(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 correct answer for this question is this one: "measuring the temperature increase of water from doing work stirring it." This experiment is generally regarded as being first carried out by James Joule is this one, <span>measuring the temperature increase of water from doing work stirring it.</span>
Answer:
the intensity of the sun on the other planet is a hundredth of that of the intensity of the sun on earth.
That is,
Intensity of sun on the other planet, Iₒ = (intensity of the sun on earth, Iₑ)/100
Explanation:
Let the intensity of light be represented by I
Let the distance of the star be d
I ∝ (1/d²)
I = k/d²
For the earth,
Iₑ = k/dₑ²
k = Iₑdₑ²
For the other planet, let intensity be Iₒ and distance be dₒ
Iₒ = k/dₒ²
But dₒ = 10dₑ
Iₒ = k/(10dₑ)²
Iₒ = k/100dₑ²
But k = Iₑdₑ²
Iₒ = Iₑdₑ²/100dₑ² = Iₑ/100
Iₒ = Iₑ/100
Meaning the intensity of the sun on the other planet is a hundredth of that of the intensity on earth.
Answer:
<u>Foot per second. Foot-pound-second system. Frames per second, the frequency (rate) at which consecutive images (frames) appear on a display.</u>
Explanation:
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