Answer:
a. 0.16240664737515434 moles
b. 67.5 degrees Celcius
Explanation:
a. Use Ideal Gas Equation
PV=nRT
Where P = pressure in pascals, V=Volume in cubic meters, n=number of moles, R is a constant=8.314 J/mol.K and T is temperature in Kelvin.
27C = 273+27=300Kelvin
volume 4L = 0.004m^3
Pressure = 1atm = 101325 Pascal
PV=nRT
101325Pa*0.004m^3=n*8.314J/mol.K*300K
Solving for n from the above you get n=0.16240664737515434 moles
b.Use combined gas law equation
P1*V1/T1=P2*V2/T2
P1= 1atm
V1=4L
T1=27C
P2= 4/16 L =0.25L
P=1*40 atm = 40atm
We do not know T2
USING THE FORMULA
(1atm*4L)/27C = (40atm*0.25L)/T2
(1*4)/27=(40*0.25)/T2
IF you simplify for T2, you get 67.5
Hence final temperature = 67.5 degrees Celcius
Answer:
The gravitational force exerted is same the only difference arises because of the masses of earth and moon since the mass of earth is quite large as compared to Moon so effect on Earth is quite insignificant as compared to effect on moon.
Explanation:
Answer:
velocity
Explanation:
the displacement of an object during a specific unit of time.
Answer:
The height is 
A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )
Explanation:
From the question we are told that
The height is 
The angle of the slope is 
According to the law of conservation of energy
The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

Where I is the moment of inertia which is mathematically represented as this for a sphere

The angular velocity
is mathematically represented as

So the equation for conservation of energy becomes
![mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2](https://tex.z-dn.net/?f=mgh_s%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5B%5Cfrac%7B2%7D%7B5%7D%20mr%5E2%20%5D%5B%5Cfrac%7Bv%7D%7Br%7D%20%5D%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
![mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]](https://tex.z-dn.net/?f=mgh_s%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2%20%5B%5Cfrac%7B2%7D%7B5%7D%20%2B1%20%5D)
![mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]](https://tex.z-dn.net/?f=mgh_s%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2%20%5B%5Cfrac%7B7%7D%7B5%7D%20%5D)
![gh_s =[\frac{7}{10} ] v^2](https://tex.z-dn.net/?f=gh_s%20%3D%5B%5Cfrac%7B7%7D%7B10%7D%20%5D%20v%5E2)

Considering a circular hoop
The moment of inertial is different for circle and it is mathematically represented as

Substituting this into the conservation equation above
![mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2](https://tex.z-dn.net/?f=mgh_c%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%28mr%5E2%29%5B%5Cfrac%7Bv%7D%7Br%7D%20%5D%20%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2)
Where
is the height where the circular hoop would be released to equal the speed of the sphere at the bottom



Recall that 


Substituting values
