Answer:
A.c
Explanation:
The chromosphere is above the photosphere, the visible "surface" of the Sun. It lies below the solar corona, the Sun's upper atmosphere, which extends many thousands of kilometers above the chromosphere into space. The plasma (electrically charged gas) in the chromosphere has a very low density.
In basic terms it is the 2nd one out from the core.
I think it 32, but i’m not sure
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.
Explanation:
Internal energy = heat + work
U = Q + W
Since there's no change in volume (rigid walls), W = 0.
U = Q
U = n Cᵥ ΔT
U = (4.0 mol) (2.5 × 8.314 J/mol/K) (354 C − 17 C)
U = 28,000 J
Answer:
6.9066 × 10⁻⁵ m
Explanation:
For constructive interference, the expression is:
Where, m = 1, 2, .....
d is the distance between the slits.
The formula can be written as:
....1
The location of the bright fringe is determined by :
Where, L is the distance between the slit and the screen.
For small angle , 
So,
Formula becomes:
Using 1, we get:
Thus, the distance between the central maximum is 3.00 cm
First bright fringe , m = 1 occur at 3.00 / 2 = 1.50 cm
Since,
1 cm = 0.01 m
y = 0.0150 m
Given L = 2.00 m
λ = 518 nm
Since, 1 nm = 10⁻⁹ m
So,
λ = 518 × 10⁻⁹ m
Applying the formula as:
<u>⇒ d, distance between the slits = 6.9066 × 10⁻⁵ m</u>
