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Jlenok [28]
2 years ago
11

When a car of mass 1167 kg accelerates from 10.0 m/s to some final speed, 4.00  10 5J of work are done. Find this final speed.

Physics
1 answer:
Akimi4 [234]2 years ago
8 0
  • Mass=1167kg
  • Initial velocity=u=10m/s
  • Acceleration=a=4m/s^2
  • Work done=105J=W
  • Final velocity=v=?
  • Force=F
  • Distance=d

Apply Newton's second law

\\ \tt\hookrightarrow F=ma

\\ \tt\hookrightarrow F=1167(4)=4668N

Now

\\ \tt\hookrightarrow W=Fd

\\ \tt\hookrightarrow d=\dfrac{W}{F}

\\ \tt\hookrightarrow d=\dfrac{105}{4668}

\\ \tt\hookrightarrow d=0.022m

Now

  • d be s

According to third equation of kinematics

\\ \tt\hookrightarrow v^2=u^2+2as

\\ \tt\hookrightarrow v^2=10^2+2(4)(0.022)

\\ \tt\hookrightarrow v^2=100+8(0.022)

\\ \tt\hookrightarrow v^2=100+0.176

\\ \tt\hookrightarrow v^2=100.176

\\ \tt\hookrightarrow v=10.001m/s

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Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final
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Complete question:

Consider the hypothetical reaction 4A + 2B → C + 3D

Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final concentration of A at the end of this same interval if its concentration was initially 1.600 M?

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the final concentration of A is 0.992 M.

Explanation:

Given;

time of reaction, t = 4.0 s

rate of change of the concentration of B =  -0.0760 M/s

initial concentration of A = 1.600 M

⇒Determine the rate of change of the concentration of A.

From the given reaction: 4A + 2B → C + 3D

2 moles of B ---------------> 4 moles of A

-0.0760 M/s of B -----------> x

x = \frac{4(-0.076)}{2} \\\\x = -0.152 \ M/s

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ΔA = -0.152 M/s  x 4s

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Therefore, the final concentration of A is 0.992 M.

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