Answer:
the required frequency of waves is 2.066 Hz
Explanation:
Given the data in the question;
μ = 1.50 kg/m
T = 6000 N
Amplitude A = 0.500 m
P = 2.00 kW = 2000 W
we know that, the average power transmit through the rope can be expressed as;
p = 
vμω²A²
p = √(T/μ)μω²A²
so we solve for ω
ω² = 2P / √(T/μ)μA²
we substitute
ω² = 2(2000) / √(6000/1.5)(1.5)(0.500)²
ω² = 4000 / 23.71708
ω² = 168.65
(2πf)² = ω²
so
(2πf)² = 168.65
4π²f² = 168.65
f² = 168.65 / 4π²
f² = 4.27195
f = √4.27195
f = 2.066 Hz
Therefore, the required frequency of waves is 2.066 Hz
Acceleration
due to gravity = 2.6m/s²
Length
of building = d = 88.3m
<span>Time
he needed before she hits the ground = ?
we can find the time by using the
formula;</span>
D =
1/2at²
Now
putting the value;
<span>88.3
= (1/2) a t</span>²<span>
88.3 = 1/2 x 23.6 x t</span>²
t² =
88.3 / 11.8
=
7.48
<span>t = 2.735 seconds
</span>
Answer:
d
Explanation:
distorts or shears rock as it travels through it
Complete Question
Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?
Answer:
The value is 
Explanation:
From the question we are told that
The semi - major axis of the rocky debris 
The semi - major axis of Planet D is 
The orbital period of planet D is 
Generally from Kepler third law
Here T is the orbital period while a is the semi major axis
So
=> ![T_R = T_D * [\frac{a_R}{a_D} ]^{\frac{3}{2} }](https://tex.z-dn.net/?f=T_R%20%20%3D%20T_D%20%2A%20%20%5B%5Cfrac%7Ba_R%7D%7Ba_D%7D%20%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D)
=> ![T_R = 18.164 * [\frac{ 45}{60} ]^{\frac{3}{2} }](https://tex.z-dn.net/?f=T_R%20%20%3D%2018.164%20%20%2A%20%20%5B%5Cfrac%7B%2045%7D%7B60%7D%20%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D)
=>
Answer:
Velocity- v=10m/s south Speed- v=10m/s
Explanation:
the salar measurement of how fast an object is moving.
Velocity- v=10m/s south :
The vector rate of change of position.
Speed- v=10 m/s : the salar measurement of how fast an object is moving.
