Answer:
Resultant force = (232.93î + 246.10j) N
x-component of the resultant force = (+232.93î) N
y-component of the resultant force = (+246.1j) N
Explanation:
The net external force on the statue is equal to the resultant force on the statue.
And the resuphant force is a vector sum of all the other forces acting on the statue.
Force 1 = (45î) N
Force 2 = (105j) N
Force 3 = (235cos 36.9°)î + (235 sin 36.9°)j = (187.93î + 141.10j) N
Resultant force = (Force 1) + (Force 2) + (Force 3)
Resultant force = 45î + 105j + (187.93î + 141.10j) = (232.93î + 246.10j) N
Hope this helps!!!
Somewhere in your book or your notes, you must have met the formula for the
gravitational attraction between two bodies. If you can go back and find it, you
only need to plug your numbers into that formula, and out will pop the answer.
Formula: <u>Force = G (mA x mB) / (distance)²</u>
If everything is in SI units, then G = 6.67 x10⁻¹¹ newton-meter² / kilogram²
You said that
mA = 8.1 kg
mB = 6.5 kg
distance = 0.5 m .
Force = (6.67 x 10⁻¹¹ nt-m²/kg²) (8.1kg x 6.5kg / (0.5m)² =
(6.67 x 10⁻¹¹ nt-m²/kg²) ( 52.65 kg² ) / (0.25 m²) =
<em>1.4047 x 10⁻⁸ newtons .</em>
That's roughly 5.052 x 10⁻⁸ ounce . (5% of one micro-ounce)
Answer:
Explanation:
weight of woman acting downwards = 90 x 10 = 900 N
force by bungee cord in upward direction = 500 N
net force = 900 - 500 = 400 N in downward direction
acceleration in downward direction
= 400 / 90
= 4.44 m /s²
Answer:
a)
, b)
, c) ![A = 1.355\,m](https://tex.z-dn.net/?f=A%20%3D%201.355%5C%2Cm)
Explanation:
a) The system have a simple armonic motion, whose position function is:
![x(t) = A\cdot \cos (\omega\cdot t + \phi)](https://tex.z-dn.net/?f=x%28t%29%20%3D%20A%5Ccdot%20%5Ccos%20%28%5Comega%5Ccdot%20t%20%2B%20%5Cphi%29)
The velocity function is determined by deriving the position function in terms of time:
![v(t) = -\omega \cdot A \cdot \sin(\omega\cdot t + \phi)](https://tex.z-dn.net/?f=v%28t%29%20%3D%20-%5Comega%20%5Ccdot%20A%20%5Ccdot%20%5Csin%28%5Comega%5Ccdot%20t%20%2B%20%5Cphi%29)
The acceleration function is found by deriving again:
![a(t) = -\omega^{2} \cdot A \cdot \cos (\omega\cdot t + \phi)](https://tex.z-dn.net/?f=a%28t%29%20%3D%20-%5Comega%5E%7B2%7D%20%5Ccdot%20A%20%5Ccdot%20%5Ccos%20%28%5Comega%5Ccdot%20t%20%2B%20%5Cphi%29)
Let assume that
. The following nonlinear system is built:
![A\cdot \cos \phi = 0.660\,m](https://tex.z-dn.net/?f=A%5Ccdot%20%5Ccos%20%5Cphi%20%3D%200.660%5C%2Cm)
![-\omega \cdot A \cdot \sin \phi = -12.3\,\frac{m}{s}](https://tex.z-dn.net/?f=-%5Comega%20%5Ccdot%20A%20%5Ccdot%20%5Csin%20%5Cphi%20%3D%20-12.3%5C%2C%5Cfrac%7Bm%7D%7Bs%7D)
![-\omega^{2}\cdot A \cdot \sin \phi = -128\,\frac{m}{s^{2}}](https://tex.z-dn.net/?f=-%5Comega%5E%7B2%7D%5Ccdot%20A%20%5Ccdot%20%5Csin%20%5Cphi%20%3D%20-128%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D)
System can be reduced by divinding the second and third expressions by the first expression:
![\omega \cdot \tan \phi = 18.636\,\frac{1}{s}](https://tex.z-dn.net/?f=%5Comega%20%5Ccdot%20%5Ctan%20%5Cphi%20%3D%2018.636%5C%2C%5Cfrac%7B1%7D%7Bs%7D)
![\omega^{2}\cdot \tan \phi = 193.94\,\frac{1}{s^{2}}](https://tex.z-dn.net/?f=%5Comega%5E%7B2%7D%5Ccdot%20%5Ctan%20%5Cphi%20%3D%20193.94%5C%2C%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%7D)
Now, the last expression is divided by the first one:
![\omega = 10.407\,\frac{rad}{s}](https://tex.z-dn.net/?f=%5Comega%20%3D%2010.407%5C%2C%5Cfrac%7Brad%7D%7Bs%7D)
b) The mass of the block is:
![m = \frac{k}{\omega^{2}}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7Bk%7D%7B%5Comega%5E%7B2%7D%7D)
![m = \frac{500\,\frac{N}{m} }{(10.407\,\frac{rad}{s})^{2} }](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B500%5C%2C%5Cfrac%7BN%7D%7Bm%7D%20%7D%7B%2810.407%5C%2C%5Cfrac%7Brad%7D%7Bs%7D%29%5E%7B2%7D%20%7D)
![m = 4.617\,kg](https://tex.z-dn.net/?f=m%20%3D%204.617%5C%2Ckg)
c) The phase angle is:
![\phi = \tan^{-1} \left(\frac{18.636\,\frac{1}{s} }{\omega} \right)](https://tex.z-dn.net/?f=%5Cphi%20%3D%20%5Ctan%5E%7B-1%7D%20%5Cleft%28%5Cfrac%7B18.636%5C%2C%5Cfrac%7B1%7D%7Bs%7D%20%7D%7B%5Comega%7D%20%20%5Cright%29)
![\phi \approx 0.338\pi](https://tex.z-dn.net/?f=%5Cphi%20%5Capprox%200.338%5Cpi)
The amplitude is:
![A = \frac{0.660\,m}{\cos 0.338\pi}](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7B0.660%5C%2Cm%7D%7B%5Ccos%200.338%5Cpi%7D)
![A = 1.355\,m](https://tex.z-dn.net/?f=A%20%3D%201.355%5C%2Cm)
Answer:
B
Explanation:this is not physics