Answer:
Newton's third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A. Notice that the forces are exerted on different objects. However much you push your fork in the food, that much a dent it will cause.
Explanation:
Answer:
a). M = 20.392 kg
b). am = 0.56
(block), aM = 0.28
(bucket)
Explanation:
a). We got N = mg cos θ,
f = 
= 
If the block is ready to slide,
T = mg sin θ + f
T = mg sin θ +
.....(i)
2T = Mg ..........(ii)
Putting (ii) in (i), we get



M = 20.392 kg
b).
.............(iii)
Here, l = total string length
Differentiating equation (iii) double time w.r.t t, l, h and h' are constants, so


.....................(iv)
We got, N = mg cos θ

∴ 
................(v)
Mg - 2T = M

(from equation (iv))
.....................(vi)
Putting (vi) in equation (v),

![$\frac{g\left[\frac{M}{2}-m \sin \theta-\mu_K m \cos \theta\right]}{(\frac{M}{4}+m)}=a_m$](https://tex.z-dn.net/?f=%24%5Cfrac%7Bg%5Cleft%5B%5Cfrac%7BM%7D%7B2%7D-m%20%5Csin%20%5Ctheta-%5Cmu_K%20m%20%5Ccos%20%5Ctheta%5Cright%5D%7D%7B%28%5Cfrac%7BM%7D%7B4%7D%2Bm%29%7D%3Da_m%24)
![$\frac{9.8\left[\frac{20.392}{2}-10(\sin 30+0.5 \cos 30)\right]}{(\frac{20.392}{4}+10)}=a_m$](https://tex.z-dn.net/?f=%24%5Cfrac%7B9.8%5Cleft%5B%5Cfrac%7B20.392%7D%7B2%7D-10%28%5Csin%2030%2B0.5%20%5Ccos%2030%29%5Cright%5D%7D%7B%28%5Cfrac%7B20.392%7D%7B4%7D%2B10%29%7D%3Da_m%24)

Using equation (iv), we get,

Answer:
The mnemonic I can use to memorize the metric prefixes in this order is: Gigantic Monsters Killed One Million Men Napping Peacefully. All right, so again, gigantic monsters killed one million men napping peacefully.
Answer:
induced current
Explanation:
intentionally manipulated.
Answer:
Maximum height of the ball, h(t) = 27.56 m
Explanation:
It is given that, a ball is shot from the ground straight up into the air with initial velocity of 42 ft/sec.
The height of the ball as a function of time t is given by :

h₀ is initial height, h₀ = 0
So,
.........(1)
For maximum/minimum height, 
...(2)
t = 1.31 s
Differentiating equation (2) wrt t
h''(t) = -32 < 0
So, at t = 1.31 seconds we will get the maximum height.
Put the value of t in equation (1)

h(t) = 27.56 m
Hence, this is the required solution.