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Ber [7]
3 years ago
12

Calculate the activation energy, E a Ea , in kilojoules per mole for a reaction at 57.0 ∘ C 57.0 ∘C that has a rate constant of

0.235 s − 1 0.235 s−1 and a frequency factor of 3.41 × 10 11 s − 1 3.41×1011 s−1 .

Physics
1 answer:
Alecsey [184]3 years ago
3 0

Explanation:

Below is an attachment containing the solution

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What force is required to accelerate to 10 kg object to 5.9 m/s/s?
g100num [7]

Force required to accelerate 10 kg object to 5.9 m/s/s ?

Mass = 10 kg

Acceleration = 5.9 m/s^2

Force = Mass * Acceleration

Force = 10 kg * 5.9 m/s^2

Force = 59 kg m /s^2 = 59 N

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3 years ago
A train accelerated from 5 km/hr in 0.5 hours. what was the rate of acceleration?
Bezzdna [24]
Acceleration is the simply rate of change in velocity, how much faster or slower is the object changing speed with respect to time.

A = v/t = 5 km/hr/0.5 hr

5/1/2 = 5 • 2 = 10 km/hr^2.

This would be the acceleration.
5 0
3 years ago
In August 2011, the Juno spacecraft was launched from Earth with the mission of orbiting Jupiter in 2016. The closest distance b
Vikentia [17]

(a) 8927 mi/h

In order to calculate the average speed, we need to convert the time (t=5.0 y) into hours first. In 1 year, we have 365 days, each day consisting of 24 hours, so the time taken is:

t=(5.0 y)(365 d/y)(24 h/d)=43,800 h

The distance covered by the spacecraft is

d=391 mil. mi = 391\cdot 10^6 mi

Therefore, the average speed is just the ratio between the distance covered and the time taken:

v=\frac{d}{t}=\frac{391\cdot 10^6 mi}{43,800 h}=8,927 mi/h

(b) 35 minutes (2097 seconds)

The transmitted signals (which is a radio wave, which is an electromagnetic wave) travels back to the Earth at the speed of light:

c=3.0\cdot 10^8 m/s

Since 1 miles = 1609 metres, the distance covered  by the signal is

d=391\cdot 10^6 mi \cdot (1609 m/mi)=6.29\cdot 10^{11} m

So, the time taken by the signal will be

t=\frac{d}{v}=\frac{6.29\cdot 10^{11} m}{3.0\cdot 10^8 m/s}=2097 s

And since 1 minute = 60 sec, the time taken is

t=2097 s \cdot \frac{1}{60 s/min}\sim 35 min

7 0
3 years ago
Two equally charged, 2.807 g spheres are placed with 3.711 cm between their centers. When released, each begins to accelerate at
statuscvo [17]
\begin{gathered} m=\text{ 2.807 g} \\ d=\text{ 3.711cm} \\ a=260.125\text{ m/s}^2 \\ m=\text{ mass of  both the spheres} \\ d=\text{ distance between the centers of sphere.} \\ a=\text{ acceleration of spheres.} \end{gathered}\begin{gathered} force\text{ due to the sphere having charge q, outside its surface is given by } \\ \vec{F}=\frac{1}{4\pi\epsilon_o}\frac{q_1q_2}{r^2}\hat{r} \\ q_1=charge\text{ on the source object.} \\ q_2=charge\text{ of the object in which we are observing the force.} \\ F=\text{ the force on the charged particle outside the sphere} \\ r=\text{ distance of the charged particle from the center of the sphere} \\ \hat{r}\text{= direction of the force acting on the charged particle} \end{gathered}\begin{gathered} from\text{ Newton's second law} \\ F=ma \\ F=\text{ force acting on the particle.} \\ m=\text{ mass of the object.} \\ a=\text{ acceleration of the object.} \end{gathered}\begin{gathered} from\text{ both the equation } \\ ma=\frac{1}{4\pi\epsilon_o}\frac{q_1q_{\frac{2}{}}}{r^2}\hat{r} \\ here\text{ q}_1\text{ and q}_2\text{ are the same, according to the question.} \end{gathered}\begin{gathered} converting\text{ all the values in s.i. unit} \\ m=2.807*10^{-3}kg \\ d=3.711*10^{-2}m \\ according\text{ to the question q}_1=\text{ q}_2 \\ value\text{ of }\frac{1}{4\pi\epsilon_o}=9*10^9\text{ Nm}^2\text{/C}^2 \\ now\text{ put all the values in the above equation } \\ 2.807*10^{-3}kg*260.125\text{ m/s\textasciicircum2}=9*10^9Nm^2\text{/C}^2*\frac{q^2}{3.711*10^{-2}m} \\  \end{gathered}\begin{gathered} by\text{ trasformation} \\ q=\sqrt{\frac{2.807*10^{-3}kg*260.125m/s^2*3.711*^10^{-2}m}{9*10^9Nm^2\text{/C}^2}} \\ by\text{ solving this we get } \\ q=17.3514*10^{-7}C \\ q=1.73514\text{ micro coulombs.} \end{gathered}Hence the correct answer is q= 1.73514 micro coulombs.
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4 years ago
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