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Vlada [557]
3 years ago
11

What's the main term used to locate in the alphabetic index for acute perforated gastrojejunal ulcer

Physics
1 answer:
nasty-shy [4]3 years ago
8 0

Answer:

the main term used to locate in the alphabetic index for acute perforated gastrojejunal ulcer is ULCER

Explanation:

because the main is a term identifying disease conditions or injuries

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The car has slowed down because of the deceleration or decelerating force acting on it. The decelerating force acting on it is generated by the application of brakes or the frictional force acting in the opposite direction of car's motion.
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What can directly lead to unconformity on an exposed rock?
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A 20-ton truck collides with a 1500-lb car and causes a lot of damage to the car. Since a lot of damage is done on the car : Sel
Anni [7]

Answer:

d)

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Newton's 3rd Law tells us that for every action we will have a reaction equal in magnitude and opposite in direction. In a collision between obects A and B this means that the force on A applied by B will be equal in magnitude than the force on B applied by A, which leaves d as the true answer and the rest as false.

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In the ballistic pendulum experiment (A.K.A conservation of momentum & energy), the velocity of the projectile was measured
Fed [463]

Answer: The combined velocity = 0.413m/a

Maximum angle Ø = 54.7

Explanation: Please find the attached files for the solution

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3 years ago
Three people pull simultaneously on a stubborn donkey. Jack pulls directly ahead of the donkey with a force of 63.9 N63.9 N , Ji
Greeley [361]

Answer:

a) F = (137.4 i ^ + 185 j ^) N

b)    F = 230.2 N  ,  θ = 53.5º

Explanation:

In this exercise we ask to find the net force, for which we will define a coordinate system fix the donkey and use trigonometry to decompose the forces

Jack       F₁ₓ = 63.9 N

Jill          F₂ = 79.1 N with direction 45º to the left

              cos (180 -45) = F₂ₓ / F₂

              sin 135 = F_{2y} / F₂

              F₂ₓ = F₂ cos 135

              F_{2y} = F₂ sin 135

              F₂ₓ = 79.1 cos 135 = -55.9 N

              F_{2y} = 79.1 sin 135 = 55.9 N

Jane      F₃ = 183 N direction 45th to the right

             cos 45 = F₃ₓ / F3

             sin 45 = F_{3y} / F3

             F₃ₓ = F₃ cos 45 = 183 cos 45

             F_{₃y} = F₃ sin 45 = 183 sin 45

             F₃ₓ = 129.4 N

             F_{3y} = 129.4 N

we add each component of the force

       Fₓ = F₁ₓ + F₂ₓ + F₃ₓ

       Fₓ = 63.9 + (-55.9) + 129.4

       Fₓ = 137.4 N

       F_{y} = F_{2y} + F_{3y}

       F_{2y} = 55.9 + 129.4

       F_{2y} = 185.3 N

we can give the result of the forms

 

a) F = (137.4 i ^ + 185 j ^) N

b) in the form of module and angle

         F = RA (Fₓ² + F_{y}²)

         F = Ra (137² + 185²)

         F = 230.2 N

         tan θ = F_{y} / Fₓ

         θ = tan⁻¹ F_{y} / Fₓ

        θ = tan⁻¹ (185/137)

        θ = 53.5º

4 0
3 years ago
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