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Agata [3.3K]
3 years ago
13

The Hubble Space Telescope is an optical imaging telescope with extremely good angular resolution. Someone discovers an object t

hat Hubble cannot see at all. How might they have found it?
Engineering
1 answer:
HACTEHA [7]3 years ago
8 0

Answer:

<h2><em><u>Through</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>use</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>Hubble</u></em><em><u> </u></em><em><u>Space</u></em><em><u> </u></em><em><u>Telescope</u></em><em><u> </u></em><em><u>that's</u></em><em><u> </u></em><em><u>why</u></em><em><u> </u></em><em><u>they</u></em><em><u> </u></em><em><u>might</u></em><em><u> </u></em><em><u>find</u></em><em><u> </u></em><em><u>it</u></em><em><u> </u></em><em><u>or</u></em><em><u> </u></em><em><u>enlarging</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>resolution</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>a</u></em><em><u> </u></em><em><u>telescope</u></em><em><u>.</u></em></h2>

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kari74 [83]

Answer:

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Explanation:

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3 0
3 years ago
A mass weighing 22 lb stretches a spring 4.5 in. The mass is also attached to a damper with Y coefficient . Determine the value
Dominik [7]

Answer:

Cc= 12.7 lb.sec/ft

Explanation:

Given that

m = 22 lb

g= 32 ft/s²

m = \dfrac{22}{32}=0.6875\ s^2/ft

x= 4.5 in

1 in = 0.083 ft

x= 0.375 ft

Spring constant ,K

K=\dfrac{m}{x}=\dfrac{22}{0.375}

K= 58.66  lb/ft

The damper coefficient for critically damped system

C_c=2\sqrt{mK}

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Cc= 12.7 lb.sec/ft

5 0
3 years ago
Select the correct answer.
Lilit [14]

Answer:

D

Explanation:

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8 0
3 years ago
Read 2 more answers
A cylindrical insulation for a steam pipe has an inside radius rt = 6 cm, outside radius r0 = 8 cm, and a thermal conductivity k
goldfiish [28.3K]

Answer:

heat loss per 1-m length of this insulation is 4368.145 W

Explanation:

given data

inside radius r1 = 6 cm

outside radius r2 = 8 cm

thermal conductivity k = 0.5 W/m°C

inside temperature t1 = 430°C

outside temperature t2 = 30°C

to find out

Determine the heat loss per 1-m length of this insulation

solution

we know thermal resistance formula for cylinder that is express as

Rth = \frac{ln\frac{r2}{r1}}{2 \pi *k * L}   .................1

here r1 is inside radius and r2 is outside radius L is length and k is thermal conductivity

so

heat loss is change in temperature divide thermal resistance

Q = \frac{t1- t2}{\frac{ln\frac{r2}{r1}}{2 \pi *k * L}}

Q = \frac{(430-30)*(2 \pi * 0.5 * 1}{ln\frac{8}{6} }

Q = 4368.145 W

so heat loss per 1-m length of this insulation is 4368.145 W

4 0
3 years ago
When an object is moving, we use the following coefficient for friction calculations a)-μk b)-μs c)-γk d)- γs
Reika [66]

Answer:\mu_{k}

Explanation:

We use kinetic friction when a body is moving i.e. \mu_{k} for calculations.

Static friction is used when a body is in rest while kinetic friction is used when a body is moving and its value is quite low as compared to static friction .

Static friction value increases as we apply more force while kinetic friction occurs when there is relative motion between bodies.

3 0
3 years ago
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