Answer:
50 N
Explanation:
Let the force in the horizontal rope be F₁ and the force in the diagonal rope be F₂:
The total force in the horizontal and vertical directions must be zero, since the object is at rest and is not accelerating.
The horizontal component of the forces:
F₁ + F₂ = -40N + F₂ = 0
F₂ = 40N
The vertical component of the forces:
F₁ + F₂ - mg = 0 + F₂ - mg = 0
F₂ = mg
If I assume the gravitational constant g = 10 m/s²:
F₂ = (3 kg) * (10 m/s²) = 30N
Adding the horizontal and vertical components of the force F₂:
F₂ = √((40N)² + (30N)²) = 50N
Suvat
we have s, u, v and we want a
the suvat equation with these values in is: v^2 = u^2 - 2as
so a = (-v^2 + u^2)/-2s
plug numbers in
a = (-85^2 + 0^2)/-2*36 = 7225/72 = 100.3... ms^-2
Answer: 29.50 m
Explanation: In order to calculate the higher accelation to stop a train without moving the crates inside the wagon which is traveling at constat speed we have to use the second Newton law so that:
f=μ*N the friction force is equal to coefficient of static friction multiply the normal force (m*g).
f=m.a=μ*N= m*a= μ*m*g= m*a
then
a=μ*g=0.32*9.8m/s^2= 3.14 m/s^2
With this value we can determine the short distance to stop the train
as follows:
x= vo*t- (a/2)* t^2
Vf=0= vo-a*t then t=vo/a
Finally; x=vo*vo/a-a/2*(vo/a)^2=vo^2/2a= (49*1000/3600)^2/(2*3.14)=29.50 m
It means, <span>Acceleration increases as mass decreases.
So, option C is your answer.
Hope this helps!
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