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Dominik [7]
3 years ago
5

A firefighting crew uses a water cannon that shoots water at 25.0 m/s at a fixed angle of 53.0° above the horizontal. The fire-f

ighters want to direct the water at a blaze that is 10.0 m above ground level. How far from the building should they position their cannon? There are two possibilities; can you get them both? (Hint: Start with a sketch showing the trajectory of the water.)
Physics
1 answer:
zysi [14]3 years ago
8 0

Answer:

8.8 m and 52.5 m

Explanation:

The vertical component and horizontal component of water velocity leaving the hose are

v_v = vsin(\alpha) = 25sin(53^0) = 25*0.8 = 19.97 m/s

v_h = vcos(\alpha) = 25cos(53^0) = 25*0.6 = 15 m/s

Neglect air resistance, vertically speaking, gravitational acceleration g = -9.8m/s2 is the only thing that affects water motion. We can find the time t that it takes to reach the blaze 10m above ground level

s = v_vt + gt^2/2

10 = 19.97t - 9.8t^2/2

4.9t^2 - 19.97t + 10 = 0

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

t= \frac{19.9658877511823\pm \sqrt{(-19.9658877511823)^2 - 4*(4.9)*(10)}}{2*(4.9)}

t= \frac{19.9658877511823\pm14.24}{9.8}

t = 3.49 or t = 0.58

We have 2 solutions for t, one is 0.58 when it first reach the blaze during the 1st shoot up, the other is 3.49s when it falls down

t is also the times it takes to travel across horizontally. We can use this to compute the horizontal distance between the fire-fighters and the building

s_1 = v_ht_1 = 15*0.58 = 8.8 m

s_2 = v_ht_2 = 15*3.49 = 52.5m

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Hope this helped!
5 0
3 years ago
HELP. A river flows South with a speed of 8m/s. From the shore, you see a boat moving South with a speed of 10 m/s. From their f
riadik2000 [5.3K]

Answer:

18 m/s

Explanation:

Given that,

Speed of river is 8 m/s due South

From the shore, you see a boat moving South with a speed of 10 m/s. Both boat and river are moving in same direction. The resultant velocity will add up.

v = 10 m/s + 8 m/s

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3 0
3 years ago
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8 0
3 years ago
Read 2 more answers
A plank of length L=2.200 m and mass M=4.00 kg is suspended horizontally by a thin cable at one end and to a pivot on a wall at
baherus [9]

Hello!

Let's begin by doing a summation of torques, placing the pivot point at the attachment point of the rod to the wall.

\Sigma \tau = 0

We have two torques acting on the rod:
- Force of gravity at the center of mass (d = 0.700 m)

- VERTICAL component of the tension at a distance of 'L' (L = 2.200 m)

Both of these act in opposite directions. Let's use the equation for torque:
\tau = r \times F

Doing the summation using their respective lever arms:

0 = L Tsin\theta  - dF_g

dF_g = LTsin\theta

Our unknown is 'theta' - the angle the string forms with the rod. Let's use right triangle trig to solve:

tan\theta = \frac{H}{L}\\\\tan^{-1}(\frac{H}{L}) = \theta\\\\tan^{-1}(\frac{1.70}{2.200}) = 37.69^o

Now, let's solve for 'T'.

T = \frac{dMg}{Lsin\theta}

Plugging in the values:
T = \frac{(0.700)(4.00)(9.8)}{(2.200)sin(37.69)} = \boxed{20.399 N}

3 0
1 year ago
2. What energy source produces most of the electrical energy in the United States?
Minchanka [31]

Answer:

Nat gas

Explanation:

Approx:

Natural gas 38%       coal 21%       oil  19%

6 0
2 years ago
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