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Oduvanchick [21]
3 years ago
12

You have two contentious friends, Chris and Pat, and you’ve quickly discovered that they need you to resolve arguments they have

about physics concepts. Chris says, "The spacing between two dots tells you the speed at the time of the first dot," but Pat says, "The spacing tells you the speed at the time of the second dot. How do you resolve this debate?"
Physics
1 answer:
pshichka [43]3 years ago
3 0

Answer:

v_average = (d₂-d₁) / Δt

this average velocity is not necessarily the velocity of the extreme points,

Explanation:

To resolve the debate, it must be shown that the two have part of the reason, the space or distance between the two points divided by time is the average speed between the points.

             v_average = (d₂-d₁) / Δt

this average velocity is not necessarily the velocity of the extreme points, in the only case that it is so is when there is no acceleration.

Therefore neither of them is right.

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How much energy is required to heat 70 g of water at 20°C to boiling
choli [55]

Answer:

Q=23,430J

Explanation:

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In this case, since we compute the required energy via:

Q=mC\Delta T

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A well is pumped at Q = 300 m3 /hr in a confined aquifer. The aquifer transmissivity is 25 m2 /hr and the storage coefficient is
Effectus [21]

Answer:

(a). The draw-down at a distance 200 m from the well after pumping for 50 hr is 5.383 m.

(b). The draw-down at a distance 200 m from the well after pumping for 50 hr is 6.707 m.

Explanation:

Given that,

Energy Q=300\ m^3/hr

Transmissivity T = 25\ m^2/hr

Storage coefficient S=2.5\times10^{-4}

Distance r= 200 m

We need to calculate the draw-down at a distance 200 m from the well after pumping for 50 hr

Using formula of draw-down

s= \dfrac{Q}{4\pi T}(-0.5772-ln\dfrac{r^2S}{4tT})

Put the value into the formula

s=\dfrac{300}{4\pi\times25}(-0.5772-ln\dfrac{(200)^2\times2.5\times10^{-4}}{4\times25\times50})

s=5.383\ m

We need to calculate the draw-down at a distance 200 m from the well after pumping for 200 hr

Using formula of draw-down

s= \dfrac{Q}{4\pi T}(-0.5772-ln\dfrac{r^2S}{4tT})

Put the value into the formula

s=\dfrac{300}{4\pi\times25}(-0.5772-ln\dfrac{(200)^2\times2.5\times10^{-4}}{4\times25\times200})

s=6.707\ m

Hence, (a). The draw-down at a distance 200 m from the well after pumping for 50 hr is 5.383 m.

(b). The draw-down at a distance 200 m from the well after pumping for 200 hr is 6.707 m.

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3 years ago
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