Answer: k = ma + uk×mgcosθ/ xf
Explanation: The body is placed on a frictionless inclined ramp.
The weight of the object has 2 components, horizontal component (mgsinθ) and vertical component (mgcosθ).
The horizontal component of weight is responsible for making tje object slide down the plane even with no applied force.
So from newton's second law of motion
mgsinθ - uk×R = ma
Where uk = coefficient of kinetic friction.
R = normal reaction = mgcosθ
mgsinθ - uk×mgcosθ = ma
mgsinθ = ma + uk×mgcosθ
mgsinθ is the applied force in this case. This applied force compresses a spring.
According to hooke's law,
F =ke
Where F = ma + uk×mgcosθ, e =xf
F = applied force , e = extension and k = spring constant.
k = F/e
k = ma + uk×mgcosθ/ xf
Answer:
truelol
Explanation:
same
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Answer:
please find the solution which is defined as follows:
Explanation:
Throughout the opposite direction, she not able to throw her tool-belt. In this scenario, she will be sending her floating through her ship. Unless interrupted, it could refer to Newton's The rule of it in motion remains in motion. Consequently, if they throw it one way because there are no molecules to interrupt your course, you can continue to go the other way.
Because the people in the car are attached to the vehicle, the people inside the vehicle are going the same speed as the vehicle.
Hope this helps! :)
Answer:
88m west and 50m north
Explanation:
1. Find the components of each vector in x and y:
(Please see the picture below)
- Vector 1:
Fx = 52m
Fy = 0m
- Vector 2:
Fx = 44m*cos34.6
Fy = 44m*sin34.6
- Vector 3:
Fx = 0m
Fy = 25m
2. Find the components of the final displacement acting in x and y:
For west direction:
Rx = 52m + 44m*cos34.6 + 0
Rx = 88m
For north direction:
Ry = 0 + 44m*sin34.6 + 25m
Ry = 50m