Answer:
The maximum mass that can fall on the mattress without exceeding the maximum compression distance is 16.6 kg
Explanation:
Hi there!
Due to conservation of energy, the potential energy (PE) of the mass at a height of 3.32 m will be transformed into elastic potential energy (EPE) when it falls on the mattress:
PE = EPE
m · g · h = 1/2 k · x²
Where:
m = mass.
g = acceleration due to gravity.
h = height.
k = spring constant.
x = compression distance
The maximum compression distance is 0.1289 m, then, the maximum elastic potential energy will be the following:
EPE =1/2 k · x²
EPE = 1/2 · 65144 N/m · (0.1289 m)² = 541.2 J
Then, using the equation of gravitational potential energy:
PE = m · g · h = 541.2 J
m = 541.2 J/ g · h
m = 541.2 kg · m²/s² / (9.8 m/s² · 3.32 m)
m = 16.6 kg
The maximum mass that can fall on the mattress without exceeding the maximum compression distance is 16.6 kg.
Answer:
(A). The current in the circuit is 19.25 mA.
(B). The store energy in the inductor is 7.04 μJ.
Explanation:
Given that,
Voltage = 8.2 V
Inductor = 38 mH
Resistance = 150 Ω
Time t = 0.110 ms
The battery has negligible internal resistance, so that the total resistance in the circuit is 150 ohms. Then use this equation for current at time t in terms of inductance
We need to calculate the current
Using formula of current
Put the value into the formula
(B). We need to calculate the store energy in the inductor
Using formula of energy
Put the value into the formula
{tex]E=7.04\ \mu J[/tex]
Hence, (A). The current in the circuit is 19.25 mA.
(B). The store energy in the inductor is 7.04 μJ.
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Answer:
1069.38 gallons
Explanation:
Let V₀ = 1.07 × 10³ be the initial volume of the gasoline at temperature θ₁ = 52 °F. Let V₁ be the volume at θ₂ = 97 °F.
V₁ = V₀(1 + βΔθ) β = coefficient of volume expansion for gasoline = 9.6 × 10⁻⁴ °C⁻¹
Δθ = (5/9)(97°F -52°F) °C = 25 °C.
Let V₂ be its final volume when it cools to 52°F in the tank is
V₂ = V₁(1 - βΔθ) = V₀(1 + βΔθ)(1 - βΔθ) = V₀(1 - [βΔθ]²)
= 1.07 × 10³(1 - [9.6 × 10⁻⁴ °C⁻¹ × 25 °C]²)
= 1.07 × 10³(1 - [0.024]²)
= 1.07 × 10³(1 - 0.000576)
= 1.07 × 10³(0.999424)
= 1069.38 gallons
Molecular structure of the primary cell wall in plants. Up to three strata or layers may be found in plant cell walls: The middle lamella, a layer rich in pectins. This outermost layer forms the interface between adjacent plant cells and glues them together.
