Answer:
B. 24.2 m/s
Explanation:
Given;
mass of the roller coaster, m = 450 kg
height of the roller coaster, h = 30 m
The maximum potential energy of the roller coaster due to its height is given by;



Therefore, the maximum speed of the roller coaster is 24.2 m/s.
Answer:
d) 1.2 mT
Explanation:
Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.
First of all, we observe that:
- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is
I = 15 A
- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).
Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by

where
is the vacuum permeability
I = 15 A is the current in the conductor
r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field
Substituting, we find:

Answer:
A financial manager.
Explanation:
This is because a financial manager oversees the financial operations of a company. Generally, a financial manager assumes accounting responsibilities for the company. A financial manager is responsible for planning and managing the company's financial resources.
Side note:
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Answer:
The energy of the capacitors connected in parallel is 0.27 J
Given:
C = 
C' =
Potential difference, V = 300 V
Solution:
Now, we know that the equivalent capacitance of the two parallel connected capacitors is given by:

The energy of the capacitor, E is given by;


Answer:
a) The angular acceleration of the beam is 0.5 rad/s²CW (direction clockwise due the tangential acceleration is positive)
b) The acceleration of point A is 3.25 m/s²
The acceleration of point E is 0.75 m/s²
Explanation:
a) The relative acceleration of B with respect to D is equal:

Where
aB = absolute acceleration of point B = 2.5 j (m/s²)
aD = absolute acceleration of point D = 1.5 j (m/s²)
(aB/D)n = relative acceleration of point B respect to D (normal direction BD) = 0, no angular velocity of the beam
(aB/D)t = relative acceleration of point B respect to D (tangential direction BD)


We have that
(aB/D)t = BDα
Where α = acceleration of the beam
BDα = 1 m/s²
Where
BD = 2

b) The acceleration of point A is:

(aA/D)t = ADαj

The acceleration of point E is:
(aE/D)t = -EDαj
