C) above, the Earth's core is about the melting point of iron.
The short answer is that the displacement is equal tothe area under the curve in the velocity-time graph. The region under the curve in the first 4.0 s is a triangle with height 10.0 m/s and length 4.0 s, so its area - and hence the displacement - is
1/2 • (10.0 m/s) • (4.0 s) = 20.00 m
Another way to derive this: since velocity is linear over the first 4.0 s, that means acceleration is constant. Recall that average velocity is defined as
<em>v</em> (ave) = ∆<em>x</em> / ∆<em>t</em>
and under constant acceleration,
<em>v</em> (ave) = (<em>v</em> (final) + <em>v</em> (initial)) / 2
According to the plot, with ∆<em>t</em> = 4.0 s, we have <em>v</em> (initial) = 0 and <em>v</em> (final) = 10.0 m/s, so
∆<em>x</em> / (4.0 s) = (10.0 m/s) / 2
∆<em>x</em> = ((4.0 s) • (10.0 m/s)) / 2
∆<em>x</em> = 20.00 m
Answer:
14.85 m/s
Explanation:
From the question given above, the following data were obtained:
Height (h) of tower = 45 m
Horizontal distance (s) moved by the balloon = 45 m
Horizontal velocity (u) =?
Next, we shall determine the time taken for the balloon to hit the shoe of the passerby. This is illustrated below:
Height (h) of tower = 45 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
45 = ½ × 9.8 × t²
45 = 4.8 × t²
Divide both side by 4.9
t² = 45/4.9
Take the square root of both side
t = √(45/4.9)
t = 3.03 s
Finally, we shall determine the magnitude of the horizontal velocity of the balloon as shown below:
Horizontal distance (s) moved by the balloon = 45 m
Time (t) = 3.03 s
Horizontal velocity (u) =?
s = ut
45 = u × 3.03
Divide both side by 3.03
u = 45/3.03
u = 14.85 m/s
Thus, the magnitude of the horizontal velocity of the balloon was 14.85 m/s
