Answer:
7 Electrons
Explanation:
Assuming that final energy level is that same as valence shell, we first need to find how many of the electrics will be used to fill the ones before it.
Electron shell capacities are as follows :
First shell/Ebergy level: 2 electrons
Seconds shell/energy level : 8 electrons
Third shell /energy level :8 electrons
The first two shells total to 10, and the first three shells total to 18. Since chlorine (17) fills more than 2 shells but less that 3, the third one is its final energy level. We find the number in the final energy level by subtracting the 10 in the full first two. 2+8=10
17-10=7
Chlorine has 7 electrons on its final energy level.
Hope this helped!
Answer:
W= 38.4 J
Explanation:
Given that
m = 80 g= 0.08 kg
Initial speed ,u= 22 m/s
Final speed ,v= 38 m/s
The change in the kinetic energy of the particle
ΔKE= 38.4 J
We know that
Work done by all the forces =Change in the kinetic energy
That is why net work done = 38.4 J.
W= 38.4 J
Therefore the answer will be 38.4 J.
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.
Answer:
The normal force the seat exerted on the driver is 125 N.
Explanation:
Given;
mass of the car, m = 2000 kg
speed of the car, u = 100 km/h = 27.78 m/s
radius of curvature of the hill, r = 100 m
mass of the driver, = 60 kg
The centripetal force of the driver at top of the hill is given as;
where;
Fc is the centripetal force
is downward force due to weight of the driver
is upward or normal force on the drive
Therefore, the normal force the seat exerted on the driver is 125 N.
Answer: decantation and screening
Explanation:
