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Illusion
1 month ago
9

The stomach lining is made up of deep muscular grooves.How might these structures help the stomach to break down food?

Physics
1 answer:
Sav [3.1K]1 month ago
7 0
I sincerely hope this information proves useful to you.

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In the reaction at Blood Falls, iron and oxygen combine to form iron oxide, which is called rust (water is also present). The re
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A reactant is any substance that takes part in a chemical reaction. Conversely, a product is what emerges from the chemical transformation. A familiar example of a chemical change is rust formation. In this reaction, oxygen and iron, which serve as the reactants, react to produce a substance known as iron oxide, or rust.
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Marla says that only one person was really responsible for the theory of planetary motion. Do you agree with her? Why or why not
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(this is my original response that was accepted)
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Which type radiation can be observed well from Earth's surface?
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As computer structures get smaller and smaller, quantum rules start to create difficulties. Suppose electrons move through a cha
Softa [3030]

Answer:

a) ∆x∆v = 5.78*10^-5

∆v = 1157.08 m/s

b) 4.32*10^{-11}

Explanation:

This problem can be addressed using Heisenberg's uncertainty principle, which is expressed as:

\Delta x\Delta p \geq \frac{\hbar}{2}

Where h represents Planck’s constant (6.62*10^-34 J s).

Assuming that the electron's mass remains the same, we proceed as follows:

\Delta x \Delta (m_ev)=m_e\Delta x\Delta v \geq \frac{\hbar}{2}

Utilizing the electron's mass (9.61*10^-31 kg) and the uncertainty in position (50 nm), we can compute ∆x∆v and ∆v:

\Delta x \Delta v\geq\frac{\hbar}{2m_e}=\frac{(1.055*10^{-34}Js)}{2(9.1*10^{-31}kg)}=5.78*10^{-5}\ m^2/s

\Delta v\geq\frac{5.78*10^{-5}}{50*10^{-9}m}=1157.08\frac{m}{s}

If we treat the electron like a classic particle, the time required to cross the channel is determined using the upper limit of the uncertainty in velocity:

t=\frac{x}{v}=\frac{50*10^{-9}m}{1157.08m/s}=4.32*10^{-11}s

6 0
2 months ago
A specimen of steel has a rectangular cross section 20 mm wide and 40 mm thick, an elastic modulus of 207 GPa, and a Poisson’s r
Softa [3030]

Response: The width decreases by 2.18 × 10^(-6) m

Clarification:

Given data;

Shear Modulus; E = 207 GPa = 207 × 10^(9) N/m²

Force; F = 60000 N.

Poisson’s ratio; υ = 0.30

The initial width is 20 mm, and the thickness is 40 mm.

Area = 20 × 10^(-3) × 40 × 10^(-3)

Area = 8 × 10^(-4) m²

The formula for shear modulus is;

E = σ/ε_z

where σ represents stress calculated as Force(F)/Area(A)

while ε_z stands for longitudinal strain.

Thus;

E = (F/A)/ε_z

ε_z = (F/A)/E

ε_z = (60,000/(8 × 10^(-4)))/(207 × 10^(9))

ε_z = 3.62 × 10^(-4)

Next, the lateral strain is given by;

ε_x = - υ × ε_z

ε_x = -0.3 × 3.62 × 10^(-4)

ε_x = -1.09 × 10^(-4)

The change in width can be determined as;

Δw = w_o × ε_x

Where w_o denotes the original width = 20 × 10^(-3) m

So; Δw = 20 × 10^(-3) × -1.09 × 10^(-4)

Δw = -2.18 × 10^(-6) m

A negative sign indicates a reduction in width.

Therefore, the width decreases by 2.18 × 10^(-6) m

6 0
2 months ago
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