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Dima020 [189]
3 years ago
11

For the hypothetical reaction A → B, calculate the average rate of disappearance of A if the initial concentration of A is 0.91

M and the concentration of A after 90 minutes is 0.11 M.
Physics
1 answer:
marishachu [46]3 years ago
7 0

Answer:

The right answer is 8.9 x 10^-3 M/min

Explanation:

           A → B

-d [A]/dt  = K [A]

ΔA/Δt     = - (C2 -C1)/t2 - t1

               = - (0.11 - 0.91)/90

               = 8.9 x 10^-3 M/min

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Hoochie [10]

I have a strange hunch that there's some more material or previous work
that goes along with this question, which you haven't included here.

I can't easily find the dates of Mercury's extremes, but here's some of the
other data you're looking for:

Distance at Aphelion (point in it's orbit that's farthest from the sun):
<span><span><span><span><span>69,816,900 km
0. 466 697 AU</span>

</span> </span> </span> <span> Distance at Perihelion (</span></span><span>point in it's orbit that's closest to the sun):</span>
<span><span><span><span>46,001,200 km
0.307 499 AU</span> </span>

Perihelion and aphelion are always directly opposite each other in
the orbit, so the time between them is  1/2  of the orbital period.

</span><span>Mercury's Orbital period = <span><span>87.9691 Earth days</span></span></span></span>

1/2 (50%) of that is  43.9845  Earth days

The average of the aphelion and perihelion distances is

     1/2 ( 69,816,900 + 46,001,200 ) = 57,909,050 km
or
     1/2 ( 0.466697 + 0.307499) = 0.387 098  AU
 
This also happens to be 1/2 of the major axis of the elliptical orbit.


3 0
3 years ago
The purpose of this lab is to explore the various ways to calculate projectile velocity using horizontal, vertical and angle inf
kolezko [41]

Answer: A projectile is any object in which the only force is gravity

Explanation: Equations on how to calculate projectile velocity is stated below:

The initial velocity Vo being a vector quantity, has two componentsVox and Voy  

V0x = V0 cos(θ) 

V0y = V0 sin(θ) 

The acceleration A is a also a vector with two components Axand Ay given

Ax = 0 and Ay = - g = - 9.8 m/s2 

Along the x axis the acceleration is equal to 0 and therefore the velocity Vx is constant  

Vx = Vocos(θ) 

Along the y axis, the acceleration is uniform and equal to - g and the velocity at time t is g

Vy = Vo sin(θ) - g t 

Along the x axis the velocity Vx is constant and therefore the component x of the displacement is

x = Vocos(θ) t 

Along the y axis, the motion is of uniform acceleration and the y component of the displacement is

y = Vo sin(θ) t - (1/2) g t2 

3 0
3 years ago
Two identical masses are connected to two different flywheels that are initially stationary. Flywheel A is larger and has more m
inysia [295]

Answer:

a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia

c) True. Information is missing to perform the calculation

Explanation:

Let's consider solving this exercise before seeing the final statements.

We use Newton's second law Rotational

      τ = I α

     T r = I α

     T gR = I α

     Alf = T R / I (1)

     T = α I / R

Now let's use Newton's second law in the mass that descends

     W- T = m a

     a = (m g -T) / m

The two accelerations need related

     a = R α

    α = a / R

    a = (m g - α I / R) / m

    R α = g - α I /m R

    α (R + I / mR) = g

    α = g / R (1 + I / mR²)

We can see that the angular acceleration depends on the radius and the moments of inertia of the steering wheels, the mass is constant

Let's review the claims

a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia

b) False. Missing data for calculation

c) True. Information is missing to perform the calculation

d) False. There is a dependency if the radius and moment of inertia increases angular acceleration decreases

4 0
3 years ago
A small space probe of mass 170 kg is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will
alukav5142 [94]

Answer:

The change  in momentum is  \Delta p =   kg \cdot m/s      

Explanation:

From the question we are told that  

       The mass of the probe is  m = 170 kg

       The location of the prob at time t = 22.9 s is  A  =

       The  momentum at time  t = 22.9 s is  p = < 51000, -7000, 0> kg m/s

        The net force on the probe is  F =  N

Generally the change in momentum is mathematically represented as

              \Delta p = F * \Delta t

The initial time is   22.6 s

 The final time  is  22.9 s

             Substituting values  

           \Delta p =  * (22.9 - 22.6)

            \Delta p =  * (0.3)  

              \Delta p =   kg \cdot m/s        

 

6 0
3 years ago
The resistance of 3A and 12V
liraira [26]
Resistance = V / I
= 12 / 3 = 4
6 0
3 years ago
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