Answer:
.conserves kinetic energy and momentum
.does not coalesce
.causes much lesser deformation as compared to inelastic
.the two bodies involved; after collision move with separate final velocities.
Explanation:
eg. throwing a ball at a wall
<em>collision</em><em> </em><em>between</em><em> </em><em>a</em><em> </em><em>punch</em><em> </em><em>and</em><em> </em><em>your</em><em> </em><em>nose</em><em>,</em><em>haha</em>
<em>I</em><em> </em><em>guess</em><em> </em><em>that</em><em> </em><em>will</em><em> </em><em>cause</em><em> </em><em>a lot</em><em> </em><em>deformation</em><em>.</em><em>.</em>
<em>.</em>
Answer:
decantation, distilling, freezing
Answer:
A and C
Explanation:
drag (the area of lower air pressure behind the car when moving) and mostly air resistance (the work to push the air in front of us away to move through - the faster we go, the stronger the air resists to move aside).
Answer:
a) F = 4.9 10⁴ N, b) F₁ = 122.5 N
Explanation:
To solve this problem we use that the pressure is transmitted throughout the entire fluid, being the same for the same height
1) pressure is defined by the relation
P = F / A
to lift the weight of the truck the force of the piston must be equal to the weight of the truck
∑F = 0
F-W = 0
F = W = mg
F = 5000 9.8
F = 4.9 10⁴ N
the area of the pisto is
A = pi r²
A = pi d² / 4
A = pi 1 ^ 2/4
A = 0.7854 m²
pressure is
P = 4.9 104 / 0.7854
P = 3.85 104 Pa
2) Let's find a point with the same height on the two pistons, the pressure is the same
where subscript 1 is for the small piston and subscript 2 is for the large piston
F₁ = 
the force applied must be equal to the weight of the truck
F₁ =
F₁ = (0.05 / 1) ² 5000 9.8
F₁ = 122.5 N