Answer:
1/2 Hz
Explanation:
A simple harmonic motion has an equation in the form of
![x(t) = Acos(\omega t - \phi)](https://tex.z-dn.net/?f=x%28t%29%20%3D%20Acos%28%5Comega%20t%20-%20%5Cphi%29)
where A is the amplitude,
is the angular frequency and
is the initial phase.
Since our body has an equation of x = 5cos(π t + π/3) we can equate
and solve for frequency f
![2\pi f = \pi](https://tex.z-dn.net/?f=2%5Cpi%20f%20%3D%20%5Cpi)
f = 1/2 Hz
Answer:
33.2 m
Explanation:
For the first object:
y₀ = 81.5 m
v₀ = 0 m/s
a = -9.8 m/s²
t₀ = 0 s
y = y₀ + v₀ t + ½ at²
y = 81.5 − 4.9t²
For the second object:
y₀ = 0 m
v₀ = 40.0 m/s
a = -9.8 m/s²
t₀ = 2.20 s
y = y₀ + v₀ t + ½ at²
y = 40(t−2.2) − 4.9(t−2.2)²
When they meet:
81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²
81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)
81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716
81.5 = 61.56t − 111.716
193.216 = 61.56t
t = 3.139
The position at that time is:
y = 81.5 − 4.9(3.139)²
y = 33.2
Answer:
a
Explanation:
as the copper wire is very dangerous so now if these two thing happens then it would easily help the current flows through it so it might be a little bit easy for the current to flow through it
The strength of the electric field is 5 N/C
Explanation:
The magnitude of the electric field produced by a single-point charge is given by:
![E=\frac{kQ}{r^2}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7BkQ%7D%7Br%5E2%7D)
where
is the Coulomb's constant
Q is the magnitude of the charge
r is the distance from the charge
In this problem, we have:
is the charge producing the field
r = 100 m is the distance from the charge at which we want to calculate the field
Substituting into the equation, we find the s trength of the electric field:
![E=\frac{(8.99\cdot 10^9)(6\cdot 10^{-6})}{(100)^2}=5.4 N/C \sim 5 N/C](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B%288.99%5Ccdot%2010%5E9%29%286%5Ccdot%2010%5E%7B-6%7D%29%7D%7B%28100%29%5E2%7D%3D5.4%20N%2FC%20%5Csim%205%20N%2FC)
Learn more about electric field:
brainly.com/question/8960054
brainly.com/question/4273177
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