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olchik [2.2K]
4 years ago
15

A fire hose held near the ground shoots water at a speed of 6.5 m/s. At what angle(s) should the nozzle point in order that the

water land 2.5 m away (Fig. 3–36)? Why are there two different angles? Sketch the two trajectories.
Physics
1 answer:
Mariana [72]4 years ago
8 0

Answer:

17.72° or 72.28°

Explanation:

u = 6.5 m/s

R = 2.5 m

Let the angle of projection is θ.

Use the formula for the horizontal range

R=\frac{u^{2}Sin2\theta }{g}

2.5=\frac{6.5^{2}Sin2\theta }{9.8}

Sin 2θ = 0.58

2θ = 35.5°

θ = 17.72°

As we know that the range is same for the two angles which are complementary to each other.

So, the other angle is 90° - 17.72° = 72.28°

Thus, the two angles of projection are 17.72° or 72.28°.

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0.500 kg of ethyl alcohol at 33.0°C
Bingel [31]

Answer:

26.06

Explanation:

One thing to note first is that, after the ethyl alcohol and water is mixed, the heat energy (Q) is equal for both of them. So:    

<h3>m c ΔT (water) = m c ΔT (ethyl) </h3>

Second thing to note is that ΔT is just the absolute value of the final temperature of the solution minus the initial temperature of the ethyl, or water. So the previous equation becomes:

<h3>m c (Tfin - Tinit) (water) = m c (Tfin - Tinit) (ethyl) </h3>

And since we know the mass and specific heat of the water and ethyl alcohol, we can substitute those in:

<h3>2093(Tfin - 22) = 1225(33 - Tfin) </h3>

(I wrote 33 - Tfin for ethyl because ΔT should be a positive number in this case)

Now finally we can expand out the parenthesis and use a little algebra to solve:

<h3>2093Tfin - 46046 = 40425 - 1225Tfin</h3>

Now use algebra to separate the Tfin coefficients onto one side and everything else on the other, and we get:

<h3>3318Tfin = 86471</h3>

And when you divide both sides be 3318, you get your answer of:

<h3>Tfin = 26.06118 or 26.1 for significant figures</h3>

Hope that helps!

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<h2>Answer:</h2>

<h2>3m</h2>

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