If a substance cannot be separated by physical means or chemical means, it is _an element_.
Explanation:
initial velocity U = 20m/s
Final velocity V = 35m/s
time = 15.0 secs
change in velocity = 35 - 15
= 20m/s
acceleration a = change in velocity/time V/t
a = (35-20)/15
a= 15/15
Hence, your acceleration is 1m/s^2
Answer:
Hydrogen nuclei fuse to form a larger nucleus, and a small amount of mass is converted into energy.
Answer:
a) 1.58 kg s^{-1}
b) x_m e^{-1.58t} x_m is initial amplitude
c) 5 kg s^{-1}
Explanation:
given data:
mass =0.5 kg
k = 12.5 N/m
from the data given
a) ![w_d = w_o - \frac{0.2}{100}w_o](https://tex.z-dn.net/?f=w_d%20%3D%20w_o%20-%20%5Cfrac%7B0.2%7D%7B100%7Dw_o)
![= w_o - 0.002w_o = 0.99w_o](https://tex.z-dn.net/?f=%3D%20w_o%20-%200.002w_o%20%3D%200.99w_o)
![w_d = \sqrt{\frac{k}{m} - \frac{b^2}{4m^2}](https://tex.z-dn.net/?f=w_d%20%3D%20%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%20-%20%5Cfrac%7Bb%5E2%7D%7B4m%5E2%7D)
![0.998w_o = \sqrt{w_o^2 - \frac{b^2}{4m^2}](https://tex.z-dn.net/?f=0.998w_o%20%3D%20%5Csqrt%7Bw_o%5E2%20-%20%5Cfrac%7Bb%5E2%7D%7B4m%5E2%7D)
![(0.998w_o)^2 = w_o^2 -\frac{b^2}{1}](https://tex.z-dn.net/?f=%280.998w_o%29%5E2%20%3D%20w_o%5E2%20-%5Cfrac%7Bb%5E2%7D%7B1%7D)
![b^2 = w_o^2 -(0.998w_o)^2](https://tex.z-dn.net/?f=b%5E2%20%3D%20w_o%5E2%20-%280.998w_o%29%5E2)
![b^2 = w_o^2(1-0.998^2) = 3.996 *10^{-3} w_o^2](https://tex.z-dn.net/?f=b%5E2%20%3D%20w_o%5E2%281-0.998%5E2%29%20%3D%203.996%20%2A10%5E%7B-3%7D%20w_o%5E2)
![b = w_o\sqrt{3.996*10^{-3}}](https://tex.z-dn.net/?f=b%20%3D%20w_o%5Csqrt%7B3.996%2A10%5E%7B-3%7D%7D)
![b = \frac{12.5}{0.5}\sqrt{3.996*10^{-3}} = 1.58 kg s^{-1}](https://tex.z-dn.net/?f=b%20%3D%20%5Cfrac%7B12.5%7D%7B0.5%7D%5Csqrt%7B3.996%2A10%5E%7B-3%7D%7D%20%3D%201.58%20kg%20s%5E%7B-1%7D)
b)![x = x_m e^{\frac{-bt}{2m}}](https://tex.z-dn.net/?f=%20x%20%3D%20x_m%20e%5E%7B%5Cfrac%7B-bt%7D%7B2m%7D%7D)
where x_m is initial amplitude
c) critical damping amplitude
![c_c =2\sqrt{km} = 2\sqrt{12.5*.5} = 5 kg s^{-1}](https://tex.z-dn.net/?f=c_c%20%3D2%5Csqrt%7Bkm%7D%20%3D%202%5Csqrt%7B12.5%2A.5%7D%20%3D%205%20kg%20s%5E%7B-1%7D)
Answer:
a= 1.59 m/s² : Magnitude of the acceleration
β = 65.22° (north of east) : Direction of the acceleration
Explanation:
Conceptual analysis
We apply Newton's second law:
∑F = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Problem development
The acceleration is presented in the direction of the resultant force applied .
Calculation of the resultant forces (R)
![R=\sqrt{(F_{N})^{2} +(F_{E})^{2} }](https://tex.z-dn.net/?f=R%3D%5Csqrt%7B%28F_%7BN%7D%29%5E%7B2%7D%20%2B%28F_%7BE%7D%29%5E%7B2%7D%20%7D)
![R=\sqrt{(390)^{2} +(180)^{2} }](https://tex.z-dn.net/?f=R%3D%5Csqrt%7B%28390%29%5E%7B2%7D%20%2B%28180%29%5E%7B2%7D%20%7D)
R= 429.5 N
We apply the formula (1) to calculate the magnitude of the acceleration(a) :
∑F = m*a , m= 270 kg
R= m*a
429.5 =270*a
![a=\frac{429.5}{270} \frac{m}{s^{2} }](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B429.5%7D%7B270%7D%20%20%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%20%7D)
a= 1.59 m/s²
Calculation of the direction of the acceleration (β)
![\beta = tan^{-1} (\frac{F_{N} }{F_{E}})](https://tex.z-dn.net/?f=%5Cbeta%20%3D%20tan%5E%7B-1%7D%20%28%5Cfrac%7BF_%7BN%7D%20%7D%7BF_%7BE%7D%7D%29)
![\beta = tan^{-1} (\frac{390 }{180})](https://tex.z-dn.net/?f=%5Cbeta%20%3D%20tan%5E%7B-1%7D%20%28%5Cfrac%7B390%20%7D%7B180%7D%29)
β = 65.22° (north of east)