Answer:
Explanation:
Given that,
Mass of block
M = 2kg
Spring constant k = 300N/m
Velocity v = 12m/s
At t = 0, the spring is neither stretched nor compressed. Then, it amplitude is zero at t=0
xo = 0
It velocity is 12m/s at t=0
Then, it initial velocity is
Vo = 12m/s
Then, amplitude is given as
A = √[xo + (Vo²/ω²)]
Where
xo is the initial amplitude =0
Vo is the initial velocity =12m/s
ω is the angular frequency and it can be determine using
ω = √(k/m)
Where
k is spring constant = 300N/m
m is the mass of object = 2kg
Then,
ω = √300/2 = √150
ω = 12.25 rad/s²
Then,
A = √[xo + (Vo²/ω²)]
A = √[0 + (12²/12.5²)]
A = √[0 + 0.96]
A = √0.96
A = 0.98m
Answer:
The Earth weighs about:
13,170,000,000,000,000,000,000,000 pounds
or
5.97 billion trillion metric tons
Explanation:
Have a great rest of your day
#TheWizzer
What exactly do u want me to do for u mam/sir
Answer:
Explanation:
If we assume there is a sharp boundary between the two masses of air, there will be a refraction. The refractive index of each medium will depend on the relative speeds of light.
n = c / v
If light travels faster in warmer air, it will have a lower refractive index
nh < nc
Snell's law of refraction relates angles of incidence and refracted with the indexes of refraction:
n1 * sin(θ1) = n2 * sin(θ2)
sin(θ2) = sin(θ1) * n1/n2
If blue light from the sky passing through the hot air will cross to the cold air, then
n1 = nh
n2 = nc
Then:
n1 < n2
So:
n1/n2 < 1
The refracted light will come into the cold air at angle θ2 wich will be smaller than θ1, so the light is bent upwards, creating the appearance of water in the distance, which is actually a mirror image of the sky.
Explanation:
The given data is as follows.
Length of beam, (L) = 5.50 m
Weight of the beam, (
) = 332 N
Weight of the Suki, (
) = 505 N
After crossing the left support of the beam by the suki then at some overhang distance the beam starts o tip. And, this is the maximum distance we need to calculate. Therefore, at the left support we will set up the moment and equate it to zero.
= 0
x = 
= 
= 0.986 m
Hence, the suki can come (2 - 0.986) m = 1.014 from the end before the beam begins to tip.
Thus, we can conclude that suki can come 1.014 m close to the end before the beam begins to tip.
