Answer:
g_x = 3.0 m / s^2
Explanation:
Given:
- Change in length of spring [email protected] = 22.6 cm
- Time taken for 11 oscillations t = 19.0 s
Find:
- The value of gravitational free fall g_x at plant X:
Solution:
- We will assume a simple harmonic motion of the mass for which Time is:
T = 2*pi*sqrt(k / m ) ...... 1
- Sum of forces in vertical direction @equilibrium is zero:
F_net = k*x - m*g_x = 0
(k / m) = (g_x / x) .... 2
- substitute Eq 2 into Eq 1:
2*pi / T = sqrt ( g_x / x )
g_x = (2*pi / T )^2 * x
- Evaluate g_x:
g_x = (2*pi / (19 / 11) )^2 * 0.226
g_x = 3.0 m / s^2
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
Answer:
The natural medium emanating from the Sun and other very hot sources (now recognised as electromagnetic radiation with a wavelength of 400-750 nm), within which vision is possible.
Explanation:
just the way it is
There's no way to tell. Without seeing a diagram of the circuit,
I'll need to know much more about it than you've told me.
I don't know anything about the components or power supply
that are in the circuit, and I don't know where point ' f ' is in it.
Right now, even with the copious volume of all the available
information, no answer to your question is possible.