more about Parallel plate capacitors, visit:
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To solve this exercise it is necessary to apply the concepts related to Robert Boyle's law where:

Where,
P = Pressure
V = Volume
T = Temperature
n = amount of substance
R = Ideal gas constant
We start by calculating the volume of inhaled O_2 for it:


Our values are given as
P = 1atm
T=293K 
Using the equation to find n, we have:




Number of molecules would be found through Avogadro number, then


Define
g = 9.8 m/s², acceleration due to gravity, positive downward.
Assume that wind resistance may be neglected.
Frog A:
u = 0.551 m/s, launch velocity, upward.
When the frog lands back on the pad, its vertical position is zero, and its vertical velocity will be 0.551 m/s downward.
If the time of flight is t, then
(0.551 m/s)*(t s) - 0.5*(9.8 m/s²)*(t s)² = 0
0.551t - 4.9t² = 0
t = 0, or t = 0.1124 s
t = 0 corresponds to launch, and t = 0.1124 s corresponds to landing.
Frog B:
Launch velocity is 1.75 m/s
When t = 0.1124 s, the position of the frog is
s = (1.75 m/s)(0.1124 s) - 0.5*(9.8 m/s²)*(0.1124 s)²
= 0.135 m
The velocity of frog B is
v = (1.75 m/s) - (9.8 m/s²)*(0.1124 s)
= 0.6485 m/s
Answer:
When frog A lands on the ground,
Frog B is 0.135 m above ground and its velocity is 0.649 m/s upward.