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svp [43]
3 years 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:
valina [46]3 years ago
8 0
The speed of water can be split into vertical and horizontal speed components:
v_x = 6.5 cos \theta \\ v_y = 6.5 sin \theta

Due to the force of gravity, the y component will be parabolic. The x component will be linear:
y(t) = -4.9t^2 + (6.5sin \theta) t \\  \\ x(t) = (6.5 cos \theta) t
To find when the water hits 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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How many lenses with different focal lengths can be obtained by combining two surfaces whose radii of curvature are 4.00 cm and
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Answer:

The lenses with different focal length are four.

Explanation:

Given that,

Radius of curvature R₁= 4

Radius of curvature R₂ = 8

We know ,

Refractive index of glass = 1.6

When, R₁= 4, R₂ = 8

We need to calculate the focal length of the lens

Using formula of focal length

\dfrac{1}{f}=(n-1)(\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}})

Put the value into the formula

\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}+\dfrac{1}{8})

\dfrac{1}{f}=\dfrac{9}{40}

f=4.44\ cm

When , R₁= -4, R₂ = 8

Put the value into the formula

\dfrac{1}{f}=(1.6-1)(\dfrac{1}{-4}+\dfrac{1}{8})

\dfrac{1}{f}=-\dfrac{3}{40}

f=-13.33\ cm

When , R₁= 4, R₂ = -8

Put the value into the formula

\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}-\dfrac{1}{8})

\dfrac{1}{f}=\dfrac{3}{40}

f=13.33\ cm

When , R₁= -4, R₂ = -8

Put the value into the formula

\dfrac{1}{f}=(1.6-1)(\dfrac{1}{-4}-\dfrac{1}{8})

\dfrac{1}{f}=-\dfrac{9}{40}

f=-4.44\ cm

Hence, The lenses with different focal length are four.

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An oscilloscope shows a steady sinusoidal signal of 5 Volt peak to peak, which spans 5 cm in vertical direction on the screen. B
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A car with a velocity of 22 m/s is accelerated at a rate of 1.6m/s2 for 6.8s. determine the final velocity
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A car with a velocity of 22 m/s is accelerated at a rate of 1.6 m/s^2 for 6.8s has the final velocity t be 32.88 m/s.

The acceleration means the amount of velocity changing per unit time.

The given data:

initial velocity, u = 22 m/s

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We will be using the equation of motion:

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The final velocity become 32.88 m/s.

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