Answer:
h = 9.83 cm
Explanation:
Let's analyze this interesting exercise a bit, let's start by comparing the density of the ball with that of water
let's reduce the magnitudes to the SI system
r = 10 cm = 0.10 m
m = 10 g = 0.010 kg
A = 100 cm² = 0.01 m²
the definition of density is
ρ = m / V
the volume of a sphere
V =
V =
π 0.1³
V = 4.189 10⁻³ m³
let's calculate the density of the ball
ρ =
ρ = 2.387 kg / m³
the tabulated density of water is
ρ_water = 997 kg / m³
we can see that the density of the body is less than the density of water. Consequently the body floats in the water, therefore the water level that rises corresponds to the submerged part of the body. Let's write the equilibrium equation
B - W = 0
B = W
where B is the thrust that is given by Archimedes' principle
ρ_liquid g V_submerged = m g
V_submerged = m / ρ_liquid
we calculate
V _submerged = 0.10 9.8 / 997
V_submerged = 9.83 10⁻⁴ m³
The volume increassed of the water container
V = A h
h = V / A
let's calculate
h = 9.83 10⁻⁴ / 0.01
h = 0.0983 m
this is equal to h = 9.83 cm
If we have the angle and magnitude of a vector A we can find its Cartesian components using the following formula

Where | A | is the magnitude of the vector and
is the angle that it forms with the x axis in the opposite direction to the hands of the clock.
In this problem we know the value of Ax and Ay and we need the angle
.
Vector A is in the 4th quadrant
So:

So:

So:

= -47.28 ° +360° = 313 °
= 313 °
Option 4.
Using current technology, useful parallax measurements can only be found for stars up to about 340 light years (100 parsecs) away.
Answer:
A)
0.395 m
B)
2.4 m/s
Explanation:
A)
= mass of the cart = 1.4 kg
= spring constant of the spring = 50 Nm⁻¹
= initial position of spring from equilibrium position = 0.21 m
= initial speed of the cart = 2.0 ms⁻¹
= amplitude of the oscillation = ?
Using conservation of energy
Final spring energy = initial kinetic energy + initial spring energy

B)
= mass of the cart = 1.4 kg
= spring constant of the spring = 50 Nm⁻¹
= amplitude of the oscillation = 0.395 m
= maximum speed at the equilibrium position
Using conservation of energy
Kinetic energy at equilibrium position = maximum spring potential energy at extreme stretch of the spring
