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zhenek [66]
4 years ago
10

A container of gas molecules is at a pressure of 2 atm and has amass density of 1.7 grams per liter. All of the molecules in the

container are diatomic nitrogen molecules with an atomic weight of28 grams per mole. What is the typical speed of the nitrogenmolecules in the container? Here we define the typical speed to bethe root-mean-square velocity (RMS velocity = vrms) ofthe center of mass of the molecule
Physics
1 answer:
Tpy6a [65]4 years ago
5 0

Answer:

The speed of nitrogen molecule is 1.87 m/s.

Explanation:

Given that,

Pressure = 2 atm

Density = 1.7 grams/liter

Atomic weight = 28 grams

We need to calculate the temperature

Using formula of idea gas

PV=nRT

P=\dfrac{WRT}{VM}

P=\dfrac{\rho RT}{M}

T=\dfrac{PM}{\rho R}

Put the value into the formula

T=\dfrac{2\times28}{1.7\times0.0821}

T=401.2\ K

We need to calculate the speed of nitrogen molecule

Using formula of RMS speed

V_{rms}=\sqrt{\dfrac{3RT}{M}}

V_{rms}=\sqrt{\dfrac{3\times0.0821\times401.2}{28}}

V_{rms}=1.87\ m/s

Hence, The speed of nitrogen molecule is 1.87 m/s.

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erastovalidia [21]
The body fights off an infection by a fever since the higher temperature will denature what is causing the infection
3 0
3 years ago
Water is flowing through a channel that is 12m wide with a
Alexus [3.1K]

Answer:

Velocity from second channel will be 1.6875 m/sec

Explanation:

We have given width of the channel , that is diameter of the channel 1 d_1 = 12 m

So radius r_1=\frac{d_1}{2}=\frac{12}{2}=6m

Speed through the channel 1 v_1=0.75m/sec

Width , that is diameter of the channel 2 d_2=4m

So r_2=2m

From continuity equation

A_1v_1=A_2v_2

\pi \times 12^2\times 0.75=4\times \pi\times  4^2\times v_2

v_2=1.6875m/sec

So velocity from smaller channel will be 1.6875 m /sec

6 0
3 years ago
A body moves in a straight line. At time, it's acceleration is given by a = 12t + 1. When t =0, the velocity of the body v is 2
madreJ [45]

Answer:

v = 6t² + t + 2, s = 2t³ + ½ t² + 2t

59 m/s, 64.5 m

Explanation:

a = 12t + 1

v = ∫ a dt

v = 6t² + t + C

At t = 0, v = 2.

2 = 6(0)² + (0) + C

2 = C

Therefore, v = 6t² + t + 2.

s = ∫ v dt

s = 2t³ + ½ t² + 2t + C

At t = 0, s = 0.

0 = 2(0)³ + ½ (0)² + 2(0) + C

0 = C

Therefore, s = 2t³ + ½ t² + 2t.

At t = 3:

v = 6(3)² + (3) + 2 = 59

s = 2(3)³ + ½ (3)² + 2(3) = 64.5

5 0
3 years ago
A test charge of -1.4 x 10-7 coulombs experiences a force of 5.4 x 10-1 newtons. Calculate the magnitude of the electric field c
VARVARA [1.3K]

Answer:

3.86×10⁶ Newton/coulombs

Explaination:

Applying,

E = F/q....................... Equation 1

Where E = Electric Field, F  = Force, q = charge.

From the question,

Given: F = 5.4×10⁻¹ N, q = -1.4×10⁻⁷ coulombs

Substitute these values into equation 1

E = 5.4×10⁻¹/ -1.4×10⁻⁷

E = -3.86×10⁶ Newtons/coulombs

Hence the magnitude of the electric field created by the

negative test charge is 3.86×10⁶ Newton/coulombs

5 0
3 years ago
In which case would electrical potential energy be built up and stored in the electric field? a) A positive charge is moved towa
mr_godi [17]

Answer:

The correct option is B

Explanation:

Although, it is common knowledge that in an electric field, unlike charges attract and like charges repel. However, to build up an electric potential, a positive charge is brought close to another positive charge - this causes work done to be changed to electric potential energy and stored in the electric field.

It should however be noted that when a negative charge is moved away from a positive charge, the negative charge gains electric potential energy.

5 0
3 years ago
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