A 6.0-kg object moving 5.0 m/s collides with and sticks to a 2.0-kg object. after the collision the composite object is moving 2
1 answer:
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Answer:
529.15 m/s
Explanation:
h = Maximum height = 70000 m
g = Acceleration due to gravity = 2 m/s²
m = Mass of sulfur
As the potential and kinetic energies are conserved
The speed with which the liquid sulfur left the volcano is 529.15 m/s
Answer:34 cm
Explanation:
Given
mass of meter stick m=80 gm
stick is balanced when support is placed at 51 cm mark
Let us take 5 gm tack is placed at x cm on meter stick so that balancing occurs at x=50 cm mark
balancing torque
Answer:
a) Fₓ = 23.5 N
b) Net force = Fₓ
Explanation:
An image of the question as described is attached to this solution.
From the image attached, the forces acting on the box include the weight of the box, the normal reaction of the surface on the box, the applied force on the box and the Frictional force opposing the motion of the box (which is negligible and equal to 0)
a) From the diagram, the horizontal component of the force is
Fₓ = 25 cos 20° = 23.49 N = 25 N
b) Again, from the diagram attached, doing a force balance on the box, in the horizontal direction, we obtain
Net force = Fₓ - Frictional force
But frictional force is 0 N
Net force = Fₓ
Hope this Helps!!!
