Position: x = 18t y = 4t - 4.9t²
First derivative: x' = 18 y' = 4 - 9.8t
Second derivative: x'' = 0 y'' = - 9.8
Position vector: P = (18t) i + (4t - 4.9t²) j
Velocity vector: V = (18) i + (4 - 9.8t) j
Acceleration vector A = (- 9.8) j
Answer:
The minumum speed the pail must have at its highest point if no water is to spill from it
= 2.64 m/s
Explanation:
Working with the forces acting on the water in the pail at any point.
The weight of water is always directed downwards.
The normal force exerted on the water by the pail is always directed towards the centre of the circle of the circular motion.
And the centripetal force, which keeps the system in its circular motion, is the net force as a result of those two previously mentioned force.
At the highest point of the motion, the top of the vertical circle, the weight and the normal force on the water are both directed downwards.
Net force = W + (normal force)
But the speed of this motion can be lowered enough to a point where the normal force becomes zero at the moment the pail reaches the highest point of its motion. Any speed lower than this value would result in the water spilling out of the pail. The water would not be able to resist the force of gravity.
At this point of minimum velocity,
Normal force = 0
Net force = W
Net force = centripetal force = (mv²/r)
W = mg
(mv²/r) = mg
r = 0.710 m
g = 9.8 m/s²
v² = gr = 9.8 × 0.71 = 6.958
v = √(6.958) = 2.64 m/s
Hope this Helps!!!
<span>So we want to know what statement is an accurate description of vibrations. So humans can hear sound frequencies from 20-20000 Hz. Below 20 Hz is infra sound and above 20000 Hz is ultra sound. Humans cant hear both infra sound and ultra sound so the correct answer is A.</span>
Answer:
The acceleration of
is ![a = 0.7156 m/s^2](https://tex.z-dn.net/?f=a%20%3D%20%200.7156%20m%2Fs%5E2)
Explanation:
From the question we are told that
The mass of first block is ![M_1 = 2.25 \ kg](https://tex.z-dn.net/?f=M_1%20%3D%20%202.25%20%5C%20kg)
The angle of inclination of first block is ![\theta _1 = 43.5^o](https://tex.z-dn.net/?f=%5Ctheta%20_1%20%3D%20%2043.5%5Eo)
The coefficient of kinetic friction of the first block is ![\mu_1 = 0.205](https://tex.z-dn.net/?f=%5Cmu_1%20%20%3D%200.205)
The mass of the second block is ![M_2 = 5.45 \ kg](https://tex.z-dn.net/?f=M_2%20%3D%205.45%20%5C%20kg)
The angle of inclination of the second block is ![\theta _2 = 32.5^o](https://tex.z-dn.net/?f=%5Ctheta%20_2%20%3D%20%2032.5%5Eo)
The coefficient of kinetic friction of the second block is ![\mu _2 = 0.105](https://tex.z-dn.net/?f=%5Cmu%20_2%20%3D%200.105)
The acceleration of
are same
The force acting on the mass
is mathematically represented as
![F_1 = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1](https://tex.z-dn.net/?f=F_1%20%3D%20T%20-%20%20M_1gsin%20%5Ctheta_1%20-%20%5Cmu_1%20M_1%20g%20cos%5Ctheta_1)
=> ![M_1 a = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1](https://tex.z-dn.net/?f=M_1%20a%20%3D%20T%20-%20%20M_1gsin%20%5Ctheta_1%20-%20%5Cmu_1%20M_1%20g%20cos%5Ctheta_1)
Where T is the tension on the rope
The force acting on the mass
is mathematically represented as
![F_2 = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2](https://tex.z-dn.net/?f=F_2%20%3D%20%20M_2gsin%20%5Ctheta_2%20-%20T%20-%5Cmu_2%20M_2%20g%20cos%5Ctheta_2)
![M_2 a = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2](https://tex.z-dn.net/?f=M_2%20a%20%3D%20%20M_2gsin%20%5Ctheta_2%20-%20T%20-%5Cmu_2%20M_2%20g%20cos%5Ctheta_2)
At equilibrium
![F_1 = F_2](https://tex.z-dn.net/?f=F_1%20%3D%20%20F_2)
So
![T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2](https://tex.z-dn.net/?f=T%20-%20%20M_1gsin%20%5Ctheta_1%20-%20%5Cmu_1%20M_1%20g%20cos%5Ctheta_1%20%3DM_2gsin%20%5Ctheta_2%20-%20T%20-%5Cmu_2%20M_2%20g%20cos%5Ctheta_2)
making a the subject of the formula
![a = \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}](https://tex.z-dn.net/?f=a%20%3D%20%20%5Cfrac%7BM_2%20g%20sin%20%5Ctheta_2%20-%20M_1%20g%20sin%20%5Ctheta_1%20-%20%5Cmu_1%20M_1g%20cos%20%5Ctheta%20-%20%5Cmu_2%20M_2%20g%20cos%20%5Ctheta_2%20%7D%7BM_1%20%2BM_2%7D)
substituting values ![a = \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}](https://tex.z-dn.net/?f=a%20%3D%20%20%5Cfrac%7B%285.45%29%20%289.8%29%20sin%20%2832.5%29%20-%20%282.25%29%20%289.8%29%20sin%20%2843.5%29%20-%20%280.205%29%2A%282.25%29%20%2A9.8cos%20%2843.5%29%20-%20%280.105%29%2A%285.45%29%20%2A%289.8%29%20cos%2832.5%29%20%7D%7B2.25%20%2B5.45%7D)
=> ![a = 0.7156 m/s^2](https://tex.z-dn.net/?f=a%20%3D%20%200.7156%20m%2Fs%5E2)