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3 years ago

How does gas exert pressure? Explain using the change in momentum.

2 answers:

7 0

Answer: (Here goes another answer) Gas pressure will be caused by the impact of the gas particles as they come into collison and strive a force on the container walls and once the degrees of the gas enlarge, the particles will then create kinetic energy and speed increases.

Hope this helps :)

san4es73 [151]3 years ago

4 0

Answer:

The molecules are continually colliding with each other and with the walls of the container. When a molecule collides with the wall, they exert small force on the wall The pressure exerted by the gas is due to the sum of all these collision forces. The more particles that hit the walls, the higher the pressure.

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Three point charges are placed on the x−y plane: a + 50.0-nC charge at the origin, a −50.0-nC charge on the x axis at 10.0 cm, a

butalik [34]

Answer:

(a) F = 0.00322i - 0.00793j with magnitude |F| = 0.00856N

(b) E = -42846.7 N/C

Explanation:

The diagram attached below explains some parameters.

Parameters given:

Charge Q1 = +50 nC at point (0, 0)

Charge Q2 = -50 nC at point (0.1, 0)

Charge Q3 = +150 nC at point (0.1, 0.08)

* The distances are in meters.

(a) The total electric force on the charge Q3 due to Q1 and Q2 is the vector sum of the forces due to Q1 and Q2. Mathematically,

F = F1 + F2

FORCE DUE TO Q1 i.e. F(Q1, Q3)

We have to find the x and y components.

From the diagram, we can find θ using SOHCAHTOA:

θ = tan⁻¹ (0.08/0.1)

θ = 38.66⁰

The distance between Q1 and Q3 can be found using Pythagoras theorem:

x² = 0.08² + 0.1²

x = 0.128 m

F1 = Fx(Q1, Q3)i + Fy(Q1, Q3)j

F1 = iF(Q1, Q3)cosθ + jF(Q1, Q3)sinθ

F(Q1, Q3) = (k * Q1 * Q3) / r²

k = Coulombs constant

F(Q1, Q3) = (9 * 10⁹ * 50 * 10⁻⁹ * 150 * 10⁻⁹) /(0.128)²

F(Q1, Q3) = 0.00412N

F1 = i0.00412 * cos38.66 + j0. 00412 * sin38.66

F1 = 0.00322i + 0.00257j N

FORCE DUE TO Q2 i.e. F(Q2, Q3)

We have to find the x and y components.

F2 = Fx(Q2, Q3)i + Fy(Q2, Q3)j

F2 = iF(Q2, Q3)cos90 + jF(Q2, Q3)cos0

F(Q2, Q3) = (k * Q2 * Q3) / r²

F(Q2, Q3) = (9 * 10⁹ * -50 * 10⁻⁹ * 150 * 10⁻⁹) /(0.08)²

F(Q2, Q3) = -0.0105N

F2 = -i0.0105 * cos90 - j0.0105 * cos0

F2 = - 0.0105j N

Hence, the total force will be

F = F1 + F2

F = 0.00322i + 0.00257j - 0.0105j

F = 0.00322i - 0.00793j N

The magnitude of this force is:

|F| = √(0.00322² + (-0.00793²)

|F| = 0.00856N

(b) The electric field at charge Q3 is the sum of the electric fields due to Q1 and Q2:

E = E1 + E2

E1, electric field due to Q1 = kQ1/r²

E1 = (9 * 10⁹ * 50 * 10⁻⁹) / (0.128²)

E1 = 27465.8 N/C

E2, electric field due to Q2 = (9 * 10⁹ * -50 * 10⁻⁹) / (0.08²)

E1 = -70312.5N/C

The total electric field:

E = E1 + E2

E = 27465.8 - 70312.5

E = -42846.7 N/C

3 0

3 years ago

In Fig.25-46,how much charge is stored on the parallel-plate capacitors by the 12.0 V battery? One is filled with air,and the ot

Fittoniya [83]

Answer:

$Q=7.9\times 10^{-10}\ C$

Explanation:

Given that

V= 12 V

K=3

d= 2 mm

Area=5.00 $ 10#3 m2

Assume that

$ = Multiple sign

# = Negative sign

$A=5\times 10^{-3}\ m^2$

We Capacitance given as

For air

$C_1=\dfrac{\varepsilon _oA}{d}$

$C_1=\dfrac{8.85\times 10^{-12}\times 5\times 10^{-3}}{2\times 10^{-3}}$

$C_1=2.2\times 10^{-11}\ F$

$C_2=\dfrac{K\varepsilon _oA}{d}$

$C_2=\dfrac{3\times 8.85\times 10^{-12}\times 5\times 10^{-3}}{2\times 10^{-3}}\ F$

$C_2=6.6\times 10^{-11}\ F$

Net capacitance

C=C₁+C₂

$C=8.8\times 10^{-11}\ F$

We know that charge Q given as

Q= C V

$Q=12\times 6.6\times 10^{-11}\ C$

$Q=7.9\times 10^{-10}\ C$

6 0

3 years ago

An apprentice bricklayer stabs 30cm long blocks on top of each other in such a way that 5cm of each block hangs over the lower b

NikAS [45]

Answer:

Explanation:The work done is twice as great for block B because it is moved twice the ... Equal forces are used to move blocks A and B across the floor. ... Does the normal force of the floor pushing upward on the block do any work? ... Suppose that the mass is halfway between one of the extreme points of its motion and the center point.

7 0

2 years ago

Audible beats are formed bu the interference of two waves

Liono4ka [1.6K]

<span>Audible beats are formed by the interferences of two waves in which are from the slightly different frequencies which is letter a. It is because having two waves that consist of slightly different frequency causes or produce the audible beats to be formed in which they are beats that causes sound to be either loud or soft.</span>

3 0

2 years ago

Una pelota de voleibol de 0.45 kg es impactada (golpeada) por una fuerza de 80 N. ¿ Cuál es la aceleración que experimenta la pe

KIM [24]

Answer:

a = 177.77 [m/s^2]

Explanation:

Este es un problema relacionado con la segunda ley de Newton. La cual nos dice que la sumatoria de fuerzas aplicada sobre un cuerpo debe ser igual al producto de la masa por la aceleración.

De esta manera tenemos:

F = m*a

F = fuerza = 80 [N]

m = masa = 0.45 [kg]

80 = 0.45 * a

a = 80 / 0.45

a = 177.77 [m/s^2]

5 0

3 years ago