<span>Data:
mass =
110-g bullet
d = 0.636 m
Force =
13500 + 11000x - 25750x^2, newtons.
a) Work, W
W = ∫( F* )(dx) =∫[13500+ 11000x - 25750x^2] (dx) =
W = 13500x + 5500x^2 - 8583.33 x^3 ] from 0 to 0.636 =
W = 8602.6 joule
b) x= 1.02 m
</span><span><span>W = 13500x + 5500x^2 - 8583.33 x^3 ] from</span> 0 to 1.02
W = 10383.5
c) %
[W in b / W in a] = 10383.5 / 8602.6 = 1.21 => W in b is 21% more than work in a.
</span>
Answer:
P = 86956.52 Pa
Explanation:
Data:
- F = 800 N
- A = 0.0092 m²
- P = ?
Use the formula:
Replace and solve:
The pressure it exerts on the ground is <u>86956.52 Pascal.</u>
Greetings.
Answer:
Los primeros gimnasios registrados datan de hace más de 3000 años en la antigua Persia, donde se los conocía como zurkhaneh, áreas que fomentaban la aptitud física.
A. It must provide water to certain areas of the farm
Answer: Normal force, N = 141.64 Newton
Explanation:
All the forces acting on the system and described in free body diagram are:
1) gravitational pull in downward direction
2) Normal force in upward direction
3) External force of 40 N acting at an angle of 37° with the horizontal can be resolved in two rectangular components:
i) F Cos 37° along the horizontal plane in forward direction and
ii) F Sin 37° along the vertical plane in downward direction
Applying the Newton's second law, net forces in the vertical plane are:
Net force, f = N - (mg + F Sin 37°)
As there is no acceleration in the vertical plane hence, net force f = 0.
So,
N - (mg + F Sin 37°) = 0
Adding (mg + F Sin 37°) both the sides in above equation, we get
N = mg + F Sin 37°
N = 12
9.8 + 40
0.601 because (Sin 37° = 0.601)
N = 117.6 + 24.04
N = 141.64 Newton