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3 years ago

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An airplane is fl ying with a velocity of 240 m/s at an angle of 30.0° with the horizontal, as the drawing shows. When the altit

ude of the plane is 2.4 km, a fl are is released from the plane. The fl are hits the target on the ground. What is the angle θ?

1 answer:

netineya [11]3 years ago

4 0

Answer:

$\theta=41.52^{\circ}$

Explanation:

Given that,

Velocity of the airplane, v = 240 m/s

Angle with horizontal, $\theta=30^{\circ}$

The altitude of the plane is 2.4 km, d = 2400 m

Vertical speed of the airplane, $v_y=v\ sin\theta=240\ sin(30)=120\ m/s$

Horizontal speed of the airplane, $v_x=v\ cos\theta=240\ sin(30)=207.84\ m/s$

So, the equation of the projectile for the flare is given by :

$4.9t^2+120t-2400=0$

On solving the above equation, we get the value of t as:

t = 13.04 seconds

Horizontal distance travelled,

$d=v_x\times t$

$d=207.84\times 13.04$

d = 2710.23 m

Let $\theta$ is the angle with which it hits the target. So,

$tan\theta=\dfrac{2400}{2710.23}$

$\theta=41.52^{\circ}$

Hence, this is the required solution.

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Answer:

frequency = 1475.45 Hz

Explanation:

given data

frequency f1 = 1215 Hz,

frequency f2 = 1265 Hz

police car moving vp = 25.0 m/s

solution

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speed of the other car = v

when the police car is stationary

the frequency the other car receives is

f2 = f1 × $\dfrac{u+v}{u}$ ................1

and

the frequency the police car receives is

f2 = f1 × $\dfrac{u}{u-v}$ ..................2

now from equation 1 and 2

$\frac{f2}{f1} = \dfrac{u+v}{u-v}$

$\frac{1275}{1240} = \frac{u+v}{u-v}$

$v =\frac{1275-1240}{1275+1240}\times 343$

v = 4.77 m/s

and

frequency the other car receives is

f2 = f1 × $\dfrac{u+v}{u-vp}$ ......................3

and

the frequency the police car receives is

f2 = f1 × $\dfrac{u+vp}{u - v}$ .......................4

now we get

f2 = f1 × $\dfrac{(u+v)(u + vp)}{(u-v)(u-vp)}$

f2 = $1240\times \frac{(343+4.77)(343+25)}{(343-4.77)(343-25)}$

f2 = 1475.45 Hz

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Answer:

Large binocular telescope, Keck 1 telescope, Hobby-Ebberly telescope, Subaru telescope, Gemini North telescope, Magellan 2 telescope

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How much light a telescope can collect depends on its diameter, since in a bigger area more photons will be collected.

Remember that in a circle the area is defined as:

$A = \pi r^{2}$ (1)

Where A is the area and r is its radius.

However, the radius can be determined by means of its diameter.

$d = 2r$

$r = \frac{d}{2}$ (1)

Where d is its diameter.

An example of this is when a person is collecting raindrops with a bucket and with a cup. Since the bucket has a bigger area than the cup, it will collect more raindrops by unit of time. In this scenario the raindrops represent the photons.

To determine the light collecting area of each telescope, equation 2 will be replaced in equation 1.

$A = \pi (\frac{d}{2})^{2}$ (3)

Case for Large binocular telescope:

$A_{mirror1} = \pi (\frac{8.4m}{2})^{2}$

$A_{mirror1} = 55.41m$

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$A = 55.41m+55.41m$

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$A = \pi (\frac{10m}{2})^{2}$

$A = 78.53m$

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$A = \pi (\frac{9.2m}{2})^{2}$

$A = 66.47m$

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$A = \pi (\frac{8.3m}{2})^{2}$

$A = 54.10m$

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$A = \pi (\frac{8m}{2})^{2}$

$A = 50.26m$

Case for Magellan 2 telescope:

$A = \pi (\frac{6.5m}{2})^{2}$

$A = 33.18m$

Hence, they may be rank in the following way:

Large binocular telescope, Keck 1 telescope, Hobby-Ebberly telescope, Subaru telescope, Gemini North telescope, Magellan 2 telescope.

<em>Key term:</em>

<em>Photons: particles that constitute light. </em>

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jeka94

Answer:

T = 188.5 s, correct is C

Explanation:

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L₀ = r m v₀ + I₀ w₀

the angular speed of the fan is zero w₀ = 0

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L_f = I₀ w + m r v

L₀ = L_f

m r v₀ = I₀ w + m r v

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v = r w

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m r v₀ = (I₀ / r + mr) v

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v = x / T

T = x / v

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x = 2π r

we substitute

T = 2π r / v

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T = 2π 0.6/0.02

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