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iogann1982 [59]
3 years ago
15

Which is an si metric unit of measurement that is used to record the heat transfer of a solution in a classroom investigation?

Physics
1 answer:
kumpel [21]3 years ago
4 0
The SI unit for heat energy is joule
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1.)A tank travels at a rate of 10.0 km/hr for 12.00 minutes, then at 15.0 km/hr for 8.00
rosijanka [135]

12.00 min = 0.2 hr

8.00 min = 0.15 hr

Total distance:

(10.0 km/hr) (0.2 hr) + (15.0 km/hr) (0.15 hr) + (20.0 km/hr) (0.2 hr)

= 8.25 km

Average speed:

(10.0 km/hr + 15.0 km/hr + 20.0 km/hr) / 3

= 15 km/hr

Change in position:

(10.0 km/hr) (0.2 hr) + (15.0 km/hr) (0.15 hr) - (20.0 km/hr) (0.2 hr)

= 0.25 km

Average velocity:

(10.0 km/hr + 15.0 km/hr - 20.0 km/hr) / 3

≈ 1.67 m/s

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This is false. There were many times in history when people discovered something that they didn't even know was possible or didn't even plan to discover it. Knowing tradeoffs doesn't mean that something won't surprise you or that all will go according to plan.
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A motorcyclist who is moving along an x axis directed toward the east has an acceleration given by a = (6.30 - 2.20t) m/s2 for 0
musickatia [10]
<span>(a) 12.02 m/s
 (b) 52.2 meters

   This problem is an example of integral calculus. You've been given an acceleration vector which is usually known as the 2nd derivative. From that you need to calculate the velocity function (1st derivative) and position (actual function) by successively calculating the anti-derivative. So:
   A(t) = 6.30 - 2.20t
 V(t) = 6.30t - 1.10t^2 + C

   We now have a velocity function, but need to determine C. Since we've been given the velocity at t = 0, that's fairly trivial.
 V(t) = 6.30t - 1.10t^2 + C
 3 = 6.30*0 - 1.10*0^2 + C
 3 = 0 + 0 + C
 3 = C

    So the entire velocity function is:
 V(t) = 6.30t - 1.10t^2 + 3
 V(t) = -1.10t^2 + 6.30t + 3

   Now for the location function which is the anti-derivative of the velocity function.
  V(t) = -1.10t^2 + 6.30t + 3
  L(t) = -0.366666667t^3 + 3.15t^2 + 3t + C

   Now we need to calculate C. And once again, we've been given the location for t = 0, so
 L(t) = -0.366666667t^3 + 3.15t^2 + 3t + C
 7.3 = -0.366666667*0^3 + 3.15*0^2 + 3*0 + C
 7.3 = 0 + 0 + 0 + C
 7.3 = C

   L(t) = -0.366666667t^3 + 3.15t^2 + 3t + 7.3

   Now that we have the functions, they are:
 A(t) = 6.30 - 2.20t
 V(t) = -1.10t^2 + 6.30t + 3
 L(t) = -0.366666667t^3 + 3.15t^2 + 3t + 7.3

    let's answer the questions.
 (a) What is the maximum speed achieved by the cyclist?
 This can only happen at those points that meet either of the following criteria.
  1. The derivative is undefined for the point.
  2. The value of the derivative is 0 for the point.
 As it turns out, the 1st derivative of the velocity function is the acceleration function which we have. So
 A(t) = 6.30 - 2.20t
 0 = 6.30 - 2.20t
  2.20t = 6.30
 t = 2.863636364

   So one of V(0), V(2.863636364), or V(6) will be the maximum value. Therefore: V(0) = 3
 V(2.863636364) = 12.0204545454545
 V(6) = 1.2

   So the maximum speed achieved is 12.02 m/s

   (b) Total distance traveled?
 L(0) = 7.3
 L(6) = 59.5
 Distance traveled = 59.5 m - 7.3 m = 52.2 meters</span>
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