Answer:
A pure substance (an element or a compound) is made up of only one type of atom, or molecule.
Examples: water (H2O molecules), oxygen (O2 molecules), argon (Ar atoms).
A mixture is made of up of more than one type of atom or molecule.
Examples: salt water (H2O molecules, Na+ ions, Cl- ions),
air (N2 molecules, O2 molecules, Ar atoms, CO2 molecules)
Note: the composition of a pure substance is fixed, but the composition of a mixture can vary. Two salt water solutions might not have the same concentration of salt. Or, if you prefer, one blueberry muffin might have more blueberries in it than another blueberry muffin.
make me brainlist if u think its helpful
Refer to the diagram shown below.
The initial KE (kinetic energy) of the system is
KE₁ = (1/2)mu²
After an inelastic collision, the two masses stick together.
Conservation of momentum requires that
m*u = 2m*v
Therefore
v = u/2
The final KE is
KE₂ = (1/2)(2m)v²
= m(u/2)²
= (1/4)mu²
= (1/2) KE₁
The loss in KE is
KE₁ - KE₂ = (1/2) KE₁.
Conservation of energy requires that the loss in KE be accounted for as thermal energy.
Answer: 1/2
To determine whether an object is in motion or not, you first
need to specify a reference point, because there's no such
thing as "real" motion, only motion relative to something.
Once you've named the reference point, you have to look at
the object at two different times. Each time you look at it, you
measure its distance and direction from the reference point.
If there's any difference in these measurements from one time
to the next, then the object has had average motion during the
period between the two observations.
That's the best you can do ... find average motion during some
period of time. You can never definitely tell whether or not the
object ever stopped during that time. But you can sneak up on
it by making the time period between the two observations shorter
and shorter.
I'm pretty sure they expand
Answer:
a) W₁ = 54000 Lb-ft
b) W₂ = 77760 Lb-ft
c) W₃ = 24000 Lb-ft
W₄ = 40560 Lb-ft
Step by step
W= ∫₁² ydF 1 and 2 are the levels of liquid
Where dF is the differential of weight of a thin layer
y is the height of the differential layer and
ρ*V = F
Then
dF = ρ* A*dy*g
ρ*g = 60 lb/ft³
A= Area of the base then
Area of the base is:
A(b) = 4*2 = 8 ft²
Now we have the liquid weighs 60 lb/ft³
Then the work is:
a)
W₁ = ∫₀¹⁵ 8*60*y*dy ⇒ W₁ =480* ∫₀¹⁵ y*dy
W₁ =480* y² /2 |₀¹⁵ ⇒ 480/2 [ (15)² - 0 ]
W₁ = 240*225
W₁ = 54000 Lb-ft
b) The same expression, but in this case we have to pump 3 feet higher, then:
W₂ = ∫₀¹⁸ 480*y*dy ⇒ 480*∫₀¹⁸ydy ⇒ 480* y²/2 |₀¹⁸
W₂ = 480/2 * (18)²
W₂ = 240*324
W₂ = 77760 Lb-ft
c) To pump two-thirds f the liquid we have
2/3* 15 = 10
W₃ = 480*∫₀¹⁰ y*dy ⇒ W₃ = 480* y²/2 |₀¹⁰
W₃ = 240*(10)²
W₃ = 24000 Lb-ft
d)
W₄ =480*∫₀¹³ y*dy
W₄ =480* y²/2 |₀¹³
W₄ = 240*(13)²
W₄ = 240*169
W₄ = 40560 Lb-ft
