Answer:
The angular velocity is 15.37 rad/s
Solution:
As per the question:
![\theta = 54.6^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2054.6%5E%7B%5Ccirc%7D)
Horizontal distance, x = 30.1 m
Distance of the ball from the rotation axis is its radius, R = 1.15 m
Now,
To calculate the angular velocity:
Linear velocity, v = ![\sqrt{\frac{gx}{sin2\theta}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7Bgx%7D%7Bsin2%5Ctheta%7D%7D)
v = ![\sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B9.8%5Ctimes%2030.1%7D%7Bsin2%5Ctimes%2054.6%7D%7D)
v = ![\sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B9.8%5Ctimes%2030.1%7D%7Bsin2%5Ctimes%2054.6%7D%7D)
v = ![\sqrt{\frac{294.98}{sin109.2^{\circ}}} = 17.67\ m/s](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B294.98%7D%7Bsin109.2%5E%7B%5Ccirc%7D%7D%7D%20%3D%2017.67%5C%20m%2Fs)
Now,
The angular velocity can be calculated as:
![v = \omega R](https://tex.z-dn.net/?f=v%20%3D%20%5Comega%20R)
Thus
![\omega = \frac{v}{R} = \frac{17.67}{1.15} = 15.37\ rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Cfrac%7Bv%7D%7BR%7D%20%3D%20%5Cfrac%7B17.67%7D%7B1.15%7D%20%3D%2015.37%5C%20rad%2Fs)
Answer:
15.29m/s
Explanation:
Using the law of conservation of momentum
m1u1+m2u2 = (m1+m2)v
m1 is the mass of the clown = 85kg
u1 is the velocity of the clown = 0m/s (fired out from rest)
m2 is the mass of the cannon = 480kg
u2 is the velocity of the cannon = 18m/s
Let v me the recoil speed of the cannon
Substitute into the formula
85(0)+480(18) = (85+480)v
0+8640 = 565v
v = 8640/565
v = 15.29m/s
Hence the recoil speed of the cannon is 15.29m/s
Answer:
VB = 19.9 [m/s]
Explanation:
In order to solve these problems, we must use the principle of energy conservation. Which tells us that the energy between two points after an instant of time must be equal. This is we have two points A & B, in point A is the highest point while Point B is the lowest
Now we need to identify the types of energy at each point.
<u>For point A</u>
At this point we have two energies, the kinetic energy since the roller coaster moves at a speed of 2 [m/s], in the same way there is potential energy since the roller coaster is 20 [m] above ground level.
<u>For point B</u>
At Point B we only have kinetic energy, since it is located at zero meters with respect to the ground. In this way we can determine the velocity at this point.
And the energy is expressed by means of the following expression:
![E_{A}=E_{B}\\\frac{1}{2} *m*v_{A}^{2} +m*g*h_{A}=\frac{1}{2} *m*v_{B}^{2}](https://tex.z-dn.net/?f=E_%7BA%7D%3DE_%7BB%7D%5C%5C%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av_%7BA%7D%5E%7B2%7D%20%2Bm%2Ag%2Ah_%7BA%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av_%7BB%7D%5E%7B2%7D)
where:
m = mass = 5 [kg]
VA = velocity of the roller coaster in point A = 2[m/s]
hA = elevation of the roller coaster = 20 [m]
vB = velocity of the roller coaster in point B [m/s]
Now replacing:
![\frac{1}{2}*5*(2)^{2} +(5*9.81*20)=\frac{1}{2}*(5)*v_{B}^{2}\\991=2.5*v_{B}^{2}\\v_{B}x^{2} =991/2.5\\v_{B}=\sqrt{396.4} \\v_{B}=19.9[m/s]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%2A5%2A%282%29%5E%7B2%7D%20%2B%285%2A9.81%2A20%29%3D%5Cfrac%7B1%7D%7B2%7D%2A%285%29%2Av_%7BB%7D%5E%7B2%7D%5C%5C991%3D2.5%2Av_%7BB%7D%5E%7B2%7D%5C%5Cv_%7BB%7Dx%5E%7B2%7D%20%3D991%2F2.5%5C%5Cv_%7BB%7D%3D%5Csqrt%7B396.4%7D%20%20%20%20%20%5C%5Cv_%7BB%7D%3D19.9%5Bm%2Fs%5D)
There are a lot of examples. Obvious one is human voice.
Other examples:
- seismic waves
- sea waves
- sound waves
Let's list the given information. The frictional force, denoted as Ff, is equal to 0.200 N. We have to find the normal force, denoted as Fn. The relationship between Ff and Fn is written as:
Ff = μFn
where μ is the coefficient of friction
If there is no given data for μ, we can't solve this problem. Suppose μ = 0.5, then the normal force would be:
Fn = Ff/μ = 0.2/0.5 = 0.4 N