Define principal focus of a lence

Studentka2010 [4]

I think the answer is b. easy to digest.

C. and D. are chemical properties,

7 0

3 years ago

A projectile is fired with an initial speed of 65.2 m/s at an angle of 34.5º above the horizontal on a long flat firing range. (

Margarita [4]

Answer:

A) $h = 69.58 m$

D) $v = 58.12 \frac{m}{s}$ (Speed magnitude)

α = 22.49° (Speed direction above the horizontal)

Explanation:

Conceptual analysis:

To solve this problem we consider the following concepts:

1) The projectile in its movement describes a curved line called a parabola, therefore two coordinates are required to fix the position at each instant of time, since the movement is performed in the X-Y plane.

The initial velocity (Vo) is tangent to the trajectory at the initial point and can be broken down into two components, one vertical (Voy) and one horizontal(Vox):

$V_{ox} = V_{o}Cos\alpha _{o}$ Formula (1)

$V_{oy} = V_{o}Sin\alpha _{o}$ Formula (2)

Where:

Vo: Initial velocity in m/s

$\alpha_{o}$ : Initial angle above the horizontal in grades

2) The formula to calculate its velocity at any vertical position(y) is as follows:

$v_{y}^{2} = v_{oy}^{2} -2gy$ Formula (3)

Where:

$v_{f}^{2}$ : Final speed component in vertical direction in m/s

$v_{oy}^{2}$ : Initial speed component in vertical direction in m/s

g: acceleration due to gravity in m/s2

3) The formulas to calculate the projectile velocity components at any time (t) are:

$v_{x} = v_{ox}$ Because the movement is uniform in the x direction (constant speed)

$v_{y} = v_{oy}-g*t$ Formula (4) Because the movement is uniformly accelerated in the y direction

Known information:

We know the following data:

$v_{o} = 65.2 \frac{m}{s}$

$\alpha _{o}$ = 34.5º above the horizontal

$g=9.8 \frac{m}{s^{2}}$

Development of the problem:

Initial speed components(Vox, Voy), (Formula (1), Formula (2)

$v_{ox} = 65.2*Cos34.5 = 53.7\frac{m}{s}$

$v_{oy} = 65.2*Sin34.5 = 36.93\frac{m}{s}$

A) Maximum height (h):

When the projectile reaches its maximum height (h) ,the speed component Vy = 0, then, we replace this value in the Formula (3):

$0=(36.93)^2-2*9.8*h$

$h=\frac{ (36.93)^2}{2*9.8} = 69.58 m$

D) Speed of the projectile 1.50s after firing

We replace t=1.5 s in the formula(4)

$v_{y} = 36.93-9.8*1.5 = 22.23 \frac{m}{s}$

$v_{x} = v_{ox} = 53.7 \frac{m}{s}$

$v = \sqrt{53.7^{2}+22.23^{2}} = 58.12 \frac{m}{s}$ (Speed magnitude)

$\alpha = tan^{-1} (\frac{v_{y}}{v_{x}}) = tan^{-1} (\frac{22.23}{53.7})$

α = 22.49° (Speed direction above the horizontal)

6 0

3 years ago

What wave occurs when you push a spring back and forth

kipiarov [429]

Moving a spring back and forth creates a longitudinal wave

4 0

3 years ago

Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approxim

sergiy2304 [10]

Answer:

10.53m/s²

Explanation:

Centripetal acceleration is the acceleration of an object about a circle. The formula for calculating centripetal acceleration is expressed by:

$a = \frac{v^2}{r}$

v is the velocity of the car = 24.5m/s

r is the radius of the track = 57.0m

Substitute the given values into the formula:

$a = \frac{24.5^2}{57} \\\\a = \frac{600.25}{57}\\ \\a = 10.53m/s^{2}$

Hence the centripetal acceleration of the race car is 10.53m/s²

6 0

3 years ago

. In a solid, particles do not move around each

nordsb [41]

Answer:

I am sure the awnser to this is

B. vibrate

8 0

2 years ago

Define principal focus of a lence​

Define principal focus of a lence