Answer:
the length of stretched spring in cm is 22
Explanation:
given information:
spring length, x1 = 20 cm = 0.2 m
force, F = 100 N
the length of spring streches, x2 = 22 cm = 0.22 m
According to Hooke's law
F = - kΔx
k = F/*=(x2-x1)
= 100/(0.22 - 0.20)
= 5000 N/m
if the spring is now suspended from a hook and a 10.2-kg block is attached to the bottom end
m = 10.2 kg
W = m g
= 10.2 x 9.8
= 99.96 N
F = - k Δx
Δx = F / k
= 99.96 / 5000
= 0.02
Δx = x2- x1
x2 = Δx + x1
= 0.20 + 0.02
= 0.22 m
= 22 cm
Do you speak a little English cuz I can’t help you if a can’t understand you
To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,

Here,
D is diameter of the eye


The angle that relates the distance between the lights and the distance to the lamp is given by,

For small angle, 
Here,
d = Distance between lights
L = Distance from eye to lamp
For small angle 
Therefore,



Therefore the distance is 5.367km.
The frequency doesn't change. If the wavespeed increases, then the wavelength must also increase ... It's just the distance the wave travels during each complete wiggle.
Answer:
distance cover is = 102.53 m
Explanation:
Given data:
speed of object is 17.1 m/s


from equation of motion we know that

where d_1 is distance covered in time t1
so
=


where d_2 is distance covered in time t2


distance cover is = 213.31 - 110.78 = 102.53 m