A 2.98-kg object oscillates on a spring with an amplitude of 8.05 cm. Its maximum acceleration is 3.55 m/s2. Calculate the total
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Answer:
F_scale = 20.18 N
Explanation:
The scale reading corresponds to two factors, the first the weight of the water in the container and the second the force of the liquid that is falling at the moment of reading.
* Let's find the amount of liquid in the container for a time of t = 2.93 s
Let's use a direct proportion rule. If 0.373 l falls in one second at t = 2.93 s, how many liters are there
V_{water} = 2.93 s (0.373 l / 1s) = 1.09 l
V_{water} = 1.09 10⁻³ m³
the amount of water is
ρ = m / V
m = ρ V
m = 1000 1.09 10⁻³
m = 1.09 kg
so the weight of the liquid in the container for this time is
W = mg
W = 1.09 9.8
W = 10.68 N
* Let's look for the force of the falling jet
Let's use Bernoulli's equation, where the subscript 1 is for the container and the subscript 2 is for the water at a height h
P₁ + 1/2 ρ g v₁² + ρ g y₁ = P₂ + 1/2 ρ g v₂² + ρ g y₂
In this case, the water falls freely, so the external pressure is atmospheric.
P₂ = P_{atm}
since they indicate that the water falls, we assume that its initial velocity is zero v₂ = 0
let's use kinematics to find the speed of a drop when it reaches the container y = 0
v² = v₀² - 2 g (y-y₀)
v =
let's calculate
v = √(2 9.8 40.5)
v = 28.17 m / s
this is the speed in the container v₁ = 28.17 m / s
the height from where it falls is y₂ = 40.5 and reaches the container y₁ = 0
we substitute in Bernoulli's equation
P₁ +1/2 ρ g v₁² + 0 = P_{atm} + 0 + ρ g y₂
P₁ + ½ ρ g v₁² = P_{atm} + ρ g y₂
P₁ = P_{atm} + ρ g y₂ - ½ ρ g v₁²
P₁ = 1 10⁵ + 1000 9.8 40.5 - ½ 1000 28.17²
P₁ = 1 10⁵ + 3.97 10⁵ - 3.69 10⁵
P₁ = 1.28 10⁵ Pa
The definition of Pressure is
P = F / A
F = P A
We must suppose a time to carry out the reading suppose an average time of the modern equipment t = 0.1 s, in this time how much is now arriving
m₂ = 0.373 0.2 = 0.0746 l = 0.0746 10⁻³ m³
the volume is V = A l
if the length of l = 1 m
A = 0.0746 10⁻³ m³ = 7.45 10⁻⁵ m²
the force of this jet is
F = P A
F = 1.28 10⁵ 7.46 10⁻⁵
F = 9.5 N
with these data let's use the equilibrium equation
F_ scale -W - F = 0
F_scale = W + F
F_scale = 10.682 + 9.5
F_scale = 20.18 N
The problem is solved and the questions are answered below.
Explanation:
a. To calculate the speed of the 0.66 kg ball just before the collision
V₀ + K₀ = V₁ + K₁
= mgh₀ = 1/2 mv₁²
where, h= r - r cosθ
V =
V = 2.42 m/s
b. Calculate the speed of the 0.22 kg ball immediately after the collision
y = y₀ + Vy₀t - 1/2 gt²
0 = 1.2 - 1/2 gt²
t = 0.495 s
x = x₀ + Vx₀t
1.4 = 0 + vx₀ (0.495)
Vx₀ = 2.83 m/s
C. To Calculate the speed of the 0.66 kg ball immediately after the collision
m₁ v₁ = m₁ v₃ + m₂ v₄
(0.66)(2.42) = (0.66) v₃ + (0.22)(2.83)
V₃ = 1.48 m/s
D. To Indicate the direction of motion of the 0.66 kg ball immediately after the collision is to the right.
E. To Calculate the height to which the 0.66 kg ball rises after the collision
V₀ + k₀ = V₁ + k₁
1/2 mv₀² = mgh₁
h₁ = v₀²/2 g
= 0.112 m
F. Based on your data, No the collision is not elastic.
Δk = 1/2 m₁v₃² =1/2 m₂v₄² - 1/2 m₁v₁²
= 1/2 (0.66)(1.48)² + 1/2 (0.22)(2.83)² - 1/2 (0.66)(2.42)²
= - 0.329 J
Hence, kinetic energy is not conserved.