Answer:
distance= velocity ×time
distance= 62×10
distance=620m
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Answer:
We are given x= bt +ct²
So
A. bxt= m
Because m/s*s= m
So b= m/s and c= m/s²
B.
x= bt-ct²
So at x=0 t=0
x=0 t= 2
We have
bt = ct² so t = b/c at x= 0
So b-2ct= 0
B. To find velocity we use
dx / dt = b - 2 Ct
C. At rest wen V= 0
We have t= b/2c
D. To find acceleration we use
dv / dt = - 2C
(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
The answer of this question is B.
Answer:0,002 = 2 x 10⁻³
Explanation:
0,002 = 2 / 1000 = 2 / 10³ = 2 x 10⁻³
