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nataly862011
3 months ago
12

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 2.5m away?
Physics
1 answer:
inna [3.1K]3 months ago
8 0
The velocity of water can be decomposed into its vertical and horizontal components:
v_x = 6.5 cos \theta \\ v_y = 6.5 sin \theta

The vertical component will exhibit a parabolic trajectory due to gravity, while the horizontal component will be linear:
y(t) = -4.9t^2 + (6.5sin \theta) t \\ \\ x(t) = (6.5 cos \theta) t
To determine when the water reaches the ground 2.5m away, set y= 0 and x = 2.5
-4.9t^2 + (6.5sin \theta) t=0 \\ \\ t = \frac{6.5}{4.9} sin \theta \\ \\(6.5 cos \theta)(\frac{6.5}{4.9} sin \theta) = 2.5 \\ \\ sin \theta cos \theta = 0.29 \\ \\ sin 2\theta = 0.58 \\ \\ 2\theta = 35.4, 144.6 \\ \\ \theta = 17.7,72.3
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A convex mirror with a focal length of 0.25 m forms a 0.080 m tall image of an automobile at a distance of 0.24 m behind the mir
ValentinkaMS [3465]

Answer:

The object measures 6 m in distance and 2 m in height.

It creates a virtual image that is upright.

Explanation:

Provided data includes:

Focal length = 0.25 m

Image height = 0.080 m

Image distance = 0.24 m

We are to determine the object's distance.

Using the lens formula:

\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}

Substituting values into the formula:

\dfrac{1}{0.24}=\dfrac{1}{0.25}+\dfrac{1}{u}

\dfrac{1}{u}=\dfrac{1}{0.24}-\dfrac{1}{0.25}

\dfrac{1}{u}=\dfrac{1}{6}

u=6\ m

We also need to calculate magnification:

Applying the magnification formula:

m=-\dfrac{v}{u}

Substituting values into this formula:

m=-\dfrac{0.24}{-6}

m=0.04

Next, we need to find the height of the object:

Using the magnification formula once more:

m=\dfrac{h'}{h}

h=\dfrac{0.080}{0.04}

h=2\ m

A convex mirror generates a virtual and upright image on its backside.

Consequently, the object is at a distance of 6 m and has a height of 2 m.

The image formed is virtual and upright.

6 0
3 months ago
Read 2 more answers
An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
inna [3103]

Answer:

the time it takes after impact for the puck is 2.18 seconds

Explanation:

initially given information

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thickness = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

temperature of air = 15°C

to determine

time needed for the puck to reduce its speed by 10%

solution

we note that velocity changes from 0 to v

assuming initial velocity = v

therefore final velocity = 0.9v

implying a change in velocity is du = v

and clearance dy = h

shear stress acting on the surface is expressed as

= µ \frac{du}{dy}

therefore

= µ \frac{v}{h}............1

substituting the values

= 1.75 ×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

the area between the air and puck is given by

Area = \frac{\pi }{4} d^{2}

area = \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

thus, the force on the puck can be represented as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v

now applying Newton's second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

solving for t = \frac{0.1 v * 0.03}{1.37*10^{-3} v}

the time needed after impact for the puck is 2.18 seconds

3 0
3 months ago
Which of the following four circuit diagrams best represents the experiment described in this problem?
inna [3103]

There's an absence of circuit diagrams.  

Initially, this causes worry for a moment, until we remember that we have no understanding of the experiment mentioned in the problem either, rendering such worries unnecessary.

6 0
2 months ago
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