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Paladinen [302]
3 years ago
13

The rate constant for this first‑order reaction is 0.150 s−1 at 400 ∘C. A⟶products How long, in seconds, would it take for the c

oncentration of A to decrease from 0.860 M to 0.260 M?
Physics
1 answer:
Afina-wow [57]3 years ago
6 0

Answer : The time taken for the concentration will be, 7.98 seconds

Explanation :

First order reaction : A reaction is said to be of first order if the rate is depend on the concentration of the reactants, that means the rate depends linearly on one reactant concentration.

Expression for rate law for first order kinetics is given by :

k=\frac{2.303}{t}\log\frac{[A]_o}{[A]}

where,

k = rate constant  = 0.150s^{-1}

t = time taken for the process  = ?

[A]_o = initial concentration = 0.860 M

[A] = concentration after time 't' = 0.260 M

Now put all the given values in above equation, we get:

0.150s^{-1}=\frac{2.303}{t}\log\frac{0.860}{0.260}

t=7.98s

Therefore, the time taken for the concentration will be, 7.98 seconds

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It depends how far you travel.

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2 years ago
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You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 13-m-hi
Andrew [12]

Answer:

22.15 N/m

Explanation:

As we know potential energy = m*g*h

Potential energy of spring = (1/2)kx^2

m*g*h = (1/2)kx^2

Substituting the given values, we get -  

(400)*(9.8)*(10) = (0.5)*(k)*(2.0^2)

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For safety reasons, this spring constant is increased by 13 % So the new spring constant is  

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3 0
3 years ago
A wooden block of mass M resting on a frictionless, horizontal surface is attached to a rigid rod of length ℓ and of negligible
mr Goodwill [35]

Answer:

a)

mv l

b)

\frac{M }{(M + m)}

Explanation:

Complete question statement is as follows :

A wooden block of mass M resting on a friction less, horizontal surface is attached to a rigid rod of length ℓ and of negligible mass. The rod is pivoted at the other end. A bullet of mass m traveling parallel to the horizontal surface and perpendicular to the rod with speed v hits the block and becomes embedded in it.

(a) What is the angular momentum of the bullet–block system about a vertical axis through the pivot? (Use any variable or symbol stated above as necessary.)

(b) What fraction of the original kinetic energy of the bullet is converted into internal energy in the system during the collision? (Use any variable or symbol stated above as necessary.)

a)

m = mass of the bullet

v = velocity of the bullet before collision

r = distance of the line of motion of bullet from pivot = l

L = Angular momentum of the bullet-block system

Angular momentum of the bullet-block system is given as

L = m v r

L = mv l

b)

V = final velocity of bullet block combination

Using conservation of momentum

Angular momentum of bullet block combination = Angular momentum of bullet

(M + m) V l = m v l\\V =\frac{mv}{(M + m)}

K_{o} = Initial kinetic energy of the bullet

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K_{o} = (0.5) m v^{2}

K_{f} = Final kinetic energy of bullet block combination

Final kinetic energy of bullet block combination is given as

K_{f} = (0.5) (M + m) V^{2}

Fraction of original kinetic energylost is given as

Fraction = \frac{(K_{o} - K_{f})}{K_{o}} = \frac{((0.5) m v^{2} - (0.5) (M + m) V^{2})}{(0.5) m v^{2}}

Fraction = \frac{(m v^{2} - (M + m) (\frac{mv}{(M + m)})^{2})}{m v^{2}} = \frac{(Mm v^{2} + m^{2} v^{2} - m^{2} v^{2})}{(M + m) m v^{2}}

Fraction = \frac{(Mm v^{2} + m^{2} v^{2} - m^{2} v^{2})}{(M + m) m v^{2}}\\ \frac{M }{(M + m)}

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Answer:

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we know that

υ×λ = constant = velocity

υ= frequency

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120n

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