Answer:
y₀ = 1020.3 m
Explanation:
This is a projectile launching exercise, in this case as the package is released its initial vertical velocity is zero.
y = y₀ + t - ½ g t²
when it reaches the ground its height is zero
0 = y₀ + 0 - ½ g t²
y₀ = ½ g t²
let's calculate
y₀ = ½ 9.8 14.43²
y₀ = 1020.3 m
Answer:
Explanation:
mass of string = .0125 / 9.8
= 1.275 x 10⁻³ kg
Length of string l = 1.5 m .
m = mass per unit length
= ( .1.275 / 1.5) x 10⁻³ kg/m
m = .85 x 10⁻³ kg/m
wave equation: y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)
compare with equation of wave
y(x,t) = Acos(K x − ω t)
ω ( angular velocity ) = 4830 rad/s
k = 172 rad/m
Velocity = ω / k
= 4830/172 m /s
= 28.08 m /s
velocity of wave =
28.08 =
788.48 = W / .85 X 10⁻³
W = 670 x 10⁻³ N .
c ) wave length
wave length =2π / k
= 2 x 3.14 / 172
= .0365 m
no of wave lengths over whole length of string
= 1.5 / .0365
= 41
d )
equation for waves traveling down the string
= (8.50 mm)cos(172 rad/m x + 4830 rad/s t)
Answer:
<h2>0.5 m/s²</h2>Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
From the question we have
We have the final answer as
<h3>0.5 m/s²</h3>Hope this helps you