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ANEK [815]
3 years ago
12

Does fat and soap have density? ​

Physics
2 answers:
irinina [24]3 years ago
5 0

Fat has a density of 0.92 g/cm3 and soap has a density of 0.84 g/cm3. These are all properties that make fat and soap different substances.

yawa3891 [41]3 years ago
3 0

Answer:

The density of fat is about .88 g/ml and the density of soap is about 1.18 g/ml.

Explanation:

Fat and soap are not the same substances one reason is they have different properties .Malleability, hardness, solubility, melting point, color, and density are all properties. A property is a characteristic of a substance that does not change.

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While standing on a bridge 40.0 m above the ground, you drop a stone from rest. when the stone has fallen 3.80 m, you throw a se
Norma-Jean [14]
Given: distance 1 d₁ = 40 m;  distance 2 d₂ = 3.8 m   g = -9.8 m/s²

 Initial Velocity Vi = 0  Final Velocity of stone 2 is unknown = ?

Total distance dₓ = d₁ - d₂  = 40 m - 3.8 m = 36.2 m

Formula: a = Vf² -  Vi²/2d   derive for Final Velocity Vf
 
acceleration is now due to gravity, therefore a = g

Vf = √2gd   Vf = √2(9.8 m/s²)(36.2 m)

Vf = 26.64 m/s  

Reason: The second stone will still start from rest.


3 0
3 years ago
"what information does the slope of the line give you when graphing the pressure versus 1/volume at constant temperature for 1 m
Rashid [163]
Hello!

The slope of the line given by graphing pressure vs 1/Volume at constant temp for one mole of gas will give you the value for nRT from equation PV=nRT

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6 0
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How is humidity related to air pressure?
Vlad [161]
Humid air has higher pressure because of the heaviness of the water
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A 7.5 kg cannon ball leaves a canon with a speed of 185 m/s. Find the average net force applied to the ball if the cannon muzzle
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A 65.0-Ω resistor is connected to the terminals of a battery whose emf is 12.0 V and whose internal resistance is 0.5 Ω. Calcula
Luda [366]

Answer:

a) 0.1832 A

b) 11.91 Volts

c) 2.18 Watt , 0.0168 Watt

Explanation:

(a)

R = external resistor connected to the terminals of the battery = 65 Ω

E = Emf of the battery = 12.0 Volts

r = internal resistance of the battery = 0.5 Ω

i = current flowing in the circuit

Using ohm's law

E = i (R + r)

12 = i (65 + 0.5)

i = 0.1832 A

(b)

Terminal voltage is given as

V_{ab} = i R

V_{ab} = (0.1832) (65)

V_{ab} = 11.91 Volts

(c)

Power dissipated in the resister R is given as

P_{R} = i²R

P_{R} = (0.1832)²(65)

P_{R} = 2.18 Watt

Power dissipated in the internal resistance is given as

P_{r} = i²r

P_{r} = (0.1832)²(0.5)

P_{r} = 0.0168 Watt

5 0
3 years ago
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