Answer:
(a) The spring constant is 59.23 N/m
(b) The total energy involved in the motion is 0.06 J
Explanation:
Given;
mass, m = 240 g = 0.24 kg
frequency, f = 2.5 Hz
amplitude of the oscillation, A = 4.5 cm = 0.045 m
The angular speed is calculated as;
ω = 2πf
ω = 2 x π x 2.5
ω = 15.71 rad/s
(a) The spring constant is calculated as;
(b) The total energy involved in the motion;
E = ¹/₂kA²
E = (0.5) x (59.23) x (0.045)²
E = 0.06 J
The answer is b no problem
When a cloud of gas and dust in space was disturbed, maybe by the explosion of a nearby star.This explosion made waves in space which squeezed the cloud of gas & dust.
So, first the formula of Impulse is
I = force * time
We have force but no time.
Then, find time.
Next find acceleration,
F = mass * acceleration
5 = 3 * a
1.67 m/s^2
Next find time,
Acceleration = change in velocity / time
Change in velocity is velocity final - velocity initial
1.67 = 3 - 9 / time
Time = 3.6 s (round to 2 s.f.)
Lastly,
Impulse = force * time
Impulse = 5 * 3.6
Impulse is 18 Ns
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
