(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
HCl is a strong electrolyte and when it dissolves in water it separates almost completely into positively - charged hydrogen ions and negatively - charged chloride ions. This aqueous solution is usually called hydrochloric acid.
Answer:
angular acceleration is -0.2063 rad/s²
Explanation:
Given data
mass m = 95.2 kg
radius r = 0.399 m
turning ω = 93 rpm
radial force N = 19.6 N
kinetic coefficient of friction μ = 0.2
to find out
angular acceleration
solution
we know frictional force that is = radial force × kinetic coefficient of friction
frictional force = 19.6 × 0.2
frictional force = 3.92 N
and
we know moment of inertia that is
γ = I ×α = frictional force × r
so
γ = 1/2 mr²α
α = -2f /mr
α = -2(3.92) /95.2 (0.399)
α = - 7.84 / 37.9848 = -0.2063
so angular acceleration is -0.2063 rad/s²
