Answer:

Given:
Mass of the polar bear (m) = 6.8 kg
Speed of the polar bear (v) = 5.0 m/s
To Find:
Kinetic energy of the polar bear (KE)
Explanation:
Formula:

Substituting values of m & v in the equation:





Kinetic energy of the polar bear (KE) = 23002.1 J
Yeah sure someone else in answering rn
Frequency = velocity/wavelength
Frequency = 10/20
Frequency = 0.5 Hz
Answer:
The correct option is A = 1960 N/m²
Explanation:
Given that,
Mass m= 20,000kg
Area A = 100m²
Pressure different between top and bottom
Assume the plane has reached a cruising altitude and is not changing elevation. Then sum the forces in the vertical direction is given as
∑Fy = Wp + FL = 0
where
Wp = is the weight of the plane, and
FL is the lift pushing up on the plane.
Let solve for FL since the mass of the plane is given:
Wp + FL = 0
FL = -Wp
FL = -mg
FL = -20,000× -9.81
FL = 196,200N
FL should be positive since it is opposing the weight of the plane.
Let Equate FL to the pressure differential multiplied by the area of the wings:
FL = (Pb −Pt)⋅A
where Pb and Pt are the static pressures on bottom and top of the wings, respectively
FL = ∆P • A
∆P = FL/A
∆P = 196,200 / 100
∆P = 1962 N/m²
∆P ≈ 1960 N/m²
The pressure difference between the top and bottom surface of each wing when the airplane is in flight at a constant altitude is approximately 1960 N/m². Option A is correct
Charge quantization is the principle that the charge of any object is an integer multiple of the elementary charge. Thus, an object's charge can be exactly 0 e, or exactly 1 e, −1 e, 2 e, etc., but not, say, 12 e, or −3.8 e, etc.