Answer:
v = 15 m / s
Explanation:
In this exercise we are given the position function
x = 5 t²
and we are asked for the average velocity in an interval between t = 0 and t= 3 s, which is defined by the displacement between the time interval
let's look for the displacements
t = 0 x₀ = 0 m
t = 3 
= 5 3 2
x_{f} = 45 m
we substitute
v = 15 m / s
Answer:
29.16 J
Explanation:
From Hook's law,
W = 1/2(ke²)..................... Equation 1
Where W = work done, k = Spring constant, e = extension.
Given: W = 9 J, e = 0.5 m.
Substitute into equation 1
9 = 1/2(k×0.5²)
Solve for k
k = 18/0.5²
k = 72 N/m.
The work done required to stretch the spring by additional 0.4 m is
W = 1/2(72)(0.4+0.5)²
W = 36(0.9²)
W = 29.16 J.
Answer:
The forces are balanced on both animals because they are not moving
More importantly than not moving is not <u>accelerating.</u>
Explanation:
Answer:
a) t = 0.0185 s = 18.5 ms
b) T = 874.8 N
Explanation:
a)
First we find the seed of wave:
v = fλ
where,
v = speed of wave
f = frequency = 810 Hz
λ = wavelength = 0.4 m
Therefore,
v = (810 Hz)(0.4 m)
v = 324 m/s
Now,
v = L/t
where,
L = length of wire = 6 m
t = time taken by wave to travel length of wire
Therefore,
324 m/s = 6 m/t
t = (6 m)/(324 m/s)
<u>t = 0.0185 s = 18.5 ms</u>
<u></u>
b)
From the formula of fundamental frquency, we know that:
Fundamental Frequency = v/2L = (1/2L)(√T/μ)
v = √(T/μ)
where,
T = tension in string
μ = linear mass density of wire = m/L = 0.05 kg/6 m = 8.33 x 10⁻³ k gm⁻¹
Therefore,
324 m/s = √(T/8.33 x 10⁻³ k gm⁻¹)
(324 m/s)² = T/8.33 x 10⁻³ k gm⁻¹
<u>T = 874.8 N</u>
