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3 years ago

At what phase or phases could water exist at 4.579mm pressure and 0.0098°C?

1 answer:

7 0

The correct answer is A. The water at this conditions is in the solid phase. This can be verified by using a phase diagram. A phase diagram is a figure which shows the phases of a certain substance at a specific temperature and pressure.

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A river 800m wide flows at the rate of 5km/h . A swimmer who can swim at 10km/h in still water wants to cross the river straight

LuckyWell [14K]

Answer:

At an angle of $30^{\circ}$

Explanation:

Assume the river flows from East to West so for the swimmer to cross across it, assume he crosses it from West to East.

The resultant speed will be given by

$R= \sqrt {10^{2}-5^{2}=\sqrt {75}\approx 8.66 km/h\\Direction=sin^{-1}\frac {5}{10}\approx 30^{\circ}$

6 0

2 years ago

Susan gently pushes the tip of her finger against the eraser on her pencil and the pencil does not move. Which of the

pochemuha

Answer: D. ➡️⬅️

Explanation: I just knew the answer ;)

6 0

2 years ago

A 65.0-kg runner has a speed of 5.20 m/s at one instant dur- ing a long-distance event. (a) What is the runner’s kinetic energy

vladimir2022 [97]

Answer:

a)KE=878.8 J

b)W=2636.4 J

Explanation:

Given that

mass ,m = 65 kg

Initial speed ,u = 5.2 m/s

We know that kinetic energy KE is given as follows

$KE=\dfrac{1}{2}mu^2$

m=mass

u=velocity

Now by putting the values in the above equation we get

$KE=\dfrac{1}{2}\times 65\times 5.2^2\ J$

KE=878.8 J

We know that

Work done by all forces = Change in the kinetic energy

The final velocity , v= 2 u = 2 x 5.2 m/s

v= 10.4 m/s

$W=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2$

Now by putting the values in the above equation we get

$W=\dfrac{1}{2}\times 65\times 10.4^2-\dfrac{1}{2}\times 65\times 5.2^2\ J$

W=2636.4 J

a)KE=878.8 J

b)W=2636.4 J

8 0

3 years ago

Please help me with this question, especially the angle part. I have solved the initial velocity already. I don't know what to d

Helga [31]

I think you almost got it.

At the top, the velocity only has horizontal component, so v=12 m/s is v_x, which is v*cos(theta), because v_x is constant, so the same when it was launched or now.

With the value of the initial speed (28 m/s, which is the total speed), you can set

v_x = v * cos( theta ) ---> 12 = 28*cos(theta) --> cos(theta)=12/28=3/7

or theta = 64.62 deg, it is D. Think about it. I hope you see it.

5 0

3 years ago

Which of the following statements about digestion is/are TRUE? (Select all that apply.)

MArishka [77]

1st one and fourth one are true.

7 0

3 years ago

Read 2 more answers