Answer:
a) μ = 0.475
, b) μ = 0.433
Explanation:
a) For this exercise of Newton's second law, we create a reference system with the x-axis parallel to the plane and the y-axis perpendicular to it
X axis
Wₓ - fr = m a
the friction force has the expression
fr = μ N
y Axis
N - = 0
let's use trigonometry for the components the weight
sin 27 = Wₓ / W
Wₓ = W sin 27
cos 27 = W_{y} / W
W_{y} = W cos 27
N = W cos 27
W sin 27 - μ W cos 27 = m a
mg sin 27 - μ mg cos 27 = m a
μ = (g sin 27 - a) / (g cos 27)
very = tan 27 - a / g sec 27
μ = 0.510 - 0.0344
μ = 0.475
b) now the block starts with an initial speed of 3m / s. In Newton's second law velocity does not appear, so this term does not affect the result, the change in slope does affect the result
μ = tan 25 - 0.3 / 9.8 sec 25
μ = 0.466 -0.03378
μ = 0.433
If the distance between the charged particles is deceased
Answer:
1.29 m
Explanation:
Net force acting on the dog, will be given by
where m is mass and g is acceleration due to gravity
Since the work done by the net force against the ground equals the final potential energy
Height which the dog can jump,
H=0.6+0.44=1.04 m
Total height to reach=Height to jump+ half height of the dog
=1.04 m+0.5(0.5 m)=1.29 m
Answer:8.8 m/s
Explanation:
Given
mass of train
velocity of Train
mass of another train
Final velocity of both train
Let be the velocity of before collision
As external Force is zero therefore change in momentum is zero
conserving Momentum
It decreases .
I did this problem just now on USA test prep