Answer:
The resultant force would (still) be zero.
Explanation:
Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.
In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.
By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.
When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.
However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.
Answer:
Muscle contraction thus results from an interaction between the actin and myosin filaments that generates their movement relative to one another. The molecular basis for this interaction is the binding of myosin to actin filaments, allowing myosin to function as a motor that drives filament sliding.
Given:
A cylindrical container closed of both end has a radius of 7cm and height of 6cm.
Explanation:
A.) Find the total surface area of the container.
- A = 2πrh + 2πr²
- A = 2(3.14)(7)(6) + 2(3.14)(7 × 7)
- A = 263.76 + 307.72
- A = 571.48
B.) Find the volume of the container.
- V = πr²h
- V = (3.14)(7×7)(6)
- V = 923.16
Not sure huhuness.
#CarryOnLearning
Answer:
The concentration of OH⁻ in the mixture is 0.05 M
Explanation:
The reaction of neutralization between HCl and NaOH is the following:
H⁺(aq) + OH⁻(aq) ⇄ H₂O(l)
The number of moles of HCl is:
Similarly, the number of moles of NaOH is:
Now, from the reaction of HCl and NaOH we have the following number of moles of NaOH remaining:
Finally, the concentration of OH⁻ in the mixture is:
Therefore, the concentration of OH⁻ in the mixture is 0.05 M.
I hope it helps you!
