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andreyandreev [35.5K]
3 years ago
14

Suppose that a nascar race car is moving to the right with a constant velocity of +82m/s. what is the average acceleration of th

e car?
Physics
1 answer:
Veseljchak [2.6K]3 years ago
8 0
  <span>Iacceleration which would be roughly 16.17m/s^2</span><span>
</span>
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Consider the following geometric solids.
Bad White [126]

Before answering this question, first we have to understand the effect of ratio of surface area to volume on the rate of diffusion.

The rate of diffusion for a body having larger surface area as compared to the ratio of surface area to volume will be more than a body having less surface area. Mathematically it can written as-

                           V∝ R            [ where v is the rate of diffusion and r is the ratio of surface area to volume]

As per the question,the ratio of surface area to volume for a sphere is given 0.08m^{-1}

The surface area to volume ratio for right circular cylinder is given 2.1m^{-1}

Hence, it is obvious that the ratio is more for right circular cylinder.As the rate diffusion is directly proportional to the surface area to volume ratio,hence rate of diffusion will be more for right circular cylinder.

Hence the correct option is B. The rate of diffusion would be faster for the right cylinder.


5 0
3 years ago
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A 1.5 m3 tank contains 500 kg of liquid water in equilibrium with pure water vapor, which fills the remainder of the tank. The t
Stels [109]

Answer:

Explanation:

Attached are the solutions

7 0
3 years ago
What was the long-term effect of the Northwest Ordinance of 1787?
victus00 [196]

Answer:

Established a government for the Northwest Territory, outlined the process for admitting a new state to the Union, and guaranteed that newly created states would be equal to the original thirteen states

Explanation:

Goooogle, so I hope this helps somewhat

Also, isn't this a History question? You put physics lol

4 0
3 years ago
Two positive charges q1 = q2 = 2.0 μC are located at x = 0, y = 0.30 m and x = 0, y = -0.30 m, respectively. Third point charge
Wittaler [7]

Answer:

 F = 0.111015 N

Explanation:

For this exercise the force is given by Coulomb's law

        F = k q₁q₂ / r₂₁²

we calculate the electric force of the other two particles on the charge q1

Charges q₁ and q₂

the distance between them is

          r₁₂ = y₁ -y₂

          r₁₂ = 0.30 + 0.30

          r₁₂ = 0.60 m

let's calculate

          F₁₂ = 9 10⁹ 2 10⁻⁶ 2 10⁻⁶ / 0.60 2

          F₁₂ = 1 10⁻¹ N

directed towards the positive side of the y-axis

Charges 1 and 3

Let's find the distance using the Pythagorean Theorem

             r₁₃ = RA [(0.40-0) 2 + (0-0.30) 2]

             r₁₃ = 0.50 m

            F₁₃ = 9 10⁹ 2 10⁻⁶ 4 10⁻⁶ / 0.50²

            F₁₃ = 1.697 10⁻² N

The direction of this force is on the line that joins the two charges (1 and 3), let's use trigonometry to find the components of this force

           tan θ = y / x

           θ = tan⁻¹ y / x

          θ = tan⁻¹ 0.3 / 0.4

           tea = 36.87º

    The angle from the positive side of the x-axis is

         θ ’= 180 - θ

        θ ’= 180 - 36.87

        θ ’= 143.13º

       sin143.13 = F_13y / F₁₃

           F_13y = F₁₃ sin 143.13

           F{13y} = 1.697 10⁻² sin 143.13

           F_13y = 1.0183 10⁻² N

            cos 143.13 = F_13x / F₁₃

           F₁₃ₓ = F₁₃ cos 143.13

           F₁₃ₓ = 1.697 10⁻² cos 143.13

           F₁₃ₓ = -1.357 10-2 N

Now we can find the components of the resultant force

          Fx = F13x + F12x

          Fx = -1,357 10-2 +0

          Fx = -1.357 10-2 N

          Fy = F13y + F12y

         Fy = 1.0183 10-2 + ​​1 10-1

          Fy = 0.110183 N

We use the Pythagorean theorem to find the modulus

         F = Ra (Fx2 + Fy2)

         F = RA [(1.357 10-2) 2 + 0.110183 2]

         F = 0.111015 N

Let's use trigonometry for the angles

         tan tea = Fy / Fx

          tea = tan-1 (0.110183 / -0.01357)

          tea = 1,448 rad

to find the angle about the positive side of the + x axis

           tea '= pi - 1,448

           Tea = 1.6936 rad

6 0
3 years ago
How many Joules in 41.87 kcal?
gladu [14]
The answer is 175184.08 joules
4 0
3 years ago
Read 2 more answers
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