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:
v = 895 m/s
Explanation:
Time period is given as 39 Earth Days
now the radius of the orbit is given as
so the total path length is given as
now the speed will be given as
Answer:
a) w = 4.24 rad / s
, b) α = 8.99 rad / s²
Explanation:
a) For this exercise we use the conservation of kinetic energy,
Initial. Vertical bar
Emo = U = m g h
Final. Just before touching the floor
Emf = K = ½ I w2
As there is no friction the mechanical energy is conserved
Emo = emf
mgh = ½ m w²
The moment of inertial of a point mass is
I = m L²
m g h = ½ (m L²) w²
w = √ 2gh / L²
The initial height h when the bar is vertical is equal to the length of the bar
h = L
w = √ 2g / L
Let's calculate
w = RA (2 9.8 / 1.09)
w = 4.24 rad / s
b) Let's use Newton's equation for rotational motion
τ = I α
F L = (m L²) α
The force applied is the weight of the object, which is at a distance L from the point of gro
mg L = m L² α
α = g / L
α = 9.8 / 1.09
α = 8.99 rad / s²
Answer:
A crane operator drops the cargo by turning off the electric current to the electromagnet. Electromagnets can be controlled by controlling the electric current. An electromagnet is stronger if there is more current flowing through it.
Explanation:
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