Answer:
The specific heat capacity of the zinc metal measured in this experiment is 0.427 J/g.°C
Explanation:
From the experimental data, the water loses heat because its initial temperature is greater than the final temperature of the mixture. On the other hand, the zinc metal gains heat because its initial temperature is less than the final temperature of the mixture
Heat loss by water = Heat gain by zinc metal
m1C1(T1 - T3) = m2C2(T3 - T2)
m1 is mass of water = 55.4 g
C1 is specific heat capacity of water = 4.2 J/g.°C
m2 is mass of zinc metal = 23.4 g
C2 is specific heat capacity of zinc metal
T1 is the initial temperature of water = 99.61 °C
T2 is the initial temperature of zinc metal = 21.6 °C
T3 is the final temperature of the mixture = 96.4 °C
55.4×4.2(99.61 - 96.4) = 23.4×C2(96.4 - 21.6)
746.9028 = 1750.32C2
C2 = 746.9028/1750.32 = 0.427 J/g.°C
(a) The moment of inertia of the wheel is 78.2 kgm².
(b) The mass (in kg) of the wheel is 1,436.2 kg.
(c) The angular speed (in rad/s) of the wheel at the end of this time period is 3.376 rad/s.
<h3>
Moment of inertia of the wheel</h3>
Apply principle of conservation of angular momentum;
Fr = Iα
where;
- F is applied force
- r is radius of the cylinder
- α is angular acceleration
- I is moment of inertia
I = Fr/α
I = (200 x 0.33) / (0.844)
I = 78.2 kgm²
<h3>Mass of the wheel</h3>
I = ¹/₂MR²
where;
- M is mass of the solid cylinder
- R is radius of the solid cylinder
- I is moment of inertia of the solid cylinder
2I = MR²
M = 2I/R²
M = (2 x 78.2) / (0.33²)
M = 1,436.2 kg
<h3>Angular speed of the wheel after 4 seconds</h3>
ω = αt
ω = 0.844 x 4
ω = 3.376 rad/s
Thus, the moment of inertia of the wheel is 78.2 kgm².
The mass (in kg) of the wheel is 1,436.2 kg.
The angular speed (in rad/s) of the wheel at the end of this time period is 3.376 rad/s.
Learn more about moment of inertia here: brainly.com/question/14839816
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1.49 MPH (Miles Per Hour)
Answer:
a) V = - x ( σ / 2ε₀)
c) parallel to the flat sheet of paper
Explanation:
a) For this exercise we use the relationship between the electric field and the electric potential
V = - ∫ E . dx (1)
for which we need the electric field of the sheet of paper, for this we use Gauss's law. Let us use as a Gaussian surface a cylinder with faces parallel to the sheet
Ф = ∫ E . dA = 
/ε₀
the electric field lines are perpendicular to the sheet, therefore they are parallel to the normal of the area, which reduces the scalar product to the algebraic product
E A = q_{int} /ε₀
area let's use the concept of density
σ = q_{int}/ A
q_{int} = σ A
E = σ /ε₀
as the leaf emits bonnet towards both sides, for only one side the field must be
E = σ / 2ε₀
we substitute in equation 1 and integrate
V = - σ x / 2ε₀
V = - x ( σ / 2ε₀)
if the area of the sheeta is 100 cm² = 10⁻² m²
V = - x (10⁻²/(2 8.85 10⁻¹²) = - x ( 5.6 10⁻¹⁰)
x = 1 cm V = -1 V
x = 2cm V = -2 V
This value is relative to the loaded sheet if we combine our reference system the values are inverted
V ’= V (inf) - V
x = 1 V = 5
x = 2 V = 4
x = 3 V = 3
These surfaces are perpendicular to the electric field lines, so they are parallel to the sheet.
In the attachment we can see a schematic representation of the equipotential surfaces
b) From the equation we can see that the equipotential surfaces are parallel to the sheet and equally spaced
c) parallel to the flat sheet of paper
