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3 years ago

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Which functions of water in living systems would still be possible if water was not polar and did not form hydrogen bonds

1 answer:

saveliy_v [14]3 years ago

7 0

Answer:

Water would still be used in industrial manufacture of Hydrogen,by electrolysis

Also would still be used in photosynthesis by making hydrogen ion and free electrons also oxygen by photolysis process

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A neutron at rest decays (breaks up) to a proton and an electron. Energy is released in the decay and appears as kinetic energy

SashulF [63]

Answer:

$5.444\times 10^{-4}$

Explanation:

The momentum of the neutron before and after the decay is the same since there's no external force.

$P_{sys}=const\\\\P=mv\\\\K=0.5mv^2$

#The neutron is initially at rest, so after the decay:

$P_A+P_B=0\\\\P_A=-P_B$

#After decay, the proton has +ve direction with a velocity $v_A$ while the electron moves in a negative direction with a velocity $v_B$

Therefore:

$P_A=m_Av_A, P_B=m_Bv_B\\\\\therefore m_Av_A,=m_Bv_B$

Let the energy released during the decay be Q:

$Q=K_{tot}=K_A+K_B\\\\Q=K_A+0.5m_Bv_B^2\\\\Q=K_A+0.5m_B(\frac{m_A}{m_B})^2v_A^2\\\\\ But \ K_A=0.5m_Av_A^2\\\\\therefore Q=K_A+\frac{m_A}{m_B}K_A=K_A(1+\frac{m_A}{m_B})\\\\=Q=\frac{m_A+m_B}{m_B}K_A\\\\m_A=1836m_B\\\\\frac{K_A}{Q}=\frac{m_B}{1836m_B+m_B}=\frac{1}{1837}\\\\\frac{K_A}{Q}=5.444\times10^{-4}$

Hence,Kp/Ktot is 5.444x10^(-4)

4 1

3 years ago

A car with a mass of 710 kg is traveling at 37 km/hr. It accelerates to a speed of 120 km/hr in 12.6 seconds. What is the net fo

guajiro [1.7K]

Answer: The net force acting on the car 1,299.3 N.

Explanation:

Mass of the car = 710 kg

Initial velocity of the car of the ,u= 37 km/h= 10.27 m/s $(1km\h=\frac{5}{18} m/s)$

Final velocity of the car,v = 120 km/h = 33.33 m/s

time taken b y car = 12.6 sec

v-u=at

$33.33m/s-10.27m/s=23.06 m/s=a(12.6 sec)$

$a = 1.83 m/s^2$

$Force=mass\times acceleration$

$Force=710 kg\times 1.83 m/s^2$

$Force=1,299.3 N$

The net force acting on the car 1,299.3 N.

8 0

3 years ago

A golf ball (mass 0.045 kg) is hit with a club from a tee. The plot shows the force on the ball as a function of time in millise

almond37 [142]

Answer:

16.3 m/s

Explanation:

We are given that

Mass of golf ball=m=0.045 kg

We have to find the speed right after the hit.

$b_1=(16-1)\times 10^{-3}=15\times 10^{-3} s$

$b_2=(12-6)\times 10^{-3}=6\times 10^{-3} s$

Force,F=70 N

Change in momentum=Area of trapezium

Area of trapezium,A= $\frac{1}{2}(b_1+b_2) h$

Using the formula

$m(v-u)=\frac{1}{2}(15+6)\times 10^{-3}\times 70$

$0.045(v-u)=\frac{1}{2}(21)\times 10^{-3}\times 70$

$v-u=\frac{1}{2}\times \frac{1}{0.045}\times 21\times 10^{-3}\times 70$

$v-u=16.3$

Initial velocity of golf ball=u=0

$v-0=16.3$

$v=16.3 m/s$

5 0

3 years ago

The outer layer of the Earth is called the<br> lithosphere or crust?

Annette [7]

The outer layer of the Earth is called <span>crust!</span>

6 0

4 years ago

Read 2 more answers

When a force is applied to an object, no work is being preformed on the object unless the object____.

Temka [501]

No work is being performed on the object unless the object is moving, as work done = force x distance

8 0

3 years ago