The frictional force is 218.6 N
Explanation:
The block in the problem is at rest along the inclined surface: this means that the net force acting along the direction parallel to the incline must be zero.
There are two forces acting along this direction:
- The component of the weight parallel to the incline, downward along the plane, of magnitude
where
m = 46 kg is the mass
is the acceleration of gravity
is the angle of the incline
- The (static) frictional force, acting upward, of magnitude 
Since the block is in equilibrium, we can write
And substituting, we find the force of friction:
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Answer:
a = 15.1 g
Explanation:
The relation between mass and acceleration is given by :
If a₁ = 0.80g, m₁ = 1510 kg, m₂ = 80 kg, we need to find a₂
So,
So, the car's acceleration would be 15.1 g.
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Answer:
Elastic potential energy, E = 200 J
Explanation:
It is given that,
Spring constant, K = 4 N/m
initial stretching in the spring, x = 5 m
Finally, it is stretched an additional 5 m i.e. x' = 5 m
Let E is the elastic energy in the spring after Varg stretches the spring. it is given by :
E = 200 J
So, the elastic energy in the spring after Varg stretches the spring is 200 J. hence, this is the required solution.
