<span>A spatial pattern arranges main points according to physical direction or location.</span>
Answer:
0.3 m
Explanation:
Initially, the package has both gravitational potential energy and kinetic energy. The spring has elastic energy. After the package is brought to rest, all the energy is stored in the spring.
Initial energy = final energy
mgh + ½ mv² + ½ kx₁² = ½ kx₂²
Given:
m = 50 kg
g = 9.8 m/s²
h = 8 sin 20º m
v = 2 m/s
k = 30000 N/m
x₁ = 0.05 m
(50)(9.8)(8 sin 20) + ½ (50)(2)² + ½ (30000)(0.05)² = ½ (30000)x₂²
x₂ ≈ 0.314 m
So the spring is compressed 0.314 m from it's natural length. However, we're asked to find the additional deformation from the original 50mm.
x₂ − x₁
0.314 m − 0.05 m
0.264 m
Rounding to 1 sig-fig, the spring is compressed an additional 0.3 meters.
To solve this, we use the Wien's Displacement Law as shown in the attached picture. First, convert the temperature to Kelvin.
C to F:
C = (F - 32)*5/9
C = (325 - 32)*5/9 = 162.78 °C
C to K:
K = C + 273
K = 162.78 + 273 = 435.78 K
λmax = 2898/435.78 =
<em>6</em><em>.65 μm</em>
Answer:
If a Gaussian surface is completely inside an electrostatic conductor, the electric field must always be zero at all points on that surface.
Explanation:
Option A is incorrect because, given this case, it is easier to calculate the field.
Option B is incorrect because, in a situation where the surface is placed inside a uniform field, option B is violated
Option C is also incorrect because it is possible to be a field from outside charges, but there will be an absence of net flux through the surface from these.
Hence, option D is the correct answer. "If a Gaussian surface is completely inside an electrostatic conductor, the electric field must always be zero at all points on that surface."
