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Bogdan [553]
3 years ago
11

I need some help on these problems:

Physics
2 answers:
SCORPION-xisa [38]3 years ago
5 0
Q1. The answer is 54.629 metric tons

A mass (m) of a liquid is its density (D) multiplied by its volume (V): m = D · V
Step 1. Calculate the volume (V) of the swimming pool.
Step 2. Calculate the mass (m) of the water.
Step 3. Convert the mass into the metric tones.

Step 1. 
V = w * l * h             (w - weight, l - length, h - height)
w = 3.5 m
l = 9 m
h = 1.75 m
_______
V = 3.5 * 9 * 1.75
V = 55.125 m³

Step 2:
m = D * V
D = 0.991 g/cm³
1 g = 0.001 kg
1 cm³ = 0.000001 m³
D = 0.991g / 1cm³ = 0.991 * 0.001 kg / 0.000001 m³ = 991 kg/m³
V = 55.125 m³
_________
m = 991 * 55.125
m = 54628.875 kg ≈ 54629 kg

Step 3.
Since 1 metric ton is 1000 kg, how many metric tons are 54629 kg?
1 mt : 1000 kg = x : 54629 kg
x = 1 mt : 1000 kg * 54629 kg
x = 54.629 metric tons



Q2. The answer is 3.2 l.
<span>
Step 1. Convert g/cm</span>³ into kg/dm³ (because 1 l = 1 dm³)
Step 2. Calculate how many liters of the liquid have a mass of 3.75 kg using the proportion

Step 1.
D = <span>1.17 g/cm3
</span>1 g = 0.001 kg
1 cm³ = 0.001 dm³
D = 1.17 * 0.001 kg / 0.001 dm³ = 1.17 kg/dm³

Step 2.
If 1.17 kg is in 1 dm³, 3.75 kg are in how many dm³:
1.17 kg : 1 dm³ = 3.75 kg : x
x =  3.75 kg * 1 dm³ : 1.17 kg
x = 3.2 dm³
1 dm³ = 1 l
x = 3.2 l



Q3. The answer is 2757140 g/m³

Step 1. Calculate the increased volume and convert units into liters.
Step 2. Calculate the density of the alloy sample and convert units into g/m³

Step 1.
The difference in the volume of the water in cylinder is:
V = 19.5 ml - 16 ml = 3.5 ml
1 ml = 0.001 l
V = 3.5 * 0.001 l = 0.0035 l

Step 2. 
A density (D) of an object with mass m and volume V is:
D = m/V

We have:
D = ?
m = 9.65 g
V = 0.0035 l
______
D = 9.65/0.0035 g/l = 2757.14 g/l
1 l = 1 dm³ = 0.001 m³
D = 2757.14 g/0.001 m³ = 2757140 g/m³
Elza [17]3 years ago
3 0
2.vol 3.75/1.17 = 3.205 cm^3
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The given lengths at 0 °C are 2.5 m

Let l₀ be the given lengths of the glass and steel rods at 0 °C. Let l and l' be the lengths of the glass and steel rods at 100 °C respectively.

From our expression for linear expansivity,

l = l₀ + l₀αΔθ where α = linear expansivity of glass = 0.000008/°C and Δθ = temperature change = θ - θ' where θ = 100 °C and θ' = 0 °C. So, Δθ = 100 °C - 0 °C = 100 °C.

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l' = l₀ + l₀α'Δθ where α' = linear expansivity of steel = 0.000012/°C and Δθ = temperature change = θ - θ' where θ = 100 °C and θ' = 0 °C. So, Δθ = 100 °C - 0 °C = 100 °C.

Since the difference in their lengths at 100 °C = 0.001 m, we have that

l - l' = l₀ + l₀αΔθ - (l₀ + l₀α'Δθ)

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Making l₀ subject of the formula, we have

l₀ = (l - l')/[(α- α')Δθ]

Substituting the values of the variables into the equation, we have

l₀ = (l - l')/[(α- α')Δθ]

l₀ = 0.001 m/[(0.000008/°C - 0.000012/°C)100 °C.]

l₀ = 0.001 m/[(-0.000004/°C)100 °C.]

l₀ = 0.001 m/-0.0004

l₀ = -2.5 m

Neglecting the negative sign,

l₀ = 2.5 m

So, the given lengths at 0 °C are 2.5 m

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From the question we have

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From the above description, it can be deduced that the alpha particle will stay and interact with the hand (because of its low penetrating power) as the remaining particles move through the skin.

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