Given that:
Distance , s = 18.5 m
Velocity , v = 3.85 m/s
Time , t =?
Since,
Velocity = distance/time
or
Time= distance/velocity
time= 18.5/ 3.85
time= 4.8 s
So the time elapse between the release of the ball and the ball passing home plate is 4.8 seconds.
(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:
6 days.
Explanation:
From radioactivity, The expression for half life is given as,
R/R' = 2⁽ᵃ/ᵇ)................... Equation 1
Where R = original mass of the radioactive substance, R' = Remaining mass of the radioactive substance after decay, a = Total time taken to decay, b = half life.
Given: R = 80 g, R' = 10 g, b = 2 days.
Substitute into equation 1
80/10 = 2⁽ᵃ/²⁾
8 = 2⁽ᵃ/²⁾
2³ = 2⁽ᵃ/²)
Equating the base and solving for a
3 = a/2
a = 2×3
a = 6 days.
A. Moving with constant non-zero speed
If you mean S is the distance then it is true
Velocity = Distance / time
