(a) The stone travels a vertical distance <em>y</em> of
<em>y</em> = (12.0 m/s) <em>t</em> + 1/2 <em>g t</em> ²
where <em>g</em> = 9.80 m/s² is the acceleration due to gravity. Note that this equation assume the downward direction to be positive, and that <em>y</em> = 0 corresponds to the height from which the stone is thrown.
So if it reaches the ground in <em>t</em> = 1.54 s, then the height of the building <em>y</em> is
<em>y</em> = (12.0 m/s) (1.54 s) + 1/2 (9.80 m/s²) (1.54 s)² ≈ 30.1 m
(b) The stone's (downward) velocity <em>v</em> at time <em>t </em>is
<em>v</em> = 12.0 m/s + <em>g t</em>
so that after <em>t</em> = 1.54 s, its velocity is
<em>v</em> = 12.0 m/s + (9.80 m/s²) (1.54 s) ≈ 27.1 m/s
(and of course, speed is the magnitude of velocity)
Answer: electrical energy
Explanation:
Electrical energy
Answer:
32 cm
Explanation:
f = focal length of the converging lens = 16 cm
Since the lens produce the image with same size as object, magnification is given as
m = magnification = - 1
p = distance of the object from the lens
q = distance of the image from the lens
magnification is given as
m = - q/p
- 1 = - q/p
q = p eq-1
Using the lens equation, we get
1/p + 1/q = 1/f
using eq-1
1/p + 1/p = 1/16
p = 32 cm
Using the Equation:
v² = vi² + 2 · a · s → Eq.1
where,
v = final velocity
vi = initial velocity
a = acceleration
s = distance
<span><span>We know that vi = 0 because the ball was at rest initially.
</span><span>
Therefore,
Solving Eq.1 for acceleration,
</span></span> v² = vi² + 2 · a · s
v² = 0 + 2 · a · s
v² = 2 · a · s
Rearranging for a,
a = v ²/2·<span>s
Substituting the values,
a = 46</span>²/2×1<span>
a = 1058 m/s</span>²
<span>Now applying Newton's 2nd law of motion,
</span>
<span>F = ma
= 0.145</span>×<span>1058
F = 153.4 N</span>
Answer:
this vehichle has 896,000 jules of energy
Explanation: KE=1/2mv squared, KE is Kinetic energy. m is mass and v is velocity
