Three arrows are shot horizontally. They have left the bow and are traveling parallel to the ground. Air resistance is negligibl
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Answer:
a) F = 2.66 10⁴ N, b) h = 1.55 m
Explanation:
For this fluid exercise we use that the pressure at the tap point is
Exterior
P₂ = P₀ = 1.01 105 Pa
inside
P₁ = P₀ + ρ g h
the liquid is water with a density of ρ=1000 km / m³
P₁ = 0.85 1.01 10⁵ + 1000 9.8 5
P₁ = 85850 + 49000
P₁ = 1.3485 10⁵ Pa
the net force is
ΔP = P₁- P₂
Δp = 1.3485 10⁵ - 1.01 10⁵
ΔP = 3.385 10⁴ Pa
Let's use the definition of pressure
P = Fe / A
F = P A
the area of a circle is
A = pi r² = [i d ^ 2/4
let's reduce the units to the SI system
d = 100 cm (1 m / 100 cm) = 1 m
F = 3.385 104 pi / 4 (1) ²
F = 2.66 10⁴ N
b) the height for which the pressures are in equilibrium is
P₁ = P₂
0.85 P₀ + ρ g h = P₀
h =
h =
h = 1.55 m
Answer:
Computer A is 1.41 times faster than the Computer B
Explanation:
Assume that number of instruction in the program is 1
Clock time of computer A is
Clock time of computer B is
Effective CPI of computer A is
Effective CPI of computer B is
CPU time of A is
CPU time of B is
Hence Computer A is Faster by
Computer A is 1.41 times faster than the Computer B
1. 12.75 J
Assuming that the force applied is parallel to the ramp, so it is parallel to the displacement of the cart, the work done by the force is
where
F = 15 N is the magnitude of the force
d = 85 cm = 0.85 m is the displacement of the cart
Substituting in the formula, we get
2. 10.6 N
In this part, the cart reaches the same vertical height as in part A. This means that the same work has been done (because the work done is equal to the gain in gravitational potential energy of the object: but if the vertical height reached is the same, then the gain in gravitational potential energy is the same, so the work done must be the same).
Therefore, the work done is
However, in this case the displacement is
d = 120 cm = 1.20 m
Therefore, the magnitude of the force in this case is