A force is a push or pull to an object
Answer: T= 715 N
Explanation:
The only external force (neglecting gravity) acting on the swinging mass, is the centripetal force, which. in this case, is represented by the tension in the string, so we can say:
T = mv² / r
At the moment that the mass be released, it wil continue moving in a straight line at the same tangential speed that it had just an instant before, which is the same speed included in the centripetal force expression.
So the kinetic energy will be the following:
K = 1/2 m v² = 15. 0 J
Solving for v², and replacing in the expression for T:
T = 1.9 Kg (3.97)² m²/s² / 0.042 m = 715 N
Answer:
Explanation:
Given data:
mass of sphere is 10 g
Angle between string and vertical axis is 
thickness of string 300 mm = 0.3 m
r =0.3 sin 13 = 0.067 m
q = ne
If only internal forces are doing work (no work done by external forces), then there is no change in the total amount of mechanical energy. The total mechanical energy is said to be conserved. ... In these situations, the sum of the kinetic and potential energy is everywhere the same.
Answer:
The arrow will leave the bow with a velocity of 10 m/s.
Explanation:
Hi there!
The potential energy stored in the bow can be calculated using the following equation:
U = 1/2 · k · d²
Where
U = elastic potential energy.
k = spring constant.
d = stretched distance of the bow
Then:
U = 1/2 · 112 N/m · (0.29 m)²
U = 4.7 J
When the bow is released, the potential energy is transformed into kinetic energy. Then, the kinetic energy of the arrow when it leaves the bow will be:
KE = 1/2 · m · v² = 4.7J
Where:
KE = kinetic energy.
m = mass of the arrow.
v = velocity of the arrow:
Then:
4.7 J = 1/2 ·0.094 kg · v²
2 · 4.7 J / 0.094 kg = v²
9.4 kg · m²/s² / 0.094 kg = v²
v = 10 m/s
The arrow will leave the bow with a velocity of 10 m/s.
