Cl2 is the answer. Hope this helps you.
Answer:
Explanation:
Given that,
Bathysphere radius
r = 1.5m
Mass of bathysphere
M = 1.2 × 10⁴ kg
Constant speed of descending.
v = 1.2m/s
Resistive force
Fr = 1100N upward direction
Density of water
ρ = 1.03 × 10³kg/m³
The volume of the bathysphere can be calculated using
V = 4πr³ / 3
V = 4π × 1.5³ / 3
V = 14.14 m³
The Bouyant force can be calculated using
Fb = ρgV
Fb = 1.03 × 10³ × 9.81 × 14.14
Fb = 142,846.18 N
Buoyant force is acting upward
Weight of the bathysphere
W = mg
W = 1.2 × 10⁴ × 9.81
W = 117,720 N
Weight is acting downward
The net positive buoyant using resolving
Fb+ = Fb — W
Fb+ = 142,846.18 — 117,720
Fb+ = 25,126.18 N
The force acting downward is the weight of the submarine and it is equal to the positive buoyant force and the resistive force
W = Fb+ + Fr
W = 25,126.18 + 1100
W = 26,226.18
mg = 26,226.18
m = 26,226.18 / 9.81
m = 2673.4kg
Mass of submarine is 2673.4kg
Answer:
B) Bb × Bb I think I can't promise that's right
I just figured this out now.
First you would use the formula
Ephoton= hc/λ and substitute in the value's of plank's constant, the speed of light in a vaccum and the wavelength which will give you the energy in joules. Then you go to the reference table and solve for the energy used between the different levels for Mercury making sure to convert electron volts to jules. In the end the correct answer should be energy level D.
Answer:
false 20 n x 0.32 m = 6.4 J