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Gnom [1K]
3 years ago
13

What does the stomach acid have in protein digestion

Physics
1 answer:
castortr0y [4]3 years ago
5 0
It contains protease which is the enzyme that breaks down protein
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Light shines through a single slit whose width is 5.6 × 10-4 m. A diffraction pattern is formed on a flat screen located 4.0 m a
Mrac [35]

Answer:

462 nm

Explanation:

Given: width of the slit, d = 5.6 × 10⁻⁴ m

Distance of the screen, D = 4.0 m

Fringe width, β = 3.3 mm = 3.3 × 10⁻³ m

First dark fringe means n =1

Wavelength of the light, λ = ?

\beta = \frac{\lambda D}{d}\\ \Rightarrow \lambda = \frac{d \beta}{D} =\frac{5.6\times 10^{-4} \times 3.3 \times 10^{-3}}{4.0} = 4.62 \times 10^{-7}m = 462 nm

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3 years ago
In the electromagnetic spectrum, _______ is the _______ frequency color of visible light. A. green; highest B. red; lowest C. vi
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Red is the lowest because it has the shortest wavelengths
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What is the name of the system of measurement used by most countries around the world?
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Most countries besides the USA use the metric system also known as S.I (system internationale) as a system of measurement... 
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A Pitot-static tube is used to measure the velocity of helium in a pipe. The temperature and pressure are 44oF and 24 psia. A wa
leonid [27]

Answer:

V=203 ft/s

Explanation:

answer is available in word document named "attach 1" due to some technical error in maths equation. please find the attached document

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4 years ago
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What is the speed vfinal of the electron when it is 10.0 cm from charge 1?
fgiga [73]

Answer:

Two stationary positive point charges, charge 1 of magnitude 3.45 nC and charge 2 of magnitude 1.85 nC, are separated by a distance of 50.0 cm. An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges. What is the speed v(final) of the electron when it is 10.0 cm from

The answer to the question is

The speed v_{final} of the electron when it is 10.0 cm from charge Q₁

= 7.53×10⁶ m/s

Explanation:

To solve the question we have

Q₁ = 3.45 nC = 3.45 × 10⁻⁹C

Q₂ = 1.85 nC = 1.85 × 10⁻⁹ C

2·d = 50.0 cm

a = 10.0 cm

q = -1.6×10⁻¹⁹C

Also initial kinetic energy = 0 and

Initial electric potential energy = k\frac{qQ_1}{d} + k\frac{qQ_2}{d} = kq(\frac{Q_1+Q_2}{d})

Final kinetic energy due to motion = 0.5·m·v²

Final electric potential energy = k\frac{qQ_1}{a} + k\frac{qQ_2}{2d-a} = kq(\frac{Q_1}{a}+\frac{ Q_2}{2d-a})

From the energy conservation principle we have

0+ kq(\frac{Q_1+Q_2}{d})=0.5mv^2+  kq(\frac{Q_1}{a}+\frac{ Q_2}{2d-a})

Solving for v gives

v=\sqrt{\frac{kq(\frac{Q_1+Q_2}{d})-   kq(\frac{Q_1}{a}+\frac{ Q_2}{2d-a})}{0.5m}}

where k = 9.0×10⁹ and m = 9.109×10⁻³¹ kg

gives v =7528188.32769 m/s or 7.53×10⁶ m/s

v_{final} = 7.53×10⁶ m/s

6 0
4 years ago
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