. The velocity of a mass attached to a spring is given by v = (1.5 cm/s) sin(ωt + π/2), ..... Which of the following is the motion of objects moving in two dimensions
Answer:
d²x/dt² = - 4dx/dt - 4x is the required differential equation.
Explanation:
Since the spring force F = kx where k is the spring constant and x its extension = 2.45 equals the weight of the 4 kg mass,
F = mg
kx = mg
k = mg/x
= 4 kg × 9.8 m/s²/2.45 m
= 39.2 kgm/s²/2.45 m
= 16 N/m
Now the drag force f = 16v where v is the velocity of the mass.
We now write an equation of motion for the forces on the mass. So,
F + f = ma (since both the drag force and spring force are in the same direction)where a = the acceleration of the mass
-kx - 16v = 4a
-16x - 16v = 4a
16x + 16v = -4a
4x + 4v = -a where v = dx/dt and a = d²x/dt²
4x + 4dx/dt = -d²x/dt²
d²x/dt² = - 4dx/dt - 4x which is the required differential equation
KE = 1/2 * m * v^2
KE = 1/2 * 0.135 * 40^2
KE = 1/2 * 0.135 * 1600
KE = 108 J
Answer:
(a) The absolute pressure at the bottom of the freshwater lake is 395.3 kPa
(b) The force exerted by the water on the window is 36101.5 N
Explanation:
(a)
The absolute pressure is given by the formula

Where
is the absolute pressure
is the atmospheric pressure
is the density
is the acceleration due to gravity (Take
)
h is the height
From the question
h = 30.0 m
= 1.00 × 10³ kg/m³ = 1000 kg/m³
= 101.3 kPa = 101300 Pa
Using the formula
P = 101300 + (1000×9.8×30.0)
P = 101300 + 294000
P =395300 Pa
P = 395.3 kPa
Hence, the absolute pressure at the bottom of the freshwater lake is 395.3 kPa
(b)
For the force exerted
From
P = F/A
Where P is the pressure
F is the force
and A is the area
Then, F = P × A
Here, The area will be area of the window of the underwater vehicle.
Diameter of the circular window = 34.1 cm = 0.341 m
From Area = πD²/4
Then, A = π×(0.341)²/4 = 0.0913269 m²
Now,
From F = P × A
F = 395300 × 0.0913269
F = 36101.5 N
Hence, the force exerted by the water on the window is 36101.5 N
(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.
(b) The maximum height above the ground reached by the ball is 8.6 m.
(c) The distance off course the ball would be carried is 0.38 m.
(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.
<h3>
Horizontal and vertical components of the ball's velocity</h3>
Vx = Vcosθ
Vx = 39.7 x cos(17.8)
Vx = 37.8 m/s
Vy = Vsin(θ)
Vy = 39.7 x sin(17.8)
Vy = 12.14 m/s
<h3>Maximum height reached by the ball</h3>

Maximum height above ground = 7.51 + 1.09 = 8.6 m
<h3>Distance off course after 2 second </h3>
Upward speed of the ball after 2 seconds, V = V₀y - gt
Vy = 12.14 - (2x 9.8)
Vy = - 7.46 m/s
Horizontal velocity will be constant = 37.8 m/s
Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

<h3>Resultant speed of the ball and crosswind</h3>

<h3>Distance off course the ball would be carried</h3>
d = Δvt = (38.72 - 38.53) x 2
d = 0.38 m
The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.
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