Answer:
h = 13.06 m
Explanation:
Given:
- Specific gravity of gasoline S.G = 0.739
- Density of water p_w = 997 kg/m^3
- The atmosphere pressure P_o = 101.325 KPa
- The change in height of the liquid is h m
Find:
How high would the level be in a gasoline barometer at normal atmospheric pressure?
Solution:
- When we consider a barometer setup. We dip the open mouth of an inverted test tube into a pool of fluid. Due to the pressure acting on the free surface of the pool, the fluid starts to rise into the test-tube to a height h.
- The relation with the pressure acting on the free surface and the height to which the fluid travels depends on the density of the fluid and gravitational acceleration as follows:
P = S.G*p_w*g*h
Where, h = P / S.G*p_w*g
- Input the values given:
h = 101.325 KPa / 0.739*9.81*997
h = 13.06 m
- Hence, the gasoline will rise up to the height of 13.06 m under normal atmospheric conditions at sea level.
Answer:
The amount of caffeine left after one half life of 5 hours is 15 oz.
Explanation:
Half life is the time taken for a radioactive substance to degenerate or decay to half of its original size.
The half life of caffeine is 5 hours. So ingesting a 30 oz, this would be reduced to half of its size after the first 5 hours.
So that:
After one half life of 5 hours, the value of caffeine that would be left is;
= 15 oz
The amount of caffeine left after one half life of 5 hours is 15 oz.
1)
p = 2.4 * 10^5 Pa
T = 18° C + 273.15 = 291.15 k
r = 0.25 m => V = [4/3]π(r^3) = [4/3]π(0.25m)^3 = 0.06545 m^3 = 65.45 L
Use ideal gas equation: pV = nRT => n = pV / RT = [2.4*10^5 Pa * 0.06545 m^3] / [8.31 J/k*mol * 291.15k] = 6.492 mol
Avogadro number = 1 mol = 6.022 * 10^23 atoms
Number of atoms = 6.492 mol * 6.022 *10^23 atom/mol = 39.097 * 10^23 atoms = 3.91 * 10^24 atoms
2) Double atoms => double volume
V2 / V1 = r2 ^3 / r1/3
2 = r2 ^3 / r1 ^3 => r2 ^3 = 2* r1 ^3
r2 = [∛2]r1
The factor is ∛2
Answer:
the magnitude of acceleration will be 1.50m/s^2
Explanation:
To calculate your acceleration, you can use your formula that states that the net force on an object is equal to the mass of the object multiplied by the acceleration of the object. Fnet=ma
if you draw out this situation and label the forces you will have your vector towards the right with a magnitude of 20.0N and then your friction vector will be pointing to the left (in other words, in the negative direction) (opposing the direction of movement) with a magnitude of 5.00N, with the 10.0 kg box in the middle.
The net force will be calculated using F1+F2=Fnet where your F1=20.0N and F2= -5.00N (since it is towards the negative direction).
you will find that Fnet=15.0N
With that, plug in the values you know to calculate the acceleration of the block:
Fnet=ma
(15.0N)=(10.0kg)a from her you can divide both sides by 10 to isolate a:
1.50=a (and now make sure to label the units of your answer)
a=1.50m/s^2 (which is the typical unit for acceleration)
Answer:
The number of bright fringes per unit width on the screen is, 
Explanation:
If d is the separation between slits, D is the distance between the slit and the screen and 
is the wavelength of the light. Let x is the number of bright fringes per unit width on the screen is given by :
is the wavelength
n is the order
If n = 1,
So, the the number of bright fringes per unit width on the screen is 
. Hence, the correct option is (B).
