Answer:
a= 4.14 m/s²
Explanation:
We calculate the weight component parallel to the displacement of the block:
We define the x-axis in the direction of the inclined plane , 25° to the horizontal.
W= m*g : Total block weight
Wx= W*sen25°= m*g* sen25°
We apply Newton's second law :
∑F = 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)
Problem development
We apply the formula (1) to calculate the acceleration of the block:
∑Fx = m*a
Wx = m*a
m*g* sen25° = m*a : We divide by m on both sides of the equation
g* sen25° = a
a = g* sen25° = 9.8* sen25° = 4.14 m/s²
Answer:
4.8 ton
Explanation:
We are given that
Mass,m=77 ton
Displacement,s=3500 ft
We have to find the force parallel to the incline would be required to hold the monolith on this causeway.
The force parallel to the incline would be required to hold the monolith on this causeway=
Using the formula
The force parallel to the incline would be required to hold the monolith on this causeway=
Answer:
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