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maw [93]
3 years ago
6

How much greater is the light-collecting area of a 6-meter telescope than a 3-meter telescope? How much greater is the light-col

lecting area of a 6- telescope than a 3- telescope?
A) two times
B) four times
C) six times
Physics
1 answer:
vovikov84 [41]3 years ago
7 0

Answer:

B) four times

Explanation:

Area of 6 meter telescope

A_1=\pi \frac{D^2}{4}\\\Rightarrow A_1=9\pi

Area of 3 meter telescope

A_2=\pi \frac{D^2}{4}\\\Rightarrow A_1=2.25\pi

Dividing the equations we get

\frac{A_1}{A_2}=\frac{9\pi}{2.25\pi}\\\Rightarrow A_1=4A_2

Hence, the 6-meter telescope has 4 times the light-collecting area of the 3-meter telescope.

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A rocket accelerates by burning its onboard fuel, so its mass decreases with time. Suppose the initial mass of the rocket at lif
Serggg [28]

Answer:

Height of the rocket be one minute after liftoff is 40.1382 km.

Explanation:

v(t)=-gt-v_e\times \ln \frac{m-rt}{m}

v = velocity of rocket at time t

g = Acceleration due to gravity =9.8 m/s^2

v_e = Constant velocity relative to the rocket = 2,900m/s.

m = Initial mass of the rocket at liftoff = 29000 kg

r = Rate at which fuel is consumed = 170 kg/s

Velocity of the rocket after 1 minute of the liftoff =v

t = 1 minute = 60 seconds'

Substituting all the given values in in the given equation:

v(60)=-9.8 m/s^2\times 60 s-2,900m/s\times \ln (\frac{29,000 kg-170 kg/s\times 60 s}{2,9000 kg})

v(60) = 668.97 m/s

Height of the rocket = h

Velocity=\frac{Displacement}{time}

668.97 m/s=\frac{h}{60 s}

h=668.97 m/s\times 60 s=40,138.2 m = 40.1382 km

Height of the rocket be one minute after liftoff is 40.1382 km.

4 0
3 years ago
A stone is dropped from the upper observation deck of a tower, "500" m above the ground. (Assume g = 9.8 m/s2.) (a) Find the dis
Yuri [45]

Answer:

h₍₁₎ = 495,1 meters

h₍₂₎ = 480,4 m

h₍₃₎ = 455,9 m

...

..

Explanation:

The exercise is "free fall". t = \sqrt{\frac{2h}{g} }

Solving with this formula you find the time it takes for the stone to reach the ground (T) = 102,04 s

The heights (h) according to his time (t) are found according to the formula:

h(t) = 500 - 1/2 * g * t²

Remplacing "t" with the desired time.

4 0
3 years ago
A pilot heads his jet due east. The jet has a speed of 475 mi/h relative to the air. The wind is blowing due north with a speed
oksian1 [2.3K]

Explanation:

a. The velocity of the wind as a vector in component form will be represented as v vector:

    v=30j

b.The velocity of the jet relative to the air as a vector in component form will be represented as u vector

    u=475i

c. The true velocity of the jet as a vector will be represented as w:

  w=u+v

  w=475i+30j

d.  The true speed of the jet will be calculated as:

    IwI=\sqrt{(475)^2+(30)^2}

    IwI=\sqrt{225625+900}

    IwI=\sqrt{226525}

    IwI=476 mi/h

e. The direction of the jet will be:

tita=tan^{-1}\frac{30}{475}

tita=tan^{-1}(0.0632)

tita=3.62degrees,or,N86.38degreesS

7 0
3 years ago
A 30-gram bullet is fired and a 50-gram bullet is dropped simultaneously from the same height. Which will hit the ground first?
Marina86 [1]

Answer: the 30gram will hit the ground first

Explanation:the 30gram bullet will hit the ground first because it is fired

4 0
3 years ago
Read 2 more answers
A person throws a ball straight up in the air. The ball rises to a maximum height and then falls back down so that the person ca
Lana71 [14]

Answer:

The acceleration is about 9.8 m/s2 (down) when the ball is falling.

Explanation:

The ball at maximum height has velocity zero

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s² (positive downward and negative upward)

v=u+at\\\Rightarrow 0=u-9.8\times t\\\Rightarrow u=9.8t

The accleration 9.8 m/s² will always be acting on the body in opposite direction when the body is going up and in the same direction when the body is going down. The acceleration on the body will never be zero

5 0
3 years ago
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