Answer:
1.129×10⁻⁵ N
1.295 m
Explanation:
Take right to be positive. Sum of forces on the 31.8 kg mass:
∑F = GM₁m / r₁² − GM₂m / r₂²
∑F = G (M₁ − M₂) m / r²
∑F = (6.672×10⁻¹¹ N kg²/m²) (516 kg − 207 kg) (31.8 kg) / (0.482 m / 2)²
∑F = 1.129×10⁻⁵ N
Repeating the same steps, but this time ∑F = 0 and we're solving for r.
∑F = GM₁m / r₁² − GM₂m / r₂²
0 = GM₁m / r₁² − GM₂m / r₂²
GM₁m / r₁² = GM₂m / r₂²
M₁ / r₁² = M₂ / r₂²
516 / r² = 207 / (0.482 − r)²
516 (0.482 − r)² = 207 r²
516 (0.232 − 0.964 r + r²) = 207 r²
119.9 − 497.4 r + 516 r² = 207 r²
119.9 − 497.4 r + 309 r² = 0
r = 0.295 or 1.315
r can't be greater than 0.482, so r = 0.295 m.
Answer:
1) 51 m
2) Some energy was transformed to other forms. (This question)
3) 3.24 J
4) 45 J
5) 1020 J
Explanation:
100%
Answer:
0.82 mm
Explanation:
The formula for calculation an bright fringe from the central maxima is given as:
so for the distance of the second-order fringe when wavelength = 745-nm can be calculated as:
where;
n = 2
= 745-nm
D = 1.0 m
d = 0.54 mm
substituting the parameters in the above equation; we have:
= 0.00276 m
= 2.76 × 10 ⁻³ m
The distance of the second order fringe when the wavelength = 660-nm is as follows:
= 1.94 × 10 ⁻³ m
So, the distance apart the two fringe can now be calculated as:
= 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m
= 10 ⁻³ (2.76 - 1.94)
= 10 ⁻³ (0.82)
= 0.82 × 10 ⁻³ m
= 0.82 × 10 ⁻³ m
= 0.82 mm
Thus, the distance apart the second-order fringes for these two wavelengths = 0.82 mm
Answer:
Please refer to the figure.
Explanation:
The magnitude of the magnetic field can be found by Biot-Savart Law. We should divide the loop into four components. Each component has a similar solution but their directions are quite different.
The directions can be found by right-hand rule. Point your index finger into the direction of current, point your middle finger towards the target point (0,0,a). Your thumb will show you the direction of magnetic field.
The movements of the tectonic plates