Answer: The volume of an irregularly shaped object is 0.50 ml
Explanation:
To calculate the volume, we use the equation:
![\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}](https://tex.z-dn.net/?f=%5Ctext%7BDensity%20of%20substance%7D%3D%5Cfrac%7B%5Ctext%7BMass%20of%20substance%7D%7D%7B%5Ctext%7BVolume%20of%20substance%7D%7D)
Density of object = ![6.0g/ml](https://tex.z-dn.net/?f=6.0g%2Fml)
mass of object = 3.0 g
Volume of object = ?
Putting in the values we get:
![6.0g/ml=\frac{3.0g}{\text{Volume of substance}}](https://tex.z-dn.net/?f=6.0g%2Fml%3D%5Cfrac%7B3.0g%7D%7B%5Ctext%7BVolume%20of%20substance%7D%7D)
![{\text{Volume of substance}}=0.50ml](https://tex.z-dn.net/?f=%7B%5Ctext%7BVolume%20of%20substance%7D%7D%3D0.50ml)
Thus the volume of an irregularly shaped object is 0.50 ml
The y-component of the stone's velocity when it is 8 m below the hand is 14.86 m / s
v² = u² + 2 a s
s = Displacement
u = Initial velocity
a = Acceleration
u = 8 m / s
s = 8 m
v² = 8² + 2 * 9.8 * 8
v² = 64 + 156.8
v = √ 220.8
v = 14.86 m / s
The equation used to solve the problem is an equation of motion. These equations are designed to locate an object in motion using components such as velocity, displacement, acceleration and time.
Therefore, the y-component of the stone's velocity is 14.86 m / s
To know more about Equations of motion
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Answer:
The energy of an electron in an isolated atom depends on b. n only.
Explanation:
The quantum number n, known as the principal quantum number represents the relative overall energy of each orbital.
The sets of orbitals with the same n value are often referred to as an electron shell, in an isolated atom all electrons in a subshell have exactly the same level of energy.
The principal quantum number comes from the solution of the Schrödinger wave equation, which describes energy in eigenstates
, and for the case of an hydrogen atom we have:
![E_n=-\cfrac{13.6}{n^2}\, eV](https://tex.z-dn.net/?f=E_n%3D-%5Ccfrac%7B13.6%7D%7Bn%5E2%7D%5C%2C%20eV)
Thus for each value of n we can describe the orbital and the energy corresponding to each electron on such orbital.
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