Answer:
7 m/s
Explanation:
To solve this problem you must use the conservation of energy.
That math speak for, initial kinetic energy plus initial potential energy equals final kinetic energy plus final potential energy.
The initial PE (potential energy) is 0 because it hasn't been raised in the air yet. The final KE (kinetic energy) is 0 because it isn't moving. This gives the following:
K1=U2
Solve for v
Input known values and you get 7 m/s.
Answer:
The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Explanation:
Given that,
Amplitude = 0.08190 m
Frequency = 2.29 Hz
Wavelength = 1.87 m
(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave
Using formula of distance
Where, d = distance
A = amplitude
Put the value into the formula
Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Answer:
The difference is 7.6 grams.
Explanation:
In mathematics the difference of two numbers is express as the subtraction between them:
So to find out the difference between the two measured masses, a will be represented by 123.6 grams since is the bigger number, and b by 115.972 grams.
Therefore, it is get:
<u>Hence, the difference is 7.6 grams. </u>
The result of 7.628 will be expressed as 7.6 to have the correct number of significant figures.
Notice how that can be express in units of kilograms too since there is 1000 gram in 1 kilogram:
⇒