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Misha Larkins [42]

3 years ago

10

In a first order decomposition in which the rate constant is 0.204 sec-1, how long will it take (in seconds) until 0.399 mol/L o

f the compound is left, if there was 0.991 mol/L at the start? (give answer to 3 decimal places)?

1 answer:

DENIUS [597]3 years ago

5 0

Answer:

2.902 to 3 decimal places

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Suppose that two point charges, each with a charge of +3.00 Coulomb are separated by a distance of 3.00 meters. Determine the ma

Sveta_85 [38]

Answer:

$9\cdot 10^9 N$

Explanation:

The magnitude of the electrical force between the two point charges is

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

$q_1 =q_2 = +3.0 C$ is the magnitude of each charge

r = 3.00 m is the separation between the two charges

Substituting the numbers into the formula, we find

$F=(9\cdot 10^9 Nm^2C^{-2})\frac{(+3 C)^2}{(3.00 m)^2}=9\cdot 10^9 N$

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3 years ago

An object glides on a horizontal tabletop with a coefficient of kinetic friction of 0.5. If its initial velocity is 4.3 m/s, how

Shkiper50 [21]

Answer:

Time, t = 0.87 seconds

Explanation:

Given that,

Initial velocity of the object, u = 4.3 m/s

The coefficient of kinetic friction between horizontal tabletop and the object is 0.5

We need to find the time taken by the object for the object to come to rest i.e. final velocity will be 0.

Using first equation of motion to find it as :

$v=u+at$

a is the acceleration, here, $a=\mu g$

$0=u+\mu gt$

$t=\dfrac{u}{\mu g}\\\\t=\dfrac{4.3}{0.5\times 9.8}\\\\t=0.87\ s$

So, the time taken by the object to come at rest is 0.87 seconds. Hence, this is the required solution.

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How did the space rock die??

EastWind [94]

Got shot with a pump shotgun to the head

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Which of following is typical stored energy used to power todays vehicles?

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Electrical energy.................

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4 years ago

In 1976, the SR-71A, flying at 20 km altitude (T = –56 0C), set the official jet-powered aircraft speed record of 3530 km/hr (21

Lapatulllka [165]

To solve this problem we will apply the concepts related to the calculation of the speed of sound, the calculation of the Mach number and finally the calculation of the temperature at the front stagnation point. We will calculate the speed in international units as well as the temperature. With these values we will calculate the speed of the sound and the number of Mach. Finally we will calculate the temperature at the front stagnation point.

The altitude is,

$z = 20km$

And the velocity can be written as,

$V = 3530km/h (\frac{1000m}{1km})(\frac{1h}{3600s})$

$V = 980.55m/s$

From the properties of standard atmosphere at altitude z = 20km temperature is

$T = 216.66K$

$k = 1.4$

$R = 287 J/kg$

Velocity of sound at this altitude is

$a = \sqrt{kRT}$

$a = \sqrt{(1.4)(287)(216.66)}$

$a = 295.049m/s$

Then the Mach number

$Ma = \frac{V}{a}$

$Ma = \frac{980.55}{296.049}$

$Ma = 3.312$

So front stagnation temperature

$T_0 = T(1+\frac{k-1}{2}Ma^2)$

$T_0 = (216.66)(1+\frac{1.4-1}{2}*3.312^2)$

$T_0 = 689.87K$

Therefore the temperature at its front stagnation point is 689.87K

6 0

4 years ago