Answer:
Between 2.0 s and 4.0 s (B and C)
Between 5.0 s and 8.0 s (D and E)
Between 10.0 s and 11.0 s (F and G)
Explanation:
The graph shown in the figure is a velocity-time graph, which means that:
- On the x-axis, the time is plotted
- On the y-axis, the velocity is plotted
Therefore, this means that the object is not moving when the line is horizontal (because at that moment, the velocity is constant, so the object is not moving). This occurs in the following intervals:
Between 2.0 s and 4.0 s (B and C)
Between 5.0 s and 8.0 s (D and E)
Between 10.0 s and 11.0 s (F and G)
From the graph, it would be possible to infer additional information. In particular:
- The area under the graph represents the total distance covered by the object
- The slope of the graph represents the acceleration of the object
So based on your question where there is a block of mass m1= 8.8kg in the inclined plane with an angle of 41 with respect to the horizontal. To find the spring constant of the problem were their is a coefficients of friction of 0.39 and 0.429, you must use the formula K*x^2=m*a*sin(angle). By calculating the minimum spring constant is 220.66 N/m^2
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Answer:
All adjustments in issue include changes in energy. Energy is either delivered or consumed. The energy is often in the form of heat, however it might be in the form of sound or light.
Explanation:
this is in my own words btw:)
If the primary wire's power is 10 A and one branch's power is 4 A, another branch's power will be 6A.
According to Kirchhoff's current law (KCL), the total current flowing through a parallel route circuit's junction equals the total current flowing away from it.
Provided that one of the two branches through which power exits the intersection has a flow of 4A, and also that the junction's overall flow entering it is 10A, the entire current going the junction should be 10A.
Consequently, the second wire's power may be expressed as;
I = I1+ I2 [ where I= total current (10A);
I1= current in one branch (4A) &
I2= current in another branch]
⇒I2 = I - I1
⇒I2 = 10A - 4A
⇒I2 = 6A
Therefore, it can be concluded that when the primary wire bears 10A power having 4A in one of its branches, another branch carries 6A power.
Learn more about Kirchhoff's law here:
brainly.com/question/6417513
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