What is the impulse of a 3kg object accelerating from rest to 12m/s?
1 answer:
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We can simply find components of the given directions and conveniently add them up to find the total distance.
290 km 20 deg north of east can be decomposed as :
290 x cos(20) EAST + 290 sin(20) NORTH
=272.5 E + 99.18 N
Now from this point ,
170 Km 30 deg west of north :
170 cos(30) N - 170 sin (30) E [ east = - west]
= 147.22 N - 85 E
hence coordinates of lake b are
(272.5 - 85) E + (99.18 + 147.22)N
187.5 E +264.4 N
use right triangle property to find distance D =
D = 324.13 Km
and angle = atan(b/a) = 54.65 deg
hence lake b is at 324.13 Km 54.65 deg east of north
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