Answer:
Explanation:
a ) The motion is one dimensional , so motion is along x - axis , starting from origin ( 0 , 0 )
b ) Initial velocity is 18.5 m /s when boat is situated at origin . When he displaces by 250 m along x axis and his position is ( 250 , 0 ) along x axis , his velocity becomes 36 m /s . Both his velocity and acceleration is along x - axis.
c ) Initial velocity vi = 18.5 m /s
final velocity vf = 36 m/s
Displacement x = 250 m
Acceleration a = ?
Most appropriate formula is given below .
vf² = vi² + 2 a x
2ax = vf² - vi²
x = ( vf² - vi² ) / 2 a
d )
Putting the given values
36² - 18.5² / 2 x 250
= 1296 - 342.25 / 500
= 1.9 m /s².
f ) Time interval t = ?
Required formula
vf = vi + at
t = (vf - vi ) / a
Putting the values
t = (30 - 18.5) / 1.9
= 6.05 second .
Answer:

Explanation:
The instantaneous velocity of a point mass that executes a simple harmonic movement is given by:

Where:

Express the amplitude in meters:

The angular frequency can be found using the next equation:

Using the data provided:

At the equilibrium position:


Answer:
A) ≥ 325Kpa
B) ( 265 < Pe < 325 ) Kpa
C) (94 < Pe < 265 )Kpa
D) Pe < 94 Kpa
Explanation:
Given data :
A large Tank : Pressures are at 400kPa and 450 K
Throat area = 4cm^2 , exit area = 5cm^2
<u>a) Determine the range of back pressures that the flow will be entirely subsonic</u>
The range of flow of back pressures that will make the flow entirely subsonic
will be ≥ 325Kpa
attached below is the detailed solution
<u>B) Have a shock wave</u>
The range of back pressures for there to be shock wave inside the nozzle
= ( 265 < Pe < 325 ) Kpa
attached below is a detailed solution
C) Have oblique shocks outside the exit
= (94 < Pe < 265 )Kpa
D) Have supersonic expansion waves outside the exit
= Pe < 94 Kpa
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