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3 years ago

11

A sample of nitrogen gas is inside a sealed container. The volume of the container decreases while the temperature is kept const

ant. This is a ________ process.
a) isothermal
b) constant-volume
c) adiabatic
d) isobaric

1 answer:

frez [133]3 years ago

7 0

I guess it’s d) isobaric mate correct me if I am wrong :D

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Newton laws escape room

ivann1987 [24]

I can barely see the image you put

7 0

3 years ago

I place an ice cube with a mass of 0.223 kg and a temperature of −35◦C is placed into an insulated aluminum container with a mas

Nadya [2.5K]

To solve this problem it is necessary to use the calorimetry principle. From the statement it asks about the remaining ice, that is, to the point where the final temperature is 0 ° C.

We will calculate the melted ice and in the end we will subtract the total initial mass to find out how much mass was left.

The amount of heat transferred is defined by

$Q = mc\Delta T$

Where,

m = mass

c = Specific heat

$\Delta T =$ Change in temperature

There are two states, the first is that of heat absorbed by that mass 'm' of melted ice and the second is that of heat absorbed by heat from -35 ° C until 0 ° C is reached.

Performing energy balance then we will have to

$Q_i-E_h = Q_m$

Where,

$Q_ i$ = Heat absorbed by whole ice

$Q_m$ = Heat absorbed by mass

$E_h$ = Heat energy by latent heat fusion/melting

$m_i*c_i \Delta T +m*L_f = (m_wc_w+m_{al}c_{al})\Delta T$

Replacing with our values we have that

$0.223*2108(-(-35))+m*3.34*10^5 = (0.452*4186+0.553*902)(27-0)$

$16452.9+334000m = (1892.072+498.806)*27$

Rearrange and find m,

$m = 0.144Kg$

Therefore the Ice left would be

$m' = 0.223-0.144$

$m' = 0.079Kg$

Therefore there is 0.079kg ice in the containter when it reaches equilibrium

8 0

4 years ago

Help please help me with all of it I don't know nothing. bless ur hearts <br>

Evgesh-ka [11]

Answer:

200 , 0 , 133.33333

Explanation:

velocity = change of X / change of T

so

400/2 = 200

0/2 = 0

400/3 = 133.33333

3 0

3 years ago

The weight of a 72.0 kg astronaut on the Moon, where g = 1.63 m/s2 is (5 points) Select one: a. 112 N b. 117 N c. 135 N d. 156 N

Hunter-Best [27]

Answer: The weight of a 72.0 kg astronaut on the Moon is 117.36 N.

Explanation:

Mass of the astronaut on the moon , m= 72 kg

Acceleration due to gravity on moon,g = 1.63 $m/s^2$

According to Newton second law of motion: F = ma

This will changes to = Weight = mass × g

$Weight=72 kg\times 1.63m/s^2=117.36 N$

The weight of a 72.0 kg astronaut on the Moon is 117.36 N.

5 0

4 years ago

Read 2 more answers

Physics !!!!!!!!!!!!!!!!!

madreJ [45]

Answer:

1) v = -0.89 m/s

2) v = 36.5 m/s

Explanation:

1) Given

The mass of the boy, M = 45 Kg

The mass of the skateboard, m = 2 Kg

The initial velocity of the boy and skateboard is zero

According to the law of conservation of linear momentum

MV + mv = 0

V = - mv/M

Substituting the given values in the above equation

V = -2 x 20/ 45

= -0.89 m/s

Hence, the velocity of the boy, v = -0.89 m/s

2) Given

The mass of the boy, M = 47 Kg

The mass of the skateboard, m = 8 Kg

The initial combined velocity of the boy and skateboard, u = 4.2 m/s

The boy jumped off from the skateboard with velocity, V = -1.3 m/s

According to the law of conservation of momentum,

MV + mv = (M + m)u

v = [(M + m)u - MV] / m

Substituting the given values,

v = [(47 + 8)4.2 - 47(-1.3)] / 8

= 36.5 m/s

Hence, the velocity of the skateboard, v = 36.5 m/s

3 0

4 years ago