Physical change = changes the physical properties (more commonly known as it's look)
Chemical change = changes the chemical properties into an entire new chemical form
Examples of physical change would be melting ice cubes or sugar cubes.
Examples of chemical change would be cooking eggs or burning paper because you're changing its chemical properties.
Aquifer- the layer permeable rock that has connecting pores & transmit water freely, the artesian well water rises to the surface.
Hope this helps!
<span>22.5 newtons.
First, let's determine how much energy the stone had at the moment of impact. Kinetic energy is expressed as:
E = 0.5mv^2
where
E = Energy
m = mass
v = velocity
Substituting known values and solving gives:
E = 0.5 3.06 kg (7 m/s)^2
E = 1.53 kg 49 m^2/s^2
E = 74.97 kg*m^2/s^2
Now ignoring air resistance, how much energy should the rock have had?
We have a 3.06 kg moving over a distance of 10.0 m under a force of 9.8 m/s^2. So
3.06 kg * 10.0 m * 9.8 m/s^2 = 299.88 kg*m^2/s^2
So without air friction, we would have had 299.88 Joules of energy, but due to air friction we only have 74.97 Joules. The loss of energy is
299.88 J - 74.97 J = 224.91 J
So we can claim that 224.91 Joules of work was performed over a distance of 10 meters. So let's do the division.
224.91 J / 10 m
= 224.91 kg*m^2/s^2 / 10 m
= 22.491 kg*m/s^2
= 22.491 N
Rounding to 3 significant figures gives an average force of 22.5 newtons.</span>
The Young modulus E is given by:
where
F is the force applied
A is the cross-sectional area perpendicular to the force applied
is the initial length of the object
is the increase (or decrease) in length of the object.
In our problem,
is the initial length of the column,
is the Young modulus. We can find the cross-sectional area by using the diameter of the column. In fact, its radius is:
and the cross-sectional area is
The force applied to the column is the weight of the load:
Now we have everything to calculate the compression of the column:
So, the column compresses by 1.83 millimeters.
Answer:
Object B has greater density
desity A=20/10=2 g cm^-3 . density B=50/10=5 g cm^-3
the object that has greater mass has the greater density because the volume of the those two objects are same
