Hi there!
1.
The period of a pendulum can be calculated using the following equation:
![\large\boxed{T = 2\pi \sqrt{\frac{L}{g}}}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7BT%20%3D%202%5Cpi%20%5Csqrt%7B%5Cfrac%7BL%7D%7Bg%7D%7D%7D)
T = period (s)
L = length of string (m)
g = acceleration due to gravity (m/s²)
Plug in the values:
![T = 2\pi \sqrt{\frac{4}{9.8}} = \boxed{4.014 s}](https://tex.z-dn.net/?f=T%20%3D%202%5Cpi%20%5Csqrt%7B%5Cfrac%7B4%7D%7B9.8%7D%7D%20%3D%20%5Cboxed%7B4.014%20s%7D)
2.
Calculate the period:
![T = \frac{\text{Time}}{\# of oscillations} = \frac{5}{20} = \boxed{0.4 s }](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B%5Ctext%7BTime%7D%7D%7B%5C%23%20of%20oscillations%7D%20%3D%20%5Cfrac%7B5%7D%7B20%7D%20%3D%20%5Cboxed%7B0.4%20s%20%7D)
Frequency is the reciprocal of the period, so:
![f = \frac{1}{T} = \frac{1}{0.4} = \boxed{2.5 Hz}](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7B1%7D%7BT%7D%20%3D%20%5Cfrac%7B1%7D%7B0.4%7D%20%3D%20%5Cboxed%7B2.5%20Hz%7D)
Weight = (mass) x (gravity)
Divide each side by (gravity): Mass = (weight) / (gravity) =
(55.54 N) / (9.83 m/s²) =
5.65 kilograms
Answer:
5.82x10^-3
Explanation:
If the decimal moves right I think it goes negative in the exponent because the starting number is small...or less than 1
I think
Answer:
26.6 m/s
Explanation:
Given:
Δy = 2.1 m
t = 5.35 s
a = -9.8 m/s²
Find: v₀
Δy = v₀ t + ½ at²
(2.1 m) = v₀ (5.35 s) + ½ (-9.8 m/s²) (5.35 s)²
v₀ = 26.6 m/s
We know, F = m * a
F = 10 * 5
F = 50 N
In short, Your Answer would be 50 Newtons
Hope this helps!