Work is equal to the force applied times the displacement. Since you pull the wagon at constant speed this means that there is no acceleration on the wagon as it does not change speed. F=ma. Since a=0, F=0. Therefore no work has been done in this situation
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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
A. A clastic Sedimentary rock
Explanation:
(a) We have,
Length of solenoid, l = 55 cm = 0.55 m
Diameter of the solenoid, d = 10 cm
Radius, r = 5 cm = 0.05 m
Number of loops in the solenoid is 1000.
(a) The self inductance in the solenoid is given by :
A is area
(b) The energy stored in the inductor is given by :
Hence, this is the required solution.
Answer:
C) 50 m/s
Explanation:
With the given information we can calculate the acceleration using the force and mass of the box.
Newton's 2nd Law: F = ma
- 5 N = 1 kg * a
- a = 5 m/s²
List out known variables:
- v₀ = 0 m/s
- a = 5 m/s²
- v = ?
- Δx = 250 m
Looking at the constant acceleration kinematic equations, we see that this one contains all four variables:
Substitute known values into the equation and solve for v.
- v² = (0)² + 2(5)(250)
- v² = 2500
- v = 50 m/s
The final velocity of the box is C) 50 m/s.
