Answer:
<h2>35.56 m/s²</h2>Explanation:
The acceleration of an object given it's velocity and time taken can be found by using the formula
v is the velocity
t is the time
From the question we have
We have the final answer as
<h3>35.56 m/s²</h3>Hope this helps you
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
Answer:
Power_input = 85.71 [W]
Explanation:
To be able to solve this problem we must first find the work done. Work is defined as the product of force by distance.
where:
W = work [J] (units of Joules)
F = force [N] (units of Newton)
d = distance [m]
We need to bear in mind that the force can be calculated by multiplying the mass by the gravity acceleration.
Now replacing:
Power is defined as the work done over a certain time. In this way by means of the following formula, we can calculate the required power.
where:
P = power [W] (units of watts)
W = work [J]
t = time = 40 [s]
The calculated power is the required power. Now as we have the efficiency of the machine, we can calculate the power that is introduced, to be able to do that work.