Answer:
See below
Explanation:
Normal force = m g cos 53 = 8 kg * 9.8 m/s^2 * cos 53 = 47.1823 N
no work is done by this force
Force friction = coeff friction * force normal = .4 * 47.1823 = 7.55 N
work of friction = 7.55 * 2 m = 15.1 j
Force Downplane = mg sin 53 = 62.61 N
work = 62.61 * 2 = 125.22 j
Net Force downplane = force downplane - force friction = 55.06 N
net Work = force * distance = 55.06 N * 2 M = 110.12 j
The image distance can be determined using the mirror equation: 1/f = 1/d_o + 1/d_i, where, f is the focal length, d_o is the object distance, and d_i is the image distance. Given that f = 28.2 and d_o = 33.2 cm, the value of d_i is calculated to be 187.248 cm. On the other hand, the image height is obtained using the magnification equation wherein, h_i/h_o = -d_i/d_o, where h_i is the image height and h_o is the object height. Using the given values, h_i is equal to -26.79 cm. Note that the negative sign indicates that the image is inverted.
Answers:
kinetic energy lost = 86.4J
Explanation:
let Kf be the kinetic energy after the collision and Ki be the kinetic energy before the collision. let the 3kg car be 1 and 2kg car be 2.
Kf = K1(f) + K2(f)
Ki = K1(i) + k2(i)
loss in kinetic energy = Kf - Ki
= 1/2(3)(2.20)^2 + 1/2(2)(2.20)^2 - 1/2(3)(7)^2 - 1/2(2)(-5)^2
= 12.1 - 98.5
= -86.4 J
therefore, the kinetic energy lost in the collision is 86.4 J.
Answer:
Explanation:
The shear stress due to torque can be calculed by using the following model:
The maximum torque on the section is:
The Torsion Constant for the circular tube is:
![J_{tube} = \frac{\pi}{4}\cdot [(0.053\,m)^{4}-(0.038\,m)^{4}]](https://tex.z-dn.net/?f=J_%7Btube%7D%20%3D%20%5Cfrac%7B%5Cpi%7D%7B4%7D%5Ccdot%20%5B%280.053%5C%2Cm%29%5E%7B4%7D-%280.038%5C%2Cm%29%5E%7B4%7D%5D)
Now, the require output is computed:
Ok I’ll help you I think 20 -192 = 1928172 then ur divided • by 181$1 then u get 244141551611671718181919191827337533535352526
