The initial speed of the shot is 15.02 m/s.
The Shot put is released at a height y<em> </em>from the ground with a speed u. It is released at an angle θ to the horizontal. In a time t, the shot put travels a distance <em>R</em> horizontally.
Pl refer to the attached diagram.
Resolve the velocity u into horizontal and vertical components, u ₓ=ucosθ and uy=u sinθ. The horizontal component remains constant in the absence of air resistance, while the vertical component varies due to the action of the gravitational force.
Write an expression for R.

Therefore,

In the time t, the net displacement of the shotput is y in the downward direction.
Use the equation of motion,

Substitute the value of t from equation (1).

Substitute -2.10 m for y, 24.77 m for R and 38.0° for θ and solve for u.

The shot put was thrown with a speed 15.02 m/s.
Answer:
<h2>12m/3s is the answer </h2><h2> I hope its right...</h2>
Atoms are added to crystal faces.

Since the sound travels from the submarine to the object AND back, it actually travelled 3625x2=7250m.

Speed of sound: 1450m/s
Answer:
359 g Mn
General Formulas and Concepts:
- Dimensional Analysis
- Reading the Periodic Table of Elements
Explanation:
<u>Step 1: Define</u>
6.53 mol Mn
<u>Step 2: Find conversion</u>
1 mol Mn = 54.94 g Mn
<u>Step 3: Dimensional Analysis</u>
<u />
= 358.758 g Mn
<u>Step 4: Simplify</u>
<em>We are given 3 sig figs.</em>
358.758 g Mn ≈ 359 g Mn