All categories

3 years ago

Which interactions can contribute to the intrinsic binding energy during enzymatic catalysis?

Physics

1 answer:

Pie3 years ago

4 0

Answer:electrostatic interactions, hydrogen bonding, van der waals interactions.

Explanation:the above listed are the interaction that that contribute

You might be interested in

A 1400 kg car starts from rest on a horizontal road and gains a speed of 61 km/h in 19 s. (a) what is its kinetic energy at the

lana [24]

(a) Let's convert the final speed of the car in m/s:
$v_f = 61 km/h = 16.9 m/s$
The kinetic energy of the car at t=19 s is
$K= \frac{1}{2}mv_f^2= \frac{1}{2}(1400 kg)(16.9 m/s)^2=2.00 \cdot 10^5 J$

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
$P= \frac{W}{\Delta t}$
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
$P= \frac{K}{\Delta t}= \frac{2.00 \cdot 10^5 J}{19 s}=1.05 \cdot 10^4 W$

(c) The instantaneous power is given by
$P_i = Fv_f$
where F is the force exerted by the engine, equal to F=ma.

So we need to find the acceleration first:
$a= \frac{v_f-v_i}{\Delta t}= \frac{16.9 m/s}{19 s}=0.89 m/s^2$
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
$P_i = Fv=(ma)v=(1400 kg)(0.89 m/s^2)(16.9 m/s)=2.11 \cdot 10^4 W$

5 0

3 years ago

How much tension must a rope withstand if it is used to accelerate a 6526 kg car vertically upwards at 8.9 m/s^2?

ExtremeBDS [4]

Answer:

The value is $T = 122036.2 \ N$

Explanation:

From the question we are told that

The mass of the car is $m = 6526 \ kg$

The acceleration is $a= 8.9 \ m/s^2$

Generally the net force applied on the rope is mathematically represented as

$F_{net} = T - W$

Here W is the weight of the car which is evaluated as

$W = m * g$

=> $W = 6526 * 9.8$

=> $W = 63954.8 \ N$

Generally the net force can also be mathematically represented as

$F = m * a$

$m * a = T - 63954.8$

=> $6526 * 8.9 = T - 63954.8$

=> $T = 122036.2 \ N$

7 0

2 years ago

A physicist found that a force of 1.32 N was measured between two charged spheres. The distance between the spheres was 95 cm. C

labwork [276]

The electric force (and the gravitational force too) is inversely proportional
to the square of the distance between the objects involved.

In this question, the distance is increased by a factor of (1.25/0.95) .

So the electric force will change by the factor of (0.95/1.25)² .

The new force is

(1.32 N) · (0.95/1.25)² = 0.762... newton (rounded)

5 0

2 years ago

If two equal charges are separated by a certain distance, the force of repulsion is F. Given F1, if each charge in F1 is doubled

Alisiya [41]

Answer: $F_2=16\times F_1$