Number of element atom for element,
Calcium (Ca) = 1
Carbon (C) = 1
Oxygen (O) = 3
(1) The ball is in the air for <u>1.4 seconds.</u>
(2) The horizontal velocity of the ball as it rolls off the table is<u> 6.32 m/s.</u>
(3) The vertical velocity of the ball right before it hits the ground is <u>13.72 m/s.</u>
(4) The horizontal velocity of the ball right before it hits the ground is<u> 6.32 m/s.</u>
(5) The initial vertical velocity as soon as the ball comes of the cliff is <u>13.72 m/s.</u>
<h3>What is the time of motion of the ball?</h3>
The time of motion of the ball is calculated by applying the following equation.
t = √(2h/g)
where;
- h is the height of the cliff
- g is acceleration due to gravity
t = √(2h/g)
t = √(2 x 9.63 / 9.8)
t = 1.4 seconds
The horizontal velocity of the ball is calculated as follows;
v = d/t
where;
- d is the horizontal distance travelled by the ball = 8.85 m
v = 8.85 m / 1.4 s
v = 6.32 m/s
The vertical velocity of the ball before it hits the ground is calculated as;
vf = vi + gt
vf = 0 + 9.8 x 1.4
vf = 13.72 m/s
The horizontal velocity of the ball right before it hits the ground is calculated as;
the initial velocity of a projectile = final horizontal velocity
vxf = vxi = 6.32 m/s
The initial vertical velocity as soon as the ball comes off the cliff = final vertical velocity = 13.72 m/s
Learn more about horizontal velocity here: brainly.com/question/24949996
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Answer:
E = 2,575 eV
Explanation:
For this exercise we will use the Planck equation and the relationship of the speed of light with the frequency and wavelength
E = h f
c = λ f
Where the Planck constant has a value of 6.63 10⁻³⁴ J s
Let's replace
E = h c / λ
Let's calculate for wavelengths
λ = 4.83 10-7 m (blue)
E = 6.63 10⁻³⁴ 3 10⁸ / 4.83 10⁻⁷
E = 4.12 10-19 J
The transformation from J to eV is 1 eV = 1.6 10⁻¹⁹ J
E = 4.12 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)
E = 2,575 eV
The units of G must be C. m³ / ( kg s² )
<h3>Further explanation</h3>
Newton's gravitational law states that the force of attraction between two objects can be formulated as follows:
<em>F = Gravitational Force ( Newton )</em>
<em>G = Gravitational Constant ( 6.67 × 10⁻¹¹ Nm² / kg² )</em>
<em>m = Object's Mass ( kg )</em>
<em>R = Distance Between Objects ( m )</em>
Let us now tackle the problem !
To find unit of Gravitational Constant can be carried out in the following way:
![{[N]}= G\frac{{[kg]}{[kg]}}{{[m^2]}}](https://tex.z-dn.net/?f=%7B%5BN%5D%7D%3D%20G%5Cfrac%7B%7B%5Bkg%5D%7D%7B%5Bkg%5D%7D%7D%7B%7B%5Bm%5E2%5D%7D%7D)
![{[kg ~ m / s^2]}= G \frac{{[kg^2]}}{{[m^2]}}](https://tex.z-dn.net/?f=%7B%5Bkg%20~%20m%20%2F%20s%5E2%5D%7D%3D%20G%20%5Cfrac%7B%7B%5Bkg%5E2%5D%7D%7D%7B%7B%5Bm%5E2%5D%7D%7D)
![G = \frac{{[kg ~ m / s^2]}{[m^2]}} {{[kg^2]} }](https://tex.z-dn.net/?f=G%20%3D%20%5Cfrac%7B%7B%5Bkg%20~%20m%20%2F%20s%5E2%5D%7D%7B%5Bm%5E2%5D%7D%7D%20%7B%7B%5Bkg%5E2%5D%7D%20%7D)
![G = \frac{{[kg ~ m^3 / s^2]}} {{[kg^2]} }](https://tex.z-dn.net/?f=G%20%3D%20%5Cfrac%7B%7B%5Bkg%20~%20m%5E3%20%2F%20s%5E2%5D%7D%7D%20%7B%7B%5Bkg%5E2%5D%7D%20%7D)
![G = \frac{{[m^3 / s^2]}} {{[kg]} }](https://tex.z-dn.net/?f=G%20%3D%20%5Cfrac%7B%7B%5Bm%5E3%20%2F%20s%5E2%5D%7D%7D%20%7B%7B%5Bkg%5D%7D%20%7D)
![\boxed {G = \frac{{[m^3]}} {{[kg ~ s^2]} }}](https://tex.z-dn.net/?f=%5Cboxed%20%7BG%20%3D%20%5Cfrac%7B%7B%5Bm%5E3%5D%7D%7D%20%7B%7B%5Bkg%20~%20s%5E2%5D%7D%20%7D%7D)
The unit of G must be 
<h3>Learn more</h3>
<h3>Answer details</h3>
Grade: High School
Subject: Physics
Chapter: Gravitational Fields
Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant
