1. In a single atom, no more than 2 electrons can occupy a single orbital? A. True
2. The maximum number of electrons allowed in a p sublevel of the 3rd principal level is?
B.6
3. A neutral atom has a ground state electronic configuration of 1s^2 2s^2. Which of the following statements concerning this atom is/are correct?
B. All of the above.
F=K*X,
F=M*a
M*a=K*X
2.5*9.81=K*0.0276
24.525=K*0.0276
24.525/0.0276=K
K= 888.6 N/m ---- force constant
assuming 2.5 refers to the new extension, just divide F/ 0.025
to get
981N/m
Answer:

Explanation:
<u>Frictional Force
</u>
When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:

The centripetal acceleration a_c is computed as

Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one

For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as

The normal force N is equal to the weight of the car, thus

Equating both forces

Simplifying

Substituting the values


Answer:
ω = 1.83 rad/s clockwise
Explanation:
We are given:
I1 = 3.0kg.m2
ω1 = -5.4rad/s (clockwise being negative)
I2 = 1.3kg.m2
ω2 = 6.4rad/s (counterclockwise being positive)
By conservation of the momentum:
I1 * ω1 + I2 * ω2 = (I1 + I2) * ω
Solving for ω:

Since it is negative, the direction is clockwise.