1/16........................................
One is their traits and their characterists that they have in common
Answer:
Frequency, f = 3.73Hz
Explanation:
The frequency of a simple harmonic 6is given by:
f = w/2pi
But w= Sqrt( k/m)
Where k is the spring constant
And m is the mass
Given:
Mass=0.20kg
Spring constant, k=130N/m
w= Sqrt(130/0.20)
w= Sqrt(650)
w= 25.50 m
Frequency, f = w/2pi
f = 25.50/(2×3.142)
f = 3.73Hz
Answer:
D = 527.31 Km
Explanation:
given,
angle of ship, θ = 23.5° N of W
distance travel in the direction = 575 Km
Distance of ship in west from harbor = ?
now,
Distance of the ship in the west direction
D = d cos θ
d = 575 Km
θ = 23.5°
inserting all the values
D = 575 x cos 23.5°
D = 575 x 0.91706
D = 527.31 Km
Hence, the distance travel by the ship in west from harbor is equal to D = 527.31 Km
So based on your question where there is a block of mass m1= 8.8kg in the inclined plane with an angle of 41 with respect to the horizontal. To find the spring constant of the problem were their is a coefficients of friction of 0.39 and 0.429, you must use the formula K*x^2=m*a*sin(angle). By calculating the minimum spring constant is 220.66 N/m^2
