Answer:
0N
Explanation:
When you draw a free body diagram, You have a 7N arrow pointing left, and a 3N arrow pointing right. When you subtract the two, you still have a force of 4N on the left. However, seeing as how gravity is in play at a force of 4N, the two forces cancel out and you get 0N.
Answer:
36s
Explanation:
Let the objects be A and B.
Let the initial velocity of A be U and the initial velocity of B be 3U
The height sustain by A will be;
The final velocity would be zero
V2 = U2-2gH
Hence
0^2= U2 -2gH
H = U^2/2g
Similarly for object B, the height sustain is;
V2 = (3U)^2-2gH
Hence
0^2= 3U^2 -2gH
U2-2gH
Hence
0^2= U2 -2gH
H = 3U^2/2g
By comparism. The object with higher velocity sustains more height and so should fall longer than object A.
Now object A would take;
From V=U+gt as the object falls freely, the initial velocity is zero hence and the final velocity of the object is;
V=10×12=120m/s let g be 10m/S2
Similarly for object B,
The final velocity for B when it's falling it should be 3×that of A
Meaning
3V= gt
t =3V/g = 3× 120/10 = 36s
(13.558 gm) · (1 L / 0.089 gm) = 152.34 L (rounded)
(fraction equal to ' 1 ') ^
Answer:
a) W = 46.8 J and b) v = 3.84 m/s
Explanation:
The energy work theorem states that the work done on the system is equal to the variation of the kinetic energy
W = ΔK = 
-K₀
a) work is the scalar product of force by distance
W = F . d
Bold indicates vectors. In this case the dog applies a force in the direction of the displacement, so the angle between the force and the displacement is zero, therefore, the scalar product is reduced to the ordinary product.
W = F d cos θ
W = 39.0 1.20 cos 0
W = 46.8 J
b) zero initial kinetic language because the package is stopped
W -
= -K₀
W - fr d= ½ m v² - 0
W - μ N d = ½ m v
on the horizontal surface using Newton's second law
N-W = 0
N = W = mg
W - μ mg d = ½ m v
v² = (W -μ mg d) 2/m
v = √(W -μ mg d) 2/m
v = √[(46.8 - 0.30 4.30 9.8 1.20) 2/4.3
]
v = √(31.63 2/4.3)
v = 3.84 m/s
