A 15.75-g<span> piece of iron absorbs 1086.75 </span>joules<span> of </span>heat<span> energy, and its ... </span>How many joules<span> of </span>heat<span> are </span>needed<span> to raise the temperature of 10.0 </span>g<span> of </span>aluminum<span> from 22°C to 55°C, if the specific </span>heat<span> of </span>aluminum<span> is o.90 J/</span>g<span>”C2 .</span>
Answer:
The slope of a position-time graph can be calculated as:

where
is the increment in the y-variable
is the increment in the x-variable
We can verify that the slope of this graph is actually equal to the velocity. In fact:
corresponds to the change in position, so it is the displacement, 
corresponds to the change in time
, so the time interval
Therefore the slope of the graph is equal to

which corresponds to the definition of velocity.
Answer:
10.6cm
Explanation:
We are given 5.3cm below the starting point (spring extension).
Therefore, to find static vertical equilibrium, we use the equation:
kx = mg
Where:
k = spring constant =
=mg/5.3 kg/s²
We are told the object was dropped from rest.
Therefore:
loss in potential energy = gain in spring p.e
Let's use the expression:
mgx = ½kx²
We are asked to find the stretch at maximum elongation x.
To find x, we make x subject of the formula.
Therefore, we have:
x = 2mg/k (after rearranging the equation above)
x = (2mg) / (mg/5.3)
x = 10.6cm
Answer: B. V/b
Explanation:
Answer:
Total pressure exerted at bottom = 119785.71 N/m^2
Explanation:
given data:
volume of water in bottle = 150 L = 0.35 m^3
Area of bottle = 2 ft^2
density of water = 1000 kg/m
Absolute pressure on bottom of bottle will be sum of atmospheric pressure and pressure due to water
Pressure due to water P = F/A
F, force exerted by water = mg
m, mass of water = density * volume
= 1000*0.350 = 350 kg
F = 350*9.8 = 3430 N
A = 2 ft^2 = 0.1858 m^2
so, pressure P = 3430/ 0.1858 = 18460.71 N/m^2
Atmospheric pressure
At sea level atmospheric pressure is 101325 Pa
Total pressure exerted at bottom = 18460.71 + 101325 = 119785.71 N/m^2
Total pressure exerted at bottom = 119785.71 N/m^2