Answer:
a.)alkene
b.)alkane
d.)alkana
about c..I think there is something wrong with the question
Displacement = distance and direction from the start-point
to the end-point, regardless of the route followed on the way.
From the throw to the 'plop', the displacement is 5 meters down.
<u>Inertia affects the motion of an object as follows:</u>
When an object is in motion, it will continue to be in the same state unless otherwise some outside force is being applied to it. Thus, inertia affects the motion of an object. It restricts some other force being acted upon the object.
But mass of an object is directly proportional to inertia. So when the inertia is more on an object, it means that the object has more mass. For example, if there are two similar bricks, one that is made up of mortar and the other one is made of Styrofoam.
To identify which brick is made of Styrofoam without lifting the bricks, push both the bricks with equal force, the one that has less resistance tends to move faster. This means that it has less inertia and hence less mass.
Answer:
0.8 N
Explanation:
From coulomb's law,
Formula:
F = kqq'/r²........................ Equation 1
Where F = Force of repulsion, k = coulomb's constant, q = first positive charge, q' = second positive charge, r = distance between the charge.
Given: q = 20 μC = 20×10⁻⁶ C, q' = 100 μC = 100×10⁻⁶ C, r = 150 cm = 1.5 m.
Constant: k = 9×10⁹ Nm²/C²
Substitute these values into equation 1
F = (20×10⁻⁶ )( 100×10⁻⁶)(9×10⁹)/1.5²
F = 1800×10⁻³/2.25
F = 1.8/2.25
F = 0.8 N
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
