The body of a porter carrying load on head on a level road uses more energy than is required when the porter is resting completely. In this way the porter is doing some work.
However, when we think of theoretical definition of work, there is no force resisting the horizontal movement of the load when carried on a level road, and therefore there is no work done when a porter carrying a load on his head walk along a level road.
When the porter is going down the stairs the gravitational energy acting on the load is assisting the downward movement of the load. In this way energy is released by downward movement of the load. Thus the porter is doing negative work on the load.
Density = mass/ volume
13.6g/cm^3 = m /14.6cm^3
13.6g/cm^3 *14.6 cm^3 = m
198.56g = m
Answer:
m_{p} = 0.3506 kg
Explanation:
For this exercise we use Newton's equilibrium equation
B - Wc-Wp = 0
where B is the thrust of the water, Wc is the weight of the coins and Wp is the weight of the plastic block
B = Wc + Wp
the state push for the Archimeas equation
B = ρ_water g V
the volume of the water is the area of the block times the submerged height h, which is
h´ = 8 - h
where h is the height out of the water
ρ_water g A h´ =
g +
g
ρ_water A h´ = m_{c} + m_{p}
write this equation to make the graph
h´= 1 /ρ_water A (m_{c} +m_{p})
h´ = 1 /ρ_waterA (m_{c} + m_{p})
if we graph this expression, we get an equation of the line
y = m x + b
where
y = h´
m = 1 /ρ_water A
b = mp /ρ_water A
whereby
m_{p} = b ρ_water A
ρ_water = 1000 kg / m³
b = 0.0312 m
m = 0.0890 m / kg
we substitute the slope equation
b = m_{p} / m
calculate
m_{p}= 0.0312 / 0.0890
m_{p} = 0.3506 kg
Total distance during the time interval =
(average speed during the time interval)
times
(length of the time interval) .
Answer:
Condensation
Explanation:
Is is the stage in which gas changes to liquid