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3 years ago

A student releases a marble from the top of a ramp. The marble increases

Physics

1 answer:

Fed [463]3 years ago

7 0

Answer:

Vf = 69.56 cm/s

Explanation:

In order to find the final speed of the ramp, we will use the equations of motion. First we use second equation of motion to find out the acceleration of marble:

s = Vi t + (1/2)at²

where,

s = distance traveled = 160 cm

Vi = Initial Speed = 0 cm/s (since, marble starts from rest)

t = time interval = 4.6 s

a = acceleration = ?

Therefore,

160 cm = (0 cm/s)(4.6 s) + (1/2)(a)(4.6 s)²

a = (320 cm)/(4.6 s)²

a = 15.12 cm/s²

Now, we use first equation of motion:

Vf = Vi + at

Vf = 0 cm/s + (15.12 cm/s²)(4.6 s)

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a ball of diameter 10 cm and mass 10 grams is dropped in a container of water. the cross sectional area of the container is 100

Anastaziya [24]

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 = $\frac{4}{3} \ \pi r^{3}$

V = $\frac{4}{3}$ π 0.1³

V = 4.189 10⁻³ m³

let's calculate the density of the ball

ρ = $\frac{0.010}{4.189 \ 10^{-3} }$

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