How does melting And burning sugar follow the law of conservation of mass
2 answers:
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Answer:
The magnitude of the acceleration of the box is 2.01 m/s².
Explanation:
Hi there!
Please, see the attached figure for a graphical description of the problem.
We have the following horizontal forces:
Fr = friction force.
Fx = Horizontal component of the applied force, F.
And we have the following vertical forces:
Fy = vertical component of the applied force.
N = normal force exerted on the box.
W = weight of the box.
According to Newton´s second law:
∑F = m · a
Then, in the horizontal direction:
Fx - Fr = m · a
Where "m" is the mass of the box and "a" its acceleration.
Fx can be obtained by trigonometry (see figure):
Fx = F · cos 30°
Fx = 90.0 N · cos 30°
Fr is calculated as follows:
Fr = μ · N
Where μ is the coefficient of friction and N the normal force.
So, we have to find the magnitude of the normal force.
Using Newton´s second law in the vertical direction:
∑F = N + Fy - W = m · a
Notice that the box has no vertical acceleration, then:
N + Fy - W = 0
Solving for N:
N = W - Fy
The weight is calculated as follows:
W = m · g
Where g is the acceleration due to gravity:
W = 20.0 kg · 9.8 m/s² = 196 N
And the vertical component of the applied force can be obtained by trigonometry:
Fy = F · sin 30°
Fy = 90.0 N · sin 30°
The normal force will be:
N = W - Fy = 196 N - 90.0 N · sin 30°
N = 151 N
Now, we can calculate the friction force:
Fr = μ · N
Fr = 0.250 · 151 N
Fr = 37.8 N
And now, we can obtain the acceleration of the box:
Fx - Fr = m · a
(Fx - Fr) / m = a
(90.0 N · cos 30° - 37.8 N ) / 20.0 kg = a
a = 2.01 m/s²
The magnitude of the acceleration of the box is 2.01 m/s².
Answer:
Explanation:
As we know that we board in the car of ferris wheel at the bottom position
So we will have
final height of the car at angular displacement given as
here we know that
so we have