The answer is A. voice uses a wider range of pitch and volume as compared to speaking
Check the attached file for the solution for this problem.
Answer:
The mass of the solution is 120 g.
Explanation:
The mass of the solution is given by:

Where:
: is the mass of the solution
: is the mass of the solvent
: is the mass of the solute
In the solution, the solvent is the majority compound (in mass) and the solute is the minority (in mass), so the solvent is the water and the solute is sodium chloride.
Hence, the mass of the solution is:
I hope it helps you!
Answer:
The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s
Explanation:
We are given that
Angular acceleration, 
Diameter of the wheel, d=21 cm
Radius of wheel,
cm
Radius of wheel, 
1m=100 cm
Magnitude of total linear acceleration, a=
We have to find the linear speed of a at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.
Tangential acceleration,


Radial acceleration,
We know that

Using the formula

Squaring on both sides
we get






Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s