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3 years ago

How do you inhibitors work

Physics

1 answer:

bonufazy [111]3 years ago

6 0

Answer:

By binding to enzymes' active sites, inhibitors reduce the compatibility of substrate and enzyme and this leads to the inhibition of Enzyme-Substrate complexes' formation, preventing the catalyzation of reactions and decreasing (at times to zero) the amount of product produced by a reaction.

Explanation:

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All organisms in Kingdom Animalia are multicellular, meaning their bodies are composed of more than one cell. Which other charac

k0ka [10]

Well,

For the first one, the answer would be C, because all organisms in Kingdom Animalia must eat in order to survive.

For the second one, all of the options are in Kingdom Animalia, but worms (A) and clams (C) are invertebrates. So that leaves options B and D.

4 0

2 years ago

Read 2 more answers

(I) A novice skier, starting from rest, slides down an icy frictionless 8.0° incline whose vertical height is 105 m. How fast is

Vlad1618 [11]

Answer:

v = 45.37 m/s

Explanation:

Given,

angle of inclination = 8.0°

Vertical height, H = 105 m

Initial K.E. = 0 J

Initial P.E. = m g H

Final PE = 0 J

Final KE = $\dfrac{1}{2}mv^2$

Using Conservation of energy

$KE_i + PE_i + KE_f + PE_f$

$0 + m g H = \dfrac{1}{2}mv^2 + 0$

$v = \sqrt{2gH}$

$v = \sqrt{2\times 9.8 \times 105}$

v = 45.37 m/s

Hence, speed of the skier at the bottom is equal to v = 45.37 m/s

3 0

2 years ago

The Moon requires about 1 month (0.08 year) to orbit Earth. Its distance from us is about 400,000 km (0.0027 AU). Use Kepler’s t

dem82 [27]

Answer:

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

Explanation:

As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula

$T = 2\pi \sqrt{\frac{r^3}{GM}}$

now for the time period of moon around the earth we can say

$T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}$

here we know that

$T_1 = 0.08 year$

$r_1 = 0.0027 AU$

$M_e$ = mass of earth

Now if the same formula is used for revolution of Earth around the sun

$T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}$

here we know that

$r_2 = 1 AU$

$T_2 = 1 year$

$M_s$ = mass of Sun

now we have

$\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}$

$\frac{1}{0.08} = \sqrt{\frac{1 M_e}{(0.0027)^3M_s}}$

$12.5 = \sqrt{(5.08 \times 10^7)\frac{M_e}{M_s}}$

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

4 0

2 years ago

A projectile is fired with a velocity of 22 m/s at an angle of 25°. What is the vertical component of the velocity?

7nadin3 [17]

Answer:

Vertical component of velocity is 9.29 m/s

Explanation:

Given that,

Velocity of projection of a projectile, v = 22 m/s

It is fired at an angle of 22°

The horizontal component of velocity is v cosθ

The vertical component of velocity is v sinθ

So, vertical component is given by :

$v_y=v\ sin(25)$

$v_y=22\ m/s\times\ sin(25)$

$v_y=9.29\ m/s$

Hence, the vertical component of the velocity is 9.29 m/s

3 0

3 years ago

A man 2 m tall walks horizontally at a constant rate of 1 m/s toward the base of a tower 23 m tall. When the man is 10 m from th

Evgen [1.6K]

Answer:

$\dfrac{d\theta}{dt}=0.038\ rad/s$

Explanation:

Given that

$\dfrac{dx}{dt}= -1\ m/s$

From the diagram

$tan\theta=\dfrac{21}{x}$

By differentiating with time t

$sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}$

When x= 10 m

$tan\theta=\dfrac{21}{10}$

θ = 64.53°

Now by putting the value in equation

$sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}$

$sec^264.53^{\circ} \dfrac{d\theta}{dt}=-\dfrac{21}{10^2}\times (-1)$

$\dfrac{d\theta}{dt}=0.038\ rad/s$

Therefore rate of change in the angle is 0.038\ rad/s

8 0

2 years ago