A bowling ball weighing 71.2 N is swinging like a pendulum at the end of a 3.8 m rope.
1 answer:
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Explanation:
An endothermic reaction is a type of of chemical reaction in which energy is absorbed from the surrounding. The temperature of the surrounding decreases.
An Exothermic reaction is a type of of chemical reaction in which energy is released into the surrounding. The temperature of the surrounding increases.
a)When an endothermic reaction takes place in test tube the temperature of the surrounding(here the outside wall of the test tube) will decrease which can be felt by touching the outside of the test tube.
b) Burning a candle
,Lighting a gas stove and running a car's engine all are example of exothermic reaction because energy is releases into surroundings. Where as using an instant cold pack is an example endothermic reaction in which energy is absorbed from the surrounding.
c) The diagram is of an exothermic reaction because the energy of the reactants is higher than the energy of the products.The difference in the energies of these two is the energy which was being released on completion of reaction.
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 = / 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