Answer:
0.146 m/s
Explanation:
We can see it in the pic.
Answer:
The shortest braking distance is 35.8 m
Explanation:
To solve this problem we must use Newton's second law applied to the boxes, on the vertical axis we have the norm up and the weight vertically down
On the horizontal axis we fear the force of friction (fr) that opposes the movement and acceleration of the train, write the equation for each axis
Y axis
N- W = 0
N = W = mg
X axis
-Fr = m a
-μ N = m a
-μ mg = ma
a = μ g
a = - 0.32 9.8
a = - 3.14 m/s²
We calculate the distance using the kinematics equations
Vf² = Vo² + 2 a x
x = (Vf² - Vo²) / 2 a
When the train stops the speed is zero (Vf = 0)
Vo = 54 km/h (1000m/1km) (1 h/3600s)= 15 m/s
x = ( 0 - 15²) / 2 (-3.14)
x= 35.8 m
The shortest braking distance is 35.8 m
Answer:
1.85 J/K
Explanation:
The computation of total change in entropy is shown below:-
Change in Entropy = Sum Q ÷ T
= 
= -3.12 + 4.97
= 1.85 J/K
Therefore for computing the total change in entropy we simply applied the above formula.
As we can see that there is heat entering the reservoir so it will be negative while cold reservoir will be positive else the process would be impossible.
Explanation:
1. To graphically add vectors, use the tail-to-tip method. Draw the first vector (it doesn't matter which), then draw the second vector where the first vector ends. The resultant vector is from the tail of the first vector to the tip of the second vector.
This graph shows two ways to get the resultant: A + B or B + A.
desmos.com/calculator/bqhcclhhqc
2. To algebraically add vectors, split each vector into x and y components.
Aₓ = 5.0 cos 45 = 3.5
Aᵧ = 5.0 sin 45 = 3.5
Bₓ = 2.0 cos 180 = -2.0
Bᵧ = 5.0 sin 180 = 0
The components of the resultant vector are the sums of the components of A and B.
Cₓ = 3.5 + -2.0 = 1.5
Cᵧ = 3.5 + 0 = 3.5
The magnitude of the resultant vector is found with Pythagorean theorem, and the direction is found with tangent.
C = √(Cₓ² + Cᵧ²) ≈ 3.9 m/s
θ = atan(Cᵧ / Cₓ) ≈ 67°
