Let F = required force, N
Given:
d = 12 m, distance
W = 280 J, work done
By definition,
W = F*d,
therefore
(F N)*(12 m) = (280 J)
F = 280/12 = 23.33 N
Answer: The force is 23.3 N (nearest tenth)
In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,
KE1 = KE2
The kinetic energy of the system before the collision is solved below.
KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
KE1 = 6125 g cm²/s²
This value should also be equal to KE2, which can be calculated using the conditions after the collision.
KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)
The value of x from the equation is 17.16 cm/s.
Hence, the answer is 17.16 cm/s.
Answer:a computer , machine forcery,0,push
Explanation:
The point obviously is in the 3rs quadrant
So
စ= tan^-1( y/x)-180
စ= -89.7°