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2 years ago

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A piston-cylinder device contains 0.005 m3 of liquid water and 0.9 m3 of water vapor in equilibrium at 600 kPa. Heat is transfer

red at constant pressure until the temperature reaches 200°C. Determine (a) the initial temperature of the water; (b) the total mass of the water; (c) the final volume; and (d) show the process on a P-? diagram with respect to saturation lines.

1 answer:

olga55 [171]2 years ago

5 0

Answer:

a) 158.83 Degree Celcius

b) 7.39 Kg

c) 2.6 cubic metre

Explanation:

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you want to compare brands of paper towels to see which holds the most liquid. the independent variable in your experiment would

Kazeer [188]

Answer: the brand of paper towel

Explanation: the independent variable is the one you control in an experiment. the dependent variable would be the amount of water in the paper towel

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2 years ago

A diver shines light up to the surface of a flat glass-bottomed boat at an angle of 30° relative to the normal. If the index of

son4ous [18]

Answer:

<h2>35</h2>

Explanation:

According to snell's law which states that the ratio of the sin of incidence (i) to the angle of refraction(n) is a constant for a given pair of media.

sini/sinr = n

n is the constant = refractive index

Since the diver shines light up to the surface of a flat glass-bottomed boat, the refractive index n = nw/ng

nw is the refractive index of water and ng is that of glass

sini/sinr = nw/ng

given i = 30°, nw = 1.33, ng = 1.5, r = angle the light leave the glass

On substitution;

sin 30/sinr = 1.33/1.5

1.5sin30 = 1.33sinr

sinr = 1.5sin30/1.33

sinr = 0.75/1.33

sinr = 0.5639

r = arcsin0.5639

r ≈35°

angle the light leave the glass is 35°

7 0

2 years ago

A 0.900 kg ornament is hanging by a 1.50 m wire when the ornament is suddenly hit by a 0.400 kg missile traveling horizontally a

just olya [345]

Explanation:

The given data is as follows.

Mass of the ornament ( $m_{1}$ ) = 0.9 kg

Length of the wire (l) = 1.5 m

Mass of missile ( $m_{2}$ ) = 0.4 kg

Initial speed of missile ( $u_{2}$ ) = 12 m/s

r = 1.5 m

According to the law of conservation of momentum,

$p_{i} = p_{f}$

$m_{1}u_{1} + m_{2}u_{2} = (m_{1} + m_{2})v$

Putting the given values into the above formula as follows.

$m_{1}u_{1} + m_{2}u_{2} = (m_{1} + m_{2})v$

$0.9 \times 0 + 0.4 \times 12 = (0.9 + 0.4)v$

0 + 4.8 = 1.3v

v = 3.69 m/s

Now, the centrifugal force produced is calculated as follows.

$F_{c} = (m_{1} + m_{2}) \times \frac{v^{2}}{r}$

= $(0.9 + 0.4) \times \frac{(3.69)^{2}}{1.5}$

= 11.80 N

Hence, tension in the wire is calculated as follows.

T = $F_{c} + (m_{1} + m_{2})g$

= $11.80 N + (0.9 + 0.4) \times 9.8$

= 24.54 N

Thus, we can conclude that tension in the wire immediately after the collision is 24.54 N.

4 0

2 years ago

What mass of ice (in g) can be melted if 27.2 kJ of thermal energy are added at the freezing point? Use molar mass = 18.02 g/mol

san4es73 [151]

Answer : The mass of ice melted can be, 3.98 grams.

Explanation :

First we have to calculate the moles of ice.

$Q=\frac{\Delta H}{n}$

where,

Q = energy absorbed = 27.2 kJ

$\Delta H$ = enthalpy of fusion of ice = 6.01 kJ/mol

n = moles = ?

Now put all the given values in the above expression, we get:

$27.2kJ=\frac{6.01kJ/mol}{n}$

$n=0.221mol$

Now we have to calculate the mass of ice.

$\text{Mass of ice}=\text{Moles of ice}\times \text{Molar mass of ice}$

Molar mass of ice = 18.02 g/mol

$\text{Mass of ice}=0.221mol\times 18.02g/mol=3.98g$

Thus, the mass of ice melted can be, 3.98 grams.

3 0

2 years ago

A 5.0-kilogram box is sliding across a level floor. The box is acted upon by a force of 27 newtons east and a frictional force o

marshall27 [118]

Answer:

The magnitude of the acceleration of the box is 2 m/s².

Explanation:

Given:

Mass of the box, $m=5.0$ kg

Force acting towards east, $F=27$ N

Frictional force acting towards west, $f=17$ N

Let the acceleration be $a$ m/s².

Now, net force acting on the box towards east is given as:

$F_{net}=F-f=27-17=10\textrm{ N}$

From Newton's second law of motion,

$F_{net}=ma\\10=5.0\times a\\a=\frac{10}{5.0}=2\textrm{ }m/s^2$

Therefore, the magnitude of the acceleration of the box is 2 m/s².

6 0

2 years ago

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