It would be D) the rope is pulled to the right. This is because their is a greater force in that direction.
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Answer:</h2>
1.68 x 10⁻⁸Ωm
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Explanation:</h2>
The resistance (R) of a wire is related to its length(L), its material resistivity(ρ) and its crossectional area(A) as follows;
R = ρL/A ------------------------(i)
Where;
A = πd² / 4 [where d = diameter of the wire]
From the question;
L = 6.90m
d = 2.15mm = 0.00215m
R = 0.0320Ω
First calculate the crossectional area (A) of the wire as follows;
A = πd² / 4
[Take π = 3.142]
d = 0.00215m
∴ A = 3.142 x (0.00215)² / 4
∴ A = 0.000003631m²
Now, substitute the values of A, L, and R into equation (i) as follows;
R = ρL/A
0.0320 = ρ x 6.90 / 0.000003631
0.0320 = 1900302.95 x ρ
Solve for ρ;
=> ρ = 0.0320 / 1900302.95
=> ρ = 1.68 x 10⁻⁸Ωm
Therefore, the resistivity of the material of the wire is 1.68 x 10⁻⁸Ωm
Answer: ratio = 3.14 × 10³⁵
Explanation:
From Newtons law of Universal Gravitation;
Every object in the universe is attracted to every other object
F = (Gm₁m₂)/r²
G = 6.67 X 10⁻¹¹ N-m²/kg²
M₁= mass of one object
M₂= mass of second object
r = distance from center of objects
q = charge = 3.2 x 10⁻¹⁹C.
- Calculating the gravitational force;
Fg = (Gm₁m₂)/r² = (Gm²)/r²
which is = (6.67×10⁻¹¹ × (6.64×10⁻²⁷)²) / r²
= (2.94 × 10⁻⁶³) / r²
- Calculating the electric force;
Fe = (Kq₁q₂) / r² = K q² / r² = (9×10⁹ × (3.2 × 10⁻¹⁹)² / r²
Fe = 9.22 × 10⁻²⁸ / r²
comparing the ratio of both forces we have;
Fe/Fg = (9.22×10⁻²⁸/r²) ÷ (2.94×10⁻⁶³/r²)
r² cancels from the above expression, which gives;
ratio as 3.14 × 10³⁵
Answer:
volume of the bubble just before it reaches the surface is 5.71 cm³
Explanation:
given data
depth h = 36 m
volume v2 = 1.22 cm³ = 1.22 × 
m³
temperature bottom t2 = 5.9°C = 278.9 K
temperature top t1 = 16.0°C = 289 K
to find out
what is the volume of the bubble just before it reaches the surface
solution
we know at top atmospheric pressure is about P1 = 
Pa
so pressure at bottom P2 = pressure at top + ρ×g×h
here ρ is density and h is height and g is 9.8 m/s²
so
pressure at bottom P2 = + 1000 × 9.8 ×36
pressure at bottom P2 =4.52 × Pa
so from gas law
here p is pressure and v is volume and t is temperature
so put here value and find v1
V1 = 5.71 cm³
volume of the bubble just before it reaches the surface is 5.71 cm³
