Answer:
Explanation:
en un movimiento sísmico?
The net force acting on a box of mass 8.0kg that experiences an acceleration of 4.0m/s² is 32N. Details about net force can be found below.
<h3>How to calculate net force?</h3>
The net force of a body can be calculated by multiplying the mass of the body by its acceleration as follows:
Force = mass × acceleration
According to this question, a box with a mass of 8.0 kg is sitting on a frictionless surface and experiences an acceleration of 4.0 m/s2 to the right.
Net force = 8kg × 4m/s²
Net force = 32N
Therefore, the net force acting on a box of mass 8.0kg that experiences an acceleration of 4.0m/s² is 32N.
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Answer: This type of questions are called probing questions. The correct option is D.
Explanation:
Probing questions are the type of questions that are asked to investigate an ongoing event. It helps the investigator to know more about what is happening and how to obtain conclusive decisions through the personal opinions of the respondent . For example from the question, Liza wanted to know more about the project updates which was held in the weekly meetings. She asked her employees questions like:
- Why were you late meeting your last deadline?
-Were there external factors that delayed your work?
-Did other coworkers get their part of the assignment to you on time?
- Do you need more help from me?".
We are given that a 500 kg object is hanging from a spring. To determine the amount the spring is stretched we will use Hook's law, which states the following:
Where:
Since the object is hanging the only force acting on the spring is the weight of the object. The weight of the object is:
Where:
Plugging in the values we get:
Solving the operations:
Now we solve for "x" from Hook's law by dividing both sides by "k":
Now we plug in the known values:
Solving the operations:
Therefore, the spring is stretched by 5.4 meters.
Answer:
xcritical = d− m1
/m2
( L
/2−d)
Explanation: the precursor to this question will had been this
the precursor to the question can be found online.
ff the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal bar may become unstable (i.e., the bar may no longer remain horizontal). What is the smallest possible value of x such that the bar remains stable (call it xcritical)
. from the principle of moments which states that sum of clockwise moments must be equal to the sum of anticlockwise moments. aslo sum of upward forces is equal to sum of downward forces
smallest possible value of x such that the bar remains stable (call it xcritical)
∑τA = 0 = m2g(d− xcritical)− m1g( −d)
xcritical = d− m1
/m2
( L
/2−d)
