(a) The ball's height <em>y</em> at time <em>t</em> is given by
<em>y</em> = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²
where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve <em>y</em> = 0 for <em>t</em> :
0 = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²
0 = <em>t</em> ((20 m/s) sin(40º) - 1/2 <em>g t</em> )
<em>t</em> = 0 or (20 m/s) sin(40º) - 1/2 <em>g t</em> = 0
The first time refers to where the ball is initially launched, so we omit that solution.
(20 m/s) sin(40º) = 1/2 <em>g t</em>
<em>t</em> = (40 m/s) sin(40º) / <em>g</em>
<em>t</em> ≈ 2.6 s
(b) At its maximum height, the ball has zero vertical velocity. In the vertical direction, the ball is in free fall and only subject to the downward acceleration <em>g</em>. So
0² - ((20 m/s) sin(40º))² = 2 (-<em>g</em>) <em>y</em>
where <em>y</em> in this equation refers to the maximum height of the ball. Solve for <em>y</em> :
<em>y</em> = ((20 m/s) sin(40º))² / (2<em>g</em>)
<em>y</em> ≈ 8.4 m
Answer:
The decrease is due to the bulge at the equator (putting more distance between the rest of the planet and the surface
Explanation:
Answer:
Explanation:
Given
-- initial velocity
--- height
Required
Determine the time to hit the ground
This will be solved using the following motion equation.
Where
So, we have:
Subtract 30.2 from both sides
Solve using quadratic formula:
Where
Split the expression
or 
or 
Time can't be negative; So, we have:
Hence, the time to hit the ground is 1.82 seconds
Answer:
杰恩斯克克斯克奇沃伊斯克克斯克什德布德克什恩克恩德恩克恩茨克杰兹姆克斯恩斯姆斯姆德恩德姆德武伊乔奥斯克斯杰德布德赫德夫赫富伊什杰吉迪赫德赫夫赫德
Answer:
B
Explanation:
Potential difference has a SI Unit of Volt and its symbol is <em>V</em>. Hence answer is <u>B</u>.
A is wrong as it has the unit Joule <em>(J)</em> which is the SI unit for energy.
C is wrong as it has the unit Newton <em>(N)</em> which is the SI unit for force.
D is wrong as it has the unit Coulomb <em>(C)</em> which is the SI unit of charge.
