Answer:
A quantity that has magnitude and direction. It's usually represented by an arrow whose direction is the same direction is the same as that of the quantity and whose length is proportional to the quantity's magnitude
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
Answer:
F = 268 Hz
Explanation:
The beat frequency is given as:
|
So, for the first flute and tuning fork:
where,
F = Frequency of tuning fork
F = 248 Hz ± 20 Hz
F = 268 Hz (OR) 228 Hz
Now, for the second flute and tuning fork:
where,
F = Frequency of tuning fork
F = 288 Hz ± 20 Hz
F = 268 Hz (OR) 308 Hz
Since, 268 Hz is common from both calculations. Therefore, it will be the frequency of the tuning fork.
<u>F = 268 Hz</u>
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
#SPJ4
Answer:
D. a nonrenewable resource
Explanation:
From the description given, helium is a non-renewable resource.
- A non-renewable resource is one in which cannot be replenished by natural processes as they consumed.
- They get depleted as they are removed from nature.
- Helium formation cannot match its consumption if exploited.
