5. What's the temperature -10°C in kelvins?

A tank is filled with an ideal gas at 400 K and pressure of 1.00 atm.

bekas [8.4K]

To find the temperature it is necessary to use the expression and concepts related to the ideal gas law.

Mathematically it can be defined as

$PV=nRT$

Where

P = Pressure

V = Volume

n = Number of moles

R = Gas constant

T = Temperature

When the number of moles and volume is constant then the expression can be written as

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

Or in practical terms for this exercise depending on the final temperature:

$T_2 = \frac{P_2T_1}{P_1}$

Our values are given as

$T_1 = 400K\\P_1 = 1atm\\P_2 = 2atm$

Replacing

$T_2 = \frac{(2)(400)}{1}\\T_2 = 800K$

Therefore the final temperature of the gas is 800K

6 0

2 years ago

The displacement volume of an automobile engine is 167 in3. what is the displacement volume in liters?

forsale [732]

The displacement volume in liters is 2.74 liters.

<h3>What is displacement volume?</h3>

Displacement volume is the quantity of solvent that will be displaced by a specified quantity of a solid during dissolution.

It can also be defined as the volume displaced by the piston as it moves between top dead center and bottom dead center in a car engine.

<h3>Displacement volume in liters</h3>

1 liter = 61.02 in³

? = 167 in³

= 167/61.02

= 2.74 liters

Thus, the displacement volume in liters is 2.74 liters.

Learn more about displacement volume here: brainly.com/question/1945909

#SPJ1

4 0

7 months ago

Please Its urgent I need your HELP!!!!!!!!!!!!!!!!!!!!!!!

ozzi

Raising the temperature results in the radiator giving off photons of high-energy ultraviolet light. As heat is added, the radiator emits photons across a wide range of visible-light frequencies

4 0

2 years ago

A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

1.

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

2.

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

2 years ago

A 35 kg child goes down a playground that is inclined at an angle of 27.5 degrees above the horizontal. Find the acceleration if

tigry1 [53]

Answer:

4.16m/s²

Explanation:

According to Newtons second law;

$\sum Fx = ma_x\\Fm - Ff = ma_x\\W sin\theta - \mu R cos \theta = ma_x\\mg sin\theta - \mu mg cos\theta = ma_x\\$

Fm is the moving force

$\mu$ is the coefficient of kinetic friction between the child and the slide

m is the mass

g is the acceleration due to gravity

a is the acceleration of the child

Substitute the given values and get the acceleration as shown;

35(9.8)sin27.5 - 0.415(35)(cos27.5) = 35a

158.38-12.88 = 35a

145.49 = 35a

a = 145.49/35

a = 4.16m/s²

Hence the acceleration of the body is 4.16m/s²

4 0

2 years ago

5. What's the temperature -10°C in kelvins? ​

5. What's the temperature -10°C in kelvins?