Correct answer is:
<h2>The maximum number of orbits in an atom is <u>Seven.</u></h2><h3>Explanation:</h3>
Every energy level has a limited one orbital including two electrons. The orbits are settled in the sub-levels and there can be further than 1 sub-level as the number of energy levels rises. On energy level 1, there is 1 sub-level and 1 orbital. Energy level 2 can possess 2 sub-levels and 2 orbitals. These remain to develop as you progress from the nucleus of the atom, closing up with an infinite potential number of levels and orbits.
Answer:
a) F= 0,19 [N] according to problem statement
b) F = 0,19*10⁹ [N] using the right value of K
Explanation:
The force between two electric charges is according to Coulomb´s law is:
F = K * q₁*q₂ / d² where q₁ and q₂ are the charges on body one and body 2 respectively, d is the distance between the two bodies and K is a constant K = 8,988100*10⁹ N.m²/C². The problem establishes to use K = 8,988100 N.m²/C².
NOTE: To value of is : K = 8,988100*10⁹ N.m²/C². I am going to solve the problem using K = 8,988100 N.m²/C² if that information was an error, all we need to get the right answer is multiply the result by 10⁹
Then:
F = 8,988100 * 1,2* 0,36 / (4,5)² [ N*m²/C² ] * [ C*C*/m²]
F = 3,882859/ 20,25 [N]
F= 0,19 [N]
The force is of repulsion since the two charges are positive and in the direction of the straight line which passes through the centers of the bodies
Answer:
8050 J
Explanation:
Given:
r = 4.6 m
I = 200 kg m²
F = 26.0 N
t = 15.0 s
First, find the angular acceleration.
∑τ = Iα
Fr = Iα
α = Fr / I
α = (26.0 N) (4.6 m) / (200 kg m²)
α = 0.598 rad/s²
Now you can find the final angular velocity, then use that to find the rotational energy:
ω = αt
ω = (0.598 rad/s²) (15.0 s)
ω = 8.97 rad/s
W = ½ I ω²
W = ½ (200 kg m²) (8.97 rad/s)²
W = 8050 J
Or you can find the angular displacement and find the work done that way:
θ = θ₀ + ω₀ t + ½ αt²
θ = ½ (0.598 rad/s²) (15.0 s)²
θ = 67.3 rad
W = τθ
W = Frθ
W = (26.0 N) (4.6 m) (67.3 rad)
W = 8050 J
See attachment file below.
Hope it helped!
