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3 years ago

How can we prove air in water

2 answers:

6 0

We can show that air is dissolved in water with the help of following activity: Take some water in a glass or metal container like pan and heat it. Just before water begins to boil, you will notice some bubbles at the inner surface of the pan. These bubbles come from the air dissolved in water.

Reil [10]3 years ago

5 0

You can heat water and boil it

You might be interested in

Kyle is flying a helicopter and is rising at 5.0m/s when he releases the bag. After 2.0s

ch4aika [34]

1) v=u+gt = -5 +10*2=15 m/s
2) r=ut+gt^2/2=-5*2+10*2^2/2=-10+20=10m
3) r of heli= vt =5*2=10m
So 10m by bag from answer number 2 plus additional 10m by helicopter equals 20m

5 0

3 years ago

Read 2 more answers

It has been suggested that rotating cylinders about 9 mi long and 5.9 mi in diameter be placed in space and used as colonies. Wh

mezya [45]

Answer:

$\omega = 4.5\times 10^{-2} rad/s$

Explanation:

Given data:

Rotating cylinder length = 9 mi

diameter of cylinder is 5.9 mi

we know that linear acceleration is given as

a = r ω^2

where ω is angular velocity

so $\omega = \sqrt{\frac{a}{r}}$

$r = \frac{5.9}[2} \frac{1609 m}{1 mi} = 4.746\times 10^{3} m$

$\omega = \sqrt{\frac{9.80}{4.746\times 10^{3}}}$

$\omega = 4.5\times 10^{-2} rad/s$

7 0

2 years ago

6. And what's up with ChiN?

n200080 [17]

I don’t know what’s up with chin man ??

4 0

2 years ago

A railroad car of mass 2.50*10^4 kg is moving with a speed of 4.00 m/s. It collides and couples with three other coupled railroa

NemiM [27]

Answer:

(a) 2.5 m/s

(b) 37.5 KJ

Explanation:

(a)

From the law of conservation of momentum, Initial momentum=Final momentum

$mV_1+3mV_2=(m+3m)V_f=4mV_f$

$V_1+3V_2=4V_f$ and making $V_f$ the subject then

$V_f=\frac {V_1+3V_2}{4}$ and since $V_1$ is initial velocity of car, value given as 4 m/s, $V_2$ is the initial velocity of the three cars stuck together, value given as 2 m/s and $v_f$ is the final velocity which is unknown. By substitution

$V_f=\frac {4+(3\times2)}{4}=2.5 m/s$

(b)

Initial kinetic energy is given by

$\frac {mV_1^{2}}{2}+\frac {3mV_2^{2}}{2}=\frac {m(V_1^{2}+3V_2^{2}}{2}=\frac {2.5\times 10^{4}(4^{2}+3(2^{2}))}{2}=350\times10^{3} J= 350 KJ$

Final kinetic energy is given by

$\frac {4mV_f^{2}}{2}=\frac {4\times 2.5\times 10^{4}\times 2.5^{2}}{2}=312.5\times 10^{3} J=312.5 KJ$

The energy lost is given by subtracting the final kinetic energy from the initial kinetic energy hence

Energy lost=350-312.5=37.5 KJ

6 0

3 years ago

Read 2 more answers

Consider one such cell where the magnitude of the potential difference is 65 mV, and the inner surface of the membrane is at a h

Gelneren [198K]

Answer: W = $1.04.10^{-20}$ J

Explanation: Since the potassium ion is at the outside membrane of a cell and the potential here is lower than the potential inside the cell, the transport will need work to happen.

The work to transport an ion from a lower potential side to a higher potential side is calculated by

$W=q.\Delta V$