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19 days ago

5

Steve and Elsie are camping in the desert, but have decided to part ways. Steve heads north, at 8 AM, and walks steadily at 2 mi

les per hour. Elsie sleeps in, and starts walking west at 2.5 miles per hour starting at 10 AM. When will the distance between them be 25 miles?

1 answer:

Keith_Richards [2.2K]19 days ago

8 0

Answer:

2.57 hours

Explanation:

Let t (in hours) represent the time it takes for Elsie to walk until they are separated by 25 miles. Since Steve starts 2 hours earlier, his time will be t + 2.

The distance covered by Steve heading north is $s_s = 2(t + 2)$

The distance Elsie travels heading west is $s_e = 2.5t$

The separation between Steve and Elsie is

$\sqrt{s_s^2 + s_e^2} = \sqrt{(2(t+2))^2 + (2.5t)^2} = 25$

We can solve for t by squaring both sides.

$(2(t+2))^2 + (2.5t)^2 = 25^2 = 625$

$4(t+2)^2 + 6.25t^2 = 625$

$4(t^2 + 4t + 4) + 6.25t^2 = 625$

$10.25t^2 + 16t - 609 = 0$

$t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$t= \frac{-16\pm \sqrt{(16)^2 - 4*(10.25)*(-109)}}{2*(10.25)}$

$t= \frac{-16\pm68.74}{20.5}$

Therefore, t = 2.57 or t = -4.13.

Since t cannot be negative, we select t = 2.57 hours.

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Answer:

Speeds of 1.83 m/s and 6.83 m/s

Explanation:

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After the collision, the kinetic energy doubles, therefore:

$2m*(0.5mv_o)=0.5m(v_1^{2}+v_2^{2})$

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Substituting the initial velocity of 5 m/s provides the equation needed to proceed. $v_o$

$2*(5^{2})= v_1^{2} + v_2^{2}$ We know that $v_1+v_2=5$ leads to $v_1=5-v_2$

$50=(5-v_2)^{2}+ v_2^{2}$

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$2v_2^{2}-10v_2-25=0$

Using the quadratic formula leads us to solve for the speeds after the explosion, specifically where a=2, b=-10, and c=-25. $v_2=6.83 m/s$

By substituting the values, the solution yields results for the speeds of the blocks, which are ultimately 1.83 m/s and 6.83 m/s.

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Considering the activity series given for nonmetals, what is the result of the below reaction? Use the activity series provided.

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Response: The result would be no reaction.

Clarification:

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A student wishes to determine the heat capacity of a coffee-cup calorimeter. After she mixes 108.7 g of water at 60.2°C with 108

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Answer: The calorimeter's heat capacity is $6.72J/g^oC$

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This scenario assumes the amount of heat lost by the hot object equals the amount of heat gained by the cold object.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat capacity of water = $4.184J/g^oC$

$c_2$ = specific heat capacity of calorimeter =?

$m_1$ = mass of water = 108.7 g

$m_2$ = mass of calorimeter = 108.7 g

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Now substituting all provided values into the formula, we obtain

$(108.7g)\times (4.184J/g^oC)\times (35.0-60.2)^oC=-(108.7g)\times c_2\times (35.0-19.3)^oC$

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Answer:

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Solution

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1 month ago

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Answer:

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2F = 723 N

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