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Hitman42 [59]
3 years ago
6

explain in detail how you would test a gingerbread biscuit solution for the presence of starch, sugar, and protein

Physics
1 answer:
mixas84 [53]3 years ago
5 0
To test a food for starch, you can add a few droplets of iodine to it, if the liquid changes to a blue/black color then starch is present. To test for sugar you can use Benedict's solution, which will also have a color change from blue to yellow/red/orange. Add Biuret Reagent solution to test for protein, the solution will turn a pink or purple color... Note that this may not improve the taste or color of your gingerbread biscuit though :)
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Answer:

thermal, hot equals thermal and such

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2 years ago
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What is the energy of a photon that has the same wavelength as an electron having a kinetic energy of 15 ev?
serg [7]

Answer: 6.268(10)^{-16}J

Explanation:

The kinetic energy of an electron K_{e} is given by the following equation:

K_{e}=\frac{(p_{e})^{2} }{2m_{e}}   (1)

Where:

K_{e}=15eV=2.403^{-18}J=2.403^{-18}\frac{kgm^{2}}{s^{2}}

p_{e} is the momentum of the electron

m_{e}=9.11(10)^{-31}kg  is the mass of the electron

From (1) we can find p_{e}:

p_{e}=\sqrt{2K_{e}m_{e}}    (2)

p_{e}=\sqrt{2(2.403^{-18}J)(9.11(10)^{-31}kg)}  

p_{e}=2.091(10)^{-24}\frac{kgm}{s}   (3)

Now, in order to find the wavelength of the electron \lambda_{e}   with this given kinetic energy (hence momentum), we will use the De Broglie wavelength equation:

\lambda_{e}=\frac{h}{p_{e}}    (4)

Where:

h=6.626(10)^{-34}J.s=6.626(10)^{-34}\frac{m^{2}kg}{s} is the Planck constant

So, we will use the value of p_{e} found in (3) for equation (4):

\lambda_{e}=\frac{6.626(10)^{-34}J.s}{2.091(10)^{-24}\frac{kgm}{s}}    

\lambda_{e}=3.168(10)^{-10}m    (5)

We are told the wavelength of the photon  \lambda_{p} is the same as the wavelength of the electron:

\lambda_{e}=\lambda_{p}=3.168(10)^{-10}m    (6)

Therefore we will use this wavelength to find the energy of the photon E_{p} using the following equation:

E_{p}=\frac{hc}{lambda_{p}}    (7)

Where c=3(10)^{8}m/s  is the spped of light in vacuum

E_{p}=\frac{(6.626(10)^{-34}J.s)(3(10)^{8}m/s)}{3.168(10)^{-10}m}  

Finally:

E_{p}=6.268(10)^{-16}J    

4 0
3 years ago
" A bowl of soup at 200Á F. is placed in a room of constant temperature of 60Á F. The
Dahasolnce [82]
<span>T(t)=60+140<span>e<span>−0.075t</span></span></span> <span>T(12)=60+140<span>e<span>−0.075∗12</span></span></span> <span>T(12)=60+140<span>e<span>−0.9</span></span></span> <span><span>T(12)=60+140(0.4065696597)
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3 years ago
A plucked violin string carries a traveling wave given by the equation f(x,t)=asin[b(x−ct)+ϕi], with a = 0.00580 m , b = 33.05 m
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Answer:

A) Φ = 0 , B)  T = 7.76 s

Explanation:

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           0 = sin Φ

           Φ = sin⁻¹ 0

           Φ = 0

B) the period is defined by time or when the movement begins to repeat itself

So that the sine function is repeated when the angle passes 2pi

            b (x- ct) = 2pi

If we are at a fixed point x = 0

           b c t = 2pi

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Let's calculate

            T = 2π / (33.05 245)

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3 years ago
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Lasers emit light of a certain frequency in one, precise direction. The light that a laser emits can be tuned to have a high fre
makkiz [27]

Answer:

An ultra intense laser is one with which intensities greater than 1015 W cm-2 can be achieved.

Explanation:

This intensity, which was the upper limit of lasers until the invention of the Chirped Pulse Amplification, CPA technique, is the value around which nonlinear effects on the transport of radiation in materials begin to appear.

Currently, the most powerful lasers reach intensities of the order of 1021W cm-2 and powers of Petawatts, PW, in each pulse. This range of intensities has opened the door for lasers to a multitude of disciplines and scientific areas traditionally reserved for accelerators and nuclear reactors, applying as generators of high-energy electron, ion, neutron and photon beams, without the need for expensive infrastructure.

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3 years ago
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