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:
The value is 
Explanation:
From the question we are told that
The diameter of the pupil is 
The distance of the page from the eye 
The wavelength is 
The refractive index is 
Generally the minimum separation of adjacent dots that can be resolved is mathematically represented as
![y = [ \frac{1.22 * \lambda }{d_p * n_r } ]* d](https://tex.z-dn.net/?f=y%20%20%3D%20%5B%20%5Cfrac%7B1.22%20%2A%20%20%5Clambda%20%7D%7Bd_p%20%2A%20n_r%20%7D%20%5D%2A%20d)
![y = [ \frac{1.22 * 500 *10^{-9} }{4.2 *10^{-3} * 1.36} ]* 0.29](https://tex.z-dn.net/?f=y%20%20%3D%20%5B%20%5Cfrac%7B1.22%20%2A%20%20500%20%2A10%5E%7B-9%7D%20%7D%7B4.2%20%2A10%5E%7B-3%7D%20%2A%201.36%7D%20%5D%2A%200.29)
Answer:
C. Momentum is conserved but not kinetic energy.
Explanation:
This case represents an entirely inelastic collision, that is, a collision between the car and the truck that reduces total kinetic energy of the entire system, whereas linear momentum is conserved. Hence, correct answer is C.
Answer:
1. v = 6.67 m/s
2. d = 9.54 m
Explanation:
1. To find the horizontal velocity of the rock we need to use the following equation:
<u>Where</u>:
d: is the distance traveled by the rock
t: is the time
The time can be calculated as follows:
<u>Where:</u>
g: is gravity = 9.8 m/s²
Now, the horizontal velocity of the rock is:
Hence, the initial velocity required to barely reach the edge of the shell below you is 6.67 m/s.
2. To calculate the distance at which the projectile will land, first, we need to find the time:
So, the distance is:
Therefore, the projectile will land at 9.54 m of the second cliff.
I hope it helps you!
