Answer: 14.16
Explanation:
Given
d = 38cm
r = d/2 = 38/2 = 19cm = 0.19m
K.E = 510J
m = 10kg
I = 1/2mr²
I = 1/2*10*0.19²
I = 0.18kgm²
When it has 510J of Kinetic Energy then,
510J = 1/2Iω²
ω² = 1020/I
ω² = 1020/0.18
ω² = 5666.67
ω = √5666.67 = 75.28 rad/s
Velocity is the block, v = ωr
V = 75.28 * 0.19
V = 14.30m/s
The "effective mass" M of the system is
M = (14.0 + ½*10.0) kg = 19.0 kg
The motive force would be
F = ma
F = 14 * 9.8
F = 137.2N
so that the acceleration would be
a = F/m
a = 137.2/19
a = 7.22m/s²
Finally, using equation of motion.
V² = u² + 2as
14.3² = 0 + 2*7.22*s
204.49 = 14.44s
s = 204.49/14.44
s = 14.16m
<span>vf^2 = vi^2 + 2*a*d
---
vf = velocity final
vi = velocity initial
a = acceleration
d = distance
---
since the airplane is decelerating to zero, vf = 0
---
0 = 55*55 + 2*(-2.5)*d
d = (-55*55)/(2*(-2.5))
d = 605 meters
</span>
Answer:
Fy = 14.3 [N]
Explanation:
To be able to solve this problem we must know that the force is a vector and has magnitude and direction, therefore it can be decomposed into the force in the X & y components:
When we have the components on the horizontal and vertical axes we must use the Pythagorean theorem.
where:
F = 15 [N]
Fx = horizontal component = 4.5 [N]
Fy = vertical component [N]
![15=\sqrt{4.5^{2}+F_{y}^{2}}\\ 15^{2}= (\sqrt{4.5^{2}+F_{y}^{2}})^{2} \\225 = 4.5^{2}+F_{y} ^{2}\\ F_{y}^{2} =225 -4.5^{2}\\ F_{y}^{2}=204.75\\F_{y}=\sqrt{204.75}\\ F_{y}=14.3 [N]](https://tex.z-dn.net/?f=15%3D%5Csqrt%7B4.5%5E%7B2%7D%2BF_%7By%7D%5E%7B2%7D%7D%5C%5C%2015%5E%7B2%7D%3D%20%28%5Csqrt%7B4.5%5E%7B2%7D%2BF_%7By%7D%5E%7B2%7D%7D%29%5E%7B2%7D%20%5C%5C225%20%3D%204.5%5E%7B2%7D%2BF_%7By%7D%20%5E%7B2%7D%5C%5C%20%20F_%7By%7D%5E%7B2%7D%20%3D225%20-4.5%5E%7B2%7D%5C%5C%20F_%7By%7D%5E%7B2%7D%3D204.75%5C%5CF_%7By%7D%3D%5Csqrt%7B204.75%7D%5C%5C%20%20F_%7By%7D%3D14.3%20%5BN%5D)
Ball thrown into the air at an angle.
