Describe a ball's motion as it rolls up a slanted
1 answer:
The ball will decelerate as it moves upwards.
The magnitude of the ball's acceleration is 0.3 m/s² and it directed backwards.
The given parameters;
- initial velocity of the ball, u = 1.25 m/s
- time of motion of the ball, t = 4.22 s
As the ball rolls up the inclined plane, the velocity decreases and eventually becomes zero when the ball reaches the highest point of the plane.
Thus, the ball decelerate as it moves upwards.
The acceleration of the ball is calculate as;
<em>at the highest point on the incline plane, the final velocity </em><em> is zero</em>
Thus, the magnitude of the ball's acceleration is 0.3 m/s² and it directed backwards.
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Answer:
a) Check Explanation
b) Check Explanation
c) The rate at which water is being pumped into the tank = 2.631 cm³/min
Explanation:
Let the rate of flow of water into the tank be k cm³/min
a) The image of the conical tank is presented in the attached image
Note, the radius and height of a cone are related through the similar triangles principle.
As shown in the attached image, it is evident that
r/h = 3/10
r = 3h/10 = 0.3 h
b) The quantities given in the problem.
- Shape of the tank, conical tank, Hence volume of the tank = πr²h/3
- total height of the tank, H = 10 cm
- Radius of the tank at the top, R = D/2 = 6/2 = 3 cm
- rate at which water is leaking from the tank = 1.5 cm³/min
- water is being pumped into the tank at constant rate of k cm³/min
- As at height of water, h = 2 cm, the rate of rise in water level = 1 cm/min
c) volume of the tank at any time = πr²h/3
Rate of change in the volume of water in the tank = (rate of flow into the tank) - (Rate of water flow out of the tank)
dV/dt = k - 1.5
V = πr²h/3 and r = 0.3 h, r² = 0.09 h²
V = 0.03πh³
dV/dt = (dV/dh) × (dh/dt)
dV/dh = 0.09π h²
dV/dt = 0.09π h² (dh/dt)
dV/dt = k - 1.5
0.09π h² (dh/dt) = k - 1.5
But at h = 2 cm, (dh/dt) = 1.0 cm/min
0.09π h² (dh/dt) = k - 1.5
0.09π 2² (1) = k - 1.5
k - 1.5 = 1.131
k = 1.5 + 1.131 = 2.631 cm³/min
Answer:
dJ = 1.7 m
Explanation:
The Equation of the Balancing the moments in the center of the seesaw is like this:
∑Mo = 0
Mo = F*d
Where:
∑Mo : Algebraic sum of moments in the center(o) of the balance
Mo : moment in the o point ( N*m)
F : Force ( N)
d : distancia of the force to the the o point ( N*m)
Data
mA = 60 kg : mass of the Anna
mJ = 70 kg : mass of theJon
dA = 2 m : Distance from Anna to the center of the seesaw
g: acceleration due to gravity
Calculation of the distance from Jon to the center of the seesaw (dJ)
∑Mo = 0 WA : Ana's weight , WJ : Jon's weight
W = m*g
(WA)(dA) - (WJ) (dJ) = 0
(mA*g)(dA) - (mJ*g)(dJ) = 0
We divide by g the equation:
(mA)(dA) - (mJ)(dJ)= 0
(mA)(dA) = (mJ)(dJ)
dJ = 1.7 m