Answer:
2.61 J
Explanation:
Since potential energy U = mgy where m = mass of object, g = acceleration due to gravity = 9.8 m/s² and y = height of object above the ground.
Now for the coffee mug, m= 0.422 kg and it is 0.63 m on a table, so it is 0.63 m above the ground. Thus, y = 0.63 m.
We compute U
U = mgy
= 0.422 kg × 9.8 m/s² × 0.63 m
= 2.605 J
≅ 2.61 J
So, the potential energy of the mug with respect to the floor is 2.61 J
The wavelengths of the constituent travelling waves CANNOT be 400 cm.
The given parameters:
- <em>Length of the string, L = 100 cm</em>
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The wavelengths of the constituent travelling waves is calculated as follows;

for first mode: n = 1

for second mode: n = 2

For the third mode: n = 3

For fourth mode: n = 4

Thus, we can conclude that, the wavelengths of the constituent travelling waves CANNOT be 400 cm.
The complete question is below:
A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent travelling waves CANNOT be:
A. 400 cm
B. 200 cm
C. 100 cm
D. 67 cm
E. 50 cm
Learn more about wavelengths of travelling waves here: brainly.com/question/19249186
Answer:
each resistor is 540 Ω
Explanation:
Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance
defined by the formula:

Therefore, R/3 is the equivalent resistance of the initial circuit.
In the second circuit, two of the resistors are in parallel, so they are equivalent to:

and when this is combined with the third resistor in series, the equivalent resistance (
) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

The problem states that the difference between the equivalent resistances in both circuits is given by:

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

TRUE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!