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3 years ago

How would decreasing the volume of the reaction vessel affect each of the following equilibria?2NOBr(g)⇌2NO(g)+Br2(g)

Physics

1 answer:

mafiozo [28]3 years ago

5 0

Answer:

The equilibrium position will shift towards the left hand side or reactants side

Explanation:

Decreasing the volume (increasing the pressure) of the system will shift the equilibrium position towards the lefthand side or reactants side. This is because, decreasing the volume (increasing the pressure) implies shifting the equilibrium position towards the side having the least number of moles.

There are two moles of reactants and a total of three moles of products(total). Hence decreasing the volume and increasing the pressure of the gas phase reaction will shift the equilibrium position towards the lefthand side.

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X(????) = 5.0???? 2 − 4.0???? 3 m.

Burka [1]

Answer:

Explanation:

s of time, (b) the velocity and acceleration at t = 2.0 s, (c) the time at which the position is a maximum, (d) the time at which the velocity is zero, and (e) the maximum position. Assume all variable and constants are in SI units.

4 0

3 years ago

4. A 62.0-kg person, standing on the diving board, dives straight down into the water. Just before striking the water, her speed

Alekssandra [29.7K]

To solve this problem it is necessary to apply the concepts related to the Moment. The moment in terms of the Force and the time can be expressed as

$\Delta P = F\Delta t$

F = Force

$\Delta t = Time$

At the same time the moment can be expressed in terms of mass and velocity, mathematically it can be given as

$P = m \Delta v$

Where

m = Mass

$\Delta v =$ Change in velocity

Our values are given as

$\Delta t=1.65s$

By equating the two equations we can find the Force,

$F\Delta t = m\Delta v$

$F = \frac{m\Delta v}{\Delta t}$

$F = \frac{62(1.1-5.5)}{1.65}$

Therefore, the net average force will be:

$F = - 165N$

The negative symbol indicates that the direction of the force is upwards.

7 0

3 years ago

a 30 kg child is sitting 2 meters from the center of a merry go round. The coefficients of static and kinetic friction between t

laiz [17]

Answer: A

Explanation:

From the question, the given parameters are given.

Mass M = 30kg

Radius r = 2 m

Coefficient of static friction μ = 0.8

Coefficient of kinetic friction μ = 0.6

Kinetic friction Fk = μ × mg

Fk = 0.6 × 30 × 9.8

Fk = 176.4 N

The force acting on the merry go round is a centripetal force F.

F = MV^2/r

This force must be greater than or equal to the kinetic friction Fk. That is,

F = Fk

F = 176.4

Substitute F , M and r into the centripetal force formula above

176.4 = (30×V^2)/2

Cross multiply

352.8 = 30V^2

V^2 = 352.8/30

V = sqrt (11.76) m/s

V = 5.24 m/s

Therefore, the maximum speed of the merry go round before the child begins to slip is sqrt (12) m/s approximately

4 0

3 years ago

A 1.51 kg ball and a 1.97 kg ball are connected by a 1.63 m long rigid, massless rod. The rod is rotating clockwise about its ce

Eddi Din [679]

Answer:

$T = 1.205\,N\cdot m$

Explanation:

Needed torque can be estimated by means of the Theorem of Angular Momentum Conservation and Impact Theorem. The center of mass of the system is:

$\bar r = \frac{(0\,m)\cdot (1.51\,kg)+(1.63\,kg)\cdot (1.97\,kg)}{1.51\,kg+1.97\,kg}$

$\bar r = 0.923\,m$

Let assume that both masses can be modelled as particles, then:

$[(1.51\,kg)\cdot (0.923\,m)^{2} + (1.97\,kg)\cdot (0.707\,m)^{2}]\cdot (38\,\frac{rev}{min} )\cdot (\frac{2\pi\,rad}{1\,rev} )\cdot (\frac{1\,min}{60\,s} ) -T\cdot (7.5\,s) = 0\,\frac{kg\cdot m^{2}}{s}$

The torque needed to stop the system is:

$T = 1.205\,N\cdot m$

5 0

3 years ago

A uniform disk with radius R = 0.400 m and mass 30.0 kg rotates in a horizontal plane on a frictionless vertical axle that passe

Rom4ik [11]

Answer:

The angle of rotation given is incorrect, the correct function to have a balanced unit is

θ(t) = (1.10 rad/s)t + (6.30 rad/s²)t²

Explanation:

Given that,

Radius of disk is

R = 0.4m

Mass of disk

M = 30kg

The angle of rotation is a function of time and it is given as

θ(t) = (1.10 rad/s)t + (6.30 rad/s²)t²

What is linear angular acceleration "a" when it rotates 0.1rev

θ = 0.1 rev

1 rev = 2πrad

θ = 0.1 × 2πrad = 0.63 rad

Let calculate the time it revolves 0.63rad

θ(t) = (1.10 rad/s)t + (6.30 rad/s²)t².

0.63 = 1.1t + 6.3t²

6.3t² + 1.1t - 0.63 = 0

Solving this quadratic equation using formula method

t = [-b±√(b²-4ac)]/2a

a = 6.3 b = 1.1 c = -0.63

t = [-1.1 ± √(1.1²-4×6.3×-0.63)]/2×6.3

t = -1.1±√(1.21+15.876) / 12.6

t = -1.1 ± 4.13 / 12.6

Let discard the negative time

So, t = (-1.1 + 4.13) / 12.6

t = 0.24 s

Now,

Angular velocity can be determine using

ωz = dθ / dt

ωz = 1.1 + 12.6t, since t = 0.24

ωz = 1.1 + 12.6(0.24)

ωz = 4.124 rad/s

Angular acceleration can be determine using

αz = dωz/dt

αz = 12.6 rad/s²

The radial acceleration can be determined using

ar = ωz²•R

ar = 4.124² × 0.4

ar = 6.8 m/s²

Also, the tangential Acceleration is related to angular acceleration using

at = αz • R

at = 12.6 × 0.4

at = 5.04 m/s²

Then, the magnitude of the resultant acceleration can be calculated using

a = √(ar² + at²)

a = √(6.8² + 5.04²)

a = √71.64

a = 8.46 m/s²

Direction

β = Arctan(at/ar)

β = Arctan(5.04/6.8)

β = 36.54

This is the direction from the origin

But it has already revolve 0.1rev

1 rev = 360°

Then, θ = 0.1° = 360 × 0.1 = 36°

Direction = β — θ

Then, it direction is 36.54 — 54° = 0.54°

The direction is 0.54°

Check attachment for direction understanding

4 0

3 years ago