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!!!
According to Newton's 2nd law of motion, Jim can throw a tennis ball farther than a <span>basketball</span>. The rest of the choices do not answer the question above.
Answer:
5 m/s
Explanation:
Horizontal distance traveled, x = 2 m
vertical distance traveled, y = 4/5 m
Let the speed of cup as it leaves the counter is v and it takes time t to hit the ground.
Use second equation of motion in vertical direction
![y = ut+\frac{1}{2}at^{2}](https://tex.z-dn.net/?f=y%20%3D%20ut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E%7B2%7D)
Here acceleration in vertical direction is 9.8 m/s^2.
So,
![\frac{4}{5} = 0+\frac{1}{2}\times9.8t^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B5%7D%20%3D%200%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes9.8t%5E%7B2%7D)
t = 0.4 second
Now in horizontal direction the acceleration in zero.
Horizontal distance = horizontal velocity x time
x = v t
2 = v (0.4)
v = 5 m/s
Thus, the horizontal velocity of cup as it leaves the counter is 5 m/s.