Answer:
μk = 0.26885
Explanation:
Conceptual analysis
We apply Newton's second law:
∑Fx = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Data:
a= -0.9 m/s²,
g = 9.81 m/s² : acceleration due to gravity
W= 75 N : Block weight
W= m*g
m = W/g = 75/9.8= 7.65 kg : Block mass
Friction force : Ff
Ff= μk*N
μk: coefficient of kinetic friction
N : Normal force (N)
Problem development
We apply the formula (1)
∑Fy = m*ay , ay=0
N-W-25 = 0
N = 75
+25
N= 100N
∑Fx = m*ax
20-Ff= m*ax
20-μk*100
= 7.65*(-0.90 )
20+7.65*(0.90) = μk*100
μk = ( 20+7.65*(0.90)) / (100)
μk = 0.26885
Answer:
Yes the statement is correct
Explanation:
The de-broglie relation is mathematically represented as
where
p is momentum of the system
From the relation we can see that the wavelength is inversely proportional to the momentum of the particle.
Answer:
The answer is 100% A THE ANSWER IS A!!!!
Explanation:
i have a 100 in all of my classes including science
Answer:
Explanation:
The length, of the string , fundamental frequency and the tension on the string are related as:
#Since both E and G have the same length and tension on them:
Where are the linear densities, the fundamental frequencies.
#taking square and inverse on both sides, we have:
Hence, the linear density of the G string is 0.00393kg/m