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drek231 [11]
3 years ago
13

A car travels at a constant velocity of 70 mph for one hour. At the second hour, the car’s velocity was 60 mph. At the third hou

r, the car’s velocity was 80 mph. At what hour was their positive acceleration? At what hour was their negative acceleration?
Physics
2 answers:
katovenus [111]3 years ago
8 0

When the velocity increases, then the acceleration will be positive and when the velocity decrease then the acceleration will be negative.

During the first hour, the velocity was 70 mph and during the seconds hour the velocity was 60 mph. Hence, the velocity decrease in the seconds hour. So, the acceleration will be negative during the second hour.

Now, during the third hour the velocity increases as it is 80 mph. Hence, the acceleration will be positive during the third hour.


kati45 [8]3 years ago
5 0

<u>Answer:</u>

  Positive acceleration is in third hour and negative acceleration is in second hour.

<u>Explanation:</u>

  Velocity of car in first hour =  70 mph

  Velocity of car in second hour = 60 mph

  Velocity of car in third hour = 80 mph

   Acceleration = Change in velocity / Time

   Acceleration in second hour = (60 - 70)/1 = -10 mph²

   Acceleration in third hour = (80 - 60)/1 = 20 mph²

   So positive acceleration is in third hour and negative acceleration is in second hour.

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1) 1.11\cdot 10^{-7} J

The capacitance of a parallel-plate capacitor is given by:

C=\frac{\epsilon_0 A}{d}

where

\epsilon_0 is the vacuum permittivity

A is the area of each plate

d is the distance between the plates

Here, the radius of each plate is

r=\frac{2.0 cm}{2}=1.0 cm=0.01 m

so the area is

A=\pi r^2 = \pi (0.01 m)^2=3.14\cdot 10^{-4} m^2

While the separation between the plates is

d=0.50 mm=5\cdot 10^{-4} m

So the capacitance is

C=\frac{(8.85\cdot 10^{-12} F/m)(3.14\cdot 10^{-4} m^2)}{5\cdot 10^{-4} m}=5.56\cdot 10^{-12} F

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U=\frac{1}{2}CV^2=\frac{1}{2}(5.56\cdot 10^{-12} F/m)(200 V)^2=1.11\cdot 10^{-7} J

2) 0.71 J/m^3

The magnitude of the electric field is given by

E=\frac{V}{d}=\frac{200 V}{5\cdot 10^{-4} m}=4\cdot 10^5 V/m

and the energy density of the electric field is given by

u=\frac{1}{2}\epsilon_0 E^2

and using

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u=\frac{1}{2}(8.85\cdot 10^{-12} F/m)(4\cdot 10^5 V/m)^2=0.71 J/m^3

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Write an expression for the magnitude of charge moved, Q, in terms of N and the fundamental charge e
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We have that for the Question "Write an expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e" it can be said its equation is

Q=\frac{E}{Nr^2}

       

From the question we are told

Write an expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e

<h3>An Expression for the <em>magnitude </em>of charge moved</h3>

Generally the equation for the  <em>magnitude </em>of charge moved, Q   is mathematically given as

Q=\frac{E}{Nr^2}

Therefore

An expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e" it can be

 Q=\frac{E}{Nr^2}

 

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For the spring in this problem, we have:

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x = 0.035 m (compression)

Substituting, we find the elastic potential energy:

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