Answer:
35.6 N
Explanation:
We can consider only the forces acting along the horizontal direction to solve the problem.
There are two forces acting along the horizontal direction:
- The horizontal component of the pushing force, which is given by
![F_x = F cos \theta](https://tex.z-dn.net/?f=F_x%20%3D%20F%20cos%20%5Ctheta)
with ![\theta=41.9^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D41.9%5E%7B%5Ccirc%7D)
- The frictional force, whose magnitude is
![F_f = \mu mg](https://tex.z-dn.net/?f=F_f%20%3D%20%5Cmu%20mg)
where
, m=8.2 kg and g=9.8 m/s^2.
The two forces have opposite directions (because the frictional force is always opposite to the motion), and their resultant must be zero, because the suitcase is moving with constant velocity (which means acceleration equals zero, so according to Newton's second law: F=ma, the net force is zero). So we can write:
![F_x - F_f=0\\F_x = F_f\\F cos \theta = \mu mg\\F=\frac{\mu mg}{cos \theta}=\frac{(0.33)(8.2 kg)(9.8 m/s^2)}{cos(41.9^{\circ})}=35.6 N](https://tex.z-dn.net/?f=F_x%20-%20F_f%3D0%5C%5CF_x%20%3D%20F_f%5C%5CF%20cos%20%5Ctheta%20%3D%20%5Cmu%20mg%5C%5CF%3D%5Cfrac%7B%5Cmu%20mg%7D%7Bcos%20%5Ctheta%7D%3D%5Cfrac%7B%280.33%29%288.2%20kg%29%289.8%20m%2Fs%5E2%29%7D%7Bcos%2841.9%5E%7B%5Ccirc%7D%29%7D%3D35.6%20N)
Answer: 6.9x 107 in standard form is 69,000,000
Answer:
![\mathbf{ current(I) =1766.67 \ A}](https://tex.z-dn.net/?f=%5Cmathbf%7B%20current%28I%29%20%3D1766.67%20%5C%20A%7D)
Explanation:
Given that:
The air resistance and friction = 700 N
The gravity caused force = 716 × 9.8 = 7016.8
Total force = (7016.8 + 700) N
Total force = 7716.8 N
∴
![13 \times current(I) \times 0.84 = \dfrac{7716.8 \times 300}{2 \times 60}](https://tex.z-dn.net/?f=13%20%5Ctimes%20%20current%28I%29%20%5Ctimes%200.84%20%3D%20%5Cdfrac%7B7716.8%20%5Ctimes%20300%7D%7B2%20%5Ctimes%2060%7D)
![current(I) \times 10.92= 19292](https://tex.z-dn.net/?f=current%28I%29%20%5Ctimes%2010.92%3D%2019292)
![current(I) = \dfrac{19292}{10.92}](https://tex.z-dn.net/?f=current%28I%29%20%3D%20%5Cdfrac%7B19292%7D%7B10.92%7D)
![\mathbf{ current(I) =1766.67 \ A}](https://tex.z-dn.net/?f=%5Cmathbf%7B%20current%28I%29%20%3D1766.67%20%5C%20A%7D)
Answer:
When the image distance is positive, the image is on the same side of the mirror as the object, and it is real and inverted. When the image distance is negative, the image is behind the mirror, so the image is virtual and upright.
Explanation: