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Physics Numericals and Word Problems (Coursework Sample)

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A U-shaped tube is partially filled with seawater, whose density is known to be ρ1= 1.024g/cm3. At the right end, an oily liquid of unknown density ρ2 is added. Since oil and seawater are immiscible, an interface is formed, as shown in the figure below. The heights of the columns are h1 = 2.1cm and h2 = 2.4cm, respective Determine the density ρ2.
At a smelting factory, a metallic piece is produced using copper, whose density is ρ_Cu = 8.93g/ml. Because of an error during the production process, an air bubble was formed in its interior. Since the bubble’s volume V_bis unknown, the piece is subject to a series of tests. Its weight is W = 25N when measured dry and W^o = 20N when fully submerged in water ρ_l (= 1.000g/mL). Determine the object’s density ρ and the bubble’s volume Vb. Note: g = 9.8m/s2 and ρ_air ∼ 0.
In a square container, a lab technician carefully pours vinegar 1.049 g/mL and palm oil 0.876 g/mL .The container and the liquids are then cooled to 0°C. A block of ice 0.917 g/mL is then placed floating at the interface, as shown in the figure below. Determine the percentage of ice that will be submerged in vinegar, disregarding any melting or dissolution

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Content:

Problem 1
A U-shaped tube is partially filled with sea water, whose density is known to be ρ1= 1.024g/cm3. At the right end, an oily liquid of unknown density ρ2 is added. Since oil and sea water are immiscible, an interface is formed, as shown in the figure below. The heights of the columns are h1 = 2.1cm and h2 = 2.4cm, respective Determine the density ρ2.
center114300Given Data:
ρ1= 1.024g/cm3
h1 = 2.1cm
h2 = 2.4cm
Findings:
ρ2=?
Where ρ1 is the density of water, ρ2 is the density of oily liquid, and hi, h2 are column heads.
213360020955Patm00Patm32004001905Patm00Patmcenter7620Solution:
2190750350520.00.2200275579120B00B3238500340995 .00 .left46990We know that the atmospheric pressure, Patm on both sides is the same.Because liquids in tubes are in equilibrium. So, pressure at the same height level in the same liquid should be the same.00We know that the atmospheric pressure, Patm on both sides is the same.Because liquids in tubes are in equilibrium. So, pressure at the same height level in the same liquid should be the same.3295650569595A00A
Thus, pressure at point A = Pressure at point B
20097759207500 Patm+ ρ2gh2= Patm+ ρ1gh1 equation (1)
Thus, simplifying the equation (1)
ρ2gh2 = ρ1gh1
ρ2h2 = ρ1h1
ρ2= ρ1h1h2
By putting the values
ρ2= 1.024g/cm3×2.1cm2.4cm
-16192422288500Calculating the values, we finally get:
ρ2= 0.896g/cm3
Here, the density of oily liquid is 0.896 grams per cubic centimeter.
Problem 2
center1632585At a smelting factory, a metallic piece is produced using copper, whose density is ρCu = 8.93g/ml. Because of an error during the production process, an air bubble was formed in its interior. Since the bubble’s volume Vbis unknown, the piece is subject to a series of tests. Its weight is W = 25N when measured dry and Wo = 20N when fully submerged in water ρl (= 1.000g/mL). Determine the object’s density ρ and the bubble’s volume Vb. Note: g = 9.8m/s2 and ρair ∼ 0.
Given Data:
ρCu = 8.93g/ml.
W = 25N
Wo = 20N
ρl = 1.000g/mL
g = 9.8m/s2
ρair ∼ 0
Where ρCuthe density of copper wire, W is the weight of water in air, Wo is the weight of an object in water, ρl is water density, g is gravitational acceleration, and ρairis the density of air.
Findings:
* Object’s density ρ=?
* Bubble’s volume Vb=?
Solution:
Weight of the object in air W= 25N
147637510414000We know that W=mg eq. (I)
So mg= 25N
m= 25N9.81m/s2 calculating for m
m= 2.54kg mass of the object
Now when object is fully submerged in water then:
Wo = mg-B
Where B = Buoyant force
We know that Buoyant force B = Vin ρl g
Where Vin denotes the volume of the object within the liquid and ρl denotes the density of the liquid.
In given situation:
B= V ρw g
145732510985500Wo = mg-B eq. (II)
Solving eq. (II)
20= 25- Vρlg
V ρw g = 5
201930023812500V = 5ρwg => Volume of object eq. (III)
Now we know that, Density of object ρ= mass of objectvolume of object
58102514478000ρ= mv eq. (IV)
Now putting the values of v and m in eq. (IV)
ρ = 2.545N1×9.81
We get ρ= 5g/mL (a) 0-63500
Now we know that, mass of copper = volume of copper × Density of copper
12858759525000m= Vcopper× ρCu eq. (V)
Where Vcopper = V-Vb and ρCuis given
Now putting the values in eq. (V) and solving for Vb
Where ρCu =8.93g/mL => 8.93×103 kg/m3
Similarly, 1.000g/mL => 1000 kg/m3
-9525076835000left1063626002.54kg = (5kgm/s21000 kg/m3×9.81m/s2-Vb) ×(8.93×103) kg/m3
Vb= (0.502×10-3)-(0.2857×10-3)
Vb= 0.2245×10-3m3
224.5cm3
Problem 3
In a square container, a lab technician carefully pours vinegar 1.049 g/mL

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