Answer: The free - body diagrams for blocks A and B. frictionless surface by a constant horizontal force F = 100 N. Find the tension in the cord between the 5 kg and 10 kg blocks. The string that attaches it to the block of mass M2 passes over a frictionless pulley of negligible mass. The coefficient of kinetic friction Hk between M.
Explanation: Hope this helped :)
The name and strength of the force holding the block up is 50 N upward - Normal force.
The given parameters:
- <em>Mass of the block, m = 5 kg</em>
The weight of the block acting downwards due to gravity is calculated as follows;
W = mg
where;
- <em>g is acceleration due to gravity = 10 m/s²</em>
W = 5 x 10
W = 50 N <em>(</em><em>downwards</em><em>)</em>
Since the block is at rest, an a force equal to the weight of the block must be acting upwards. This force is known as normal reaction.
Fₙ = 50 N <em>(</em><em>upwards</em><em>)</em>
Thus, the name and strength of the force holding the block up is 50 N upward - Normal force.
Learn more about Normal force here: brainly.com/question/14486416
Answer:
Free body diagrams are used to describe situations where several forces act on an object. On the other hand Vector diagrams are used to resolve (break down) a single force into two forces acting as right angles to eachother
Explanation:
Hope this helps !
Explanation:
First, find the velocity of the projectile needed to reach a height h when fired straight up.
Given:
Δy = h
v = 0
a = -g
Find: v₀
v² = v₀² + 2aΔy
(0)² = v₀² + 2(-g)(h)
v₀ = √(2gh)
Now find the height reached if the projectile is launched at a 45° angle.
Given:
v₀ = √(2gh) sin 45° = √(2gh) / √2 = √(gh)
v = 0
a = -g
Find: Δy
v² = v₀² + 2aΔy
(0)² = √(gh)² + 2(-g)Δy
2gΔy = gh
Δy = h/2
