In a physics experiment, a ball is released from rest, and it falls toward the ground. The timer was not paying attention but es
timates that it took 1.5 s for the ball to fall the last 32 m. T/I (a) Calculate the velocity of the ball when it is 32 m above the ground. [and: 14 m/s [downward]] (b) Calculate the total displacement of the ball. [and: 42 m [downward]]
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Answer:
The tank is losing
Explanation:
According to the Bernoulli’s equation:
We are being informed that both the tank and the hole is being exposed to air :
∴ P₁ = P₂
Also as the tank is voluminous ; we take the initial volume ≅ 0 ;
then can be determined as:
h₁ = 5 + 15 = 20 m;
h₂ = 15 m
as it leaves the hole at the base.
radius r = d/2 = 4/2 = 2.0 mm
(a) From the law of continuity; its equation can be expressed as:
J =
J = πr²
J =
J =
b)
How fast is the water from the hole moving just as it reaches the ground?
In order to determine that; we use the relation of the velocity from the equation of motion which says:
v² = u² + 2gh ₂
v² = 9.9² + 2×9.81×15
v² = 392.31
The velocity of how fast the water from the hole is moving just as it reaches the ground is :
Explanation:
Given that,
Mass, m = 0.08 kg
Radius of the path, r = 2.7 cm = 0.027 m
The linear acceleration of a yo-yo, a = 5.7 m/s²
We need to find the tension magnitude in the string and the angular acceleration magnitude of the yo‑yo.
(a) Tension :
The net force acting on the string is :
ma=mg-T
T=m(g-a)
Putting all the values,
T = 0.08(9.8-5.7)
= 0.328 N
(b) Angular acceleration,
The relation between the angular and linear acceleration is given by :
(c) Moment of inertia :
The net torque acting on it is, , I is the moment of inertia
Also,
So,
Hence, this is the required solution.