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3 years ago

The ideal temperature at which to study the effect of proteases on protein is 37°C. Why are higher temperatures not advisable?

Physics

2 answers:

elena-14-01-66 [18.8K]3 years ago

7 0

Answer:

The higher temperatures are not advisable because

A) This could contribute to irreversible denaturing of protein.

D) Only at the specified temperature, the protein interacts with the proteases.

Explanation:

The higher temperatures are not advisable because

A) This could contribute to irreversible denaturing of protein.

D) Only at the specified temperature, the protein interacts with the proteases.

1. The protease at a temperature higher than 37 degree is not active

2) The protein you are researching to be destroyed by protease may be denatured and the protease may not function because the protein(s) that it targets are messed up by the structure.

loris [4]3 years ago

3 0

Answer:a

Explanation:

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An electrical motor spins at a constant 1975.0 rpm. If the armature radius is 7.112 cm, what is the acceleration of the edge of

stealth61 [152]

Answer:

Option D is the correct answer.

Explanation:

Since value of angular acceleration is constant, the body has only centripetal acceleration.

Centripetal acceleration

$a=\frac{v^2}{r}=\frac{(r\omega )^2}{r}=r\omega ^2$

We have radius = 7.112 cm = 0.07112 m

Frequency, f = 1975 rpm = 32.92 rps

Angular frequency, ω = 2πf = 2 x π x 32.92 = 206.82 rad/s

Substituting in centripetal acceleration equation,

$a=r\omega ^2=0.07112\times 206.82^2=3042.17m/s^2$

Option D is the correct answer.

3 0

3 years ago

NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA

kkurt [141]

Answer:

The answer is " $q=0.0945\,C$ ".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

$U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}$

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

$U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}$

Potential energy shifts:

$= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\$

$=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J$

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

$=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5} }{ 4,228 \times10^{5}} \right ) \\\\$

$=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\$

$\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C$

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

6 0

3 years ago

a ball of mass 16 kg on the end of a string is spun at a constant speed of 2 m/s in a horizontal circle with a radius of 1m. Wha

miss Akunina [59]

The work done by the centripetal force during om complete revolution is 401.92 J.

<h3>What is centripetal force?</h3>

Centripetal force is a force that acts on a body undergoing a circular motion and is directed towards the center of the circle in which the body is moving.

To Calculate the work done by the centripetal force during one complete revolution, we use the formula below.

Formula:

W = (mv²/r)2πr
W = 2πmv²................... Equation 1

Where:

W = Work done by the centripetal force
m = mass of the ball
v = velocity of the ball
π = pie

From the question,

Given:

m = 16 kg
v = 2 m/s
π = 3.14

Substitute these values into equation 1

W = 2×3.14×16×2²

W = 401.92 J

Hence, The work done by the centripetal force during om complete revolution is 401.92 J.
Learn more about centripetal force here: brainly.com/question/20905151

5 0

2 years ago

A student attaches a rope to his

erastova [34]

The choices are:

a. Normal Force

b. Gravity Force

c. Applied Force

d. Friction Force

e. Tension Force

f. Air Resistance Force

Answer:

The answer is letter e, Tension Force.

Explanation:

Force refers to the "push" and "pull" of an object, provided that the object has mass. This results to acceleration or a change in velocity. There are many types of forces such as <em>Normal Force, Gravity Force, Applied Force, Friction Force, Tension Force and Air Resistance Force.</em>

The situation above is an example of a "tension force." This is considered the force that is being applied to an object by strings or ropes. This is a type force that allows the body to be pulled and not pushed, since ropes are not capable of it. In the situation above, the tension force of the rope is acting on the bag and this allows the bag to be pulled.

Thus, this explains the answer.

6 0

3 years ago

The diagram below shows the stars that are nearest to our solar system.

abruzzese [7]

Answer:

no u tried of the same dam

thing

Explanation:

7 0

3 years ago