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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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Air flows steadily through a variable sized duct in a heat transfer experiment with a speed of u = 20 – 2x, where x is the dista
zysi [14]

Answer:

a) a=-28m/s^{2}

b) \frac{dT}{dx}=-5 ^{o}C/m

Explanation:

a)

In order to solve this problem, we need to start by remembering how the acceleration is related to the velocity of a particle. We have the following relation:

a=\frac{dv}{dt}

in other words, the acceleration is defined to be the derivative of the velocity function with respect to time. So let's take our speed function:

u=20-2x

if we take its derivative we get:

du=-2dx

this is the same as writting:

\frac{du}{dt}=-2\frac{dx}{dt}

we also know that velocity is defined to be:

u=\frac{dx}{dt}

so we get that:

a=-2u

when substituting we get that:

a=-2(20-2x)

when expanding we get:

a=-40+4x

and now we can use this equation to find our acceleration at x=3, so:

a=-40+4(3)

a=-40+12

a=-28 m/s^{2}

b)

the same applies to this problem with the difference that this will be the rate of change of the temperature per m. So we proceed and take the derivative of the temperature function:

T=200-5x

\frac{dT}{dx}=-5

so the rate of change is -5^{o}C/m

7 0
3 years ago
Sea water with a density of 1025 kg/m3 flows steadily through a pump at 0.21 m3 /s. The pump inlet is 0.25 m in diameter. At the
myrzilka [38]

Answer:

\dot W_{pump} = 16264.922\,W\,(16.265\,kW)

Explanation:

The pump is modelled after applying Principle of Energy Conservation, whose form is:

\frac{P_{1}}{\rho\cdot g}+ \frac{v_{1}^{2}}{2\cdot g} +z_{1} + h_{pump}=\frac{P_{2}}{\rho\cdot g}+ \frac{v_{2}^{2}}{2\cdot g} +z_{2}

The head associated with the pump is cleared:

h_{pump} = \frac{P_{2}-P_{1}}{\rho\cdot g}+\frac{v_{2}^{2}-v_{1}^{2}}{2\cdot g}+(z_{2}-z_{1})

Inlet and outlet velocities are found:

v_{1} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.25\,m)^{2} }

v_{1} \approx 4.278\,\frac{m}{s}

v_{2} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.152\,m)^{2} }

v_{2} \approx 11.573\,\frac{m}{s}

Now, the head associated with the pump is finally computed:

h_{pump} = \frac{175\,kPa-81.326\,kPa}{(1025\,\frac{kg}{m^{3}} )\cdot (9.807\,\frac{m}{s^{2}} )} +\frac{(11.573\,\frac{m}{s} )^{2}-(4.278\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )} + 1.8\,m

h_{pump} = 7.705\,m

The power that pump adds to the fluid is:

\dot W_{pump} = \dot V \cdot \rho \cdot g \cdot h_{pump}

\dot W_{pump} = (0.21\,m^{3})\cdot (1025\,\frac{kg}{m^{3}})\cdot (9.807\,\frac{m}{s^{2}})\cdot(7.705\,m)

\dot W_{pump} = 16264.922\,W\,(16.265\,kW)

4 0
3 years ago
A five legged intersection has safety issues. in 2018, the number of crashes reported was 48 and the average 24 hour volume ente
sashaice [31]

Answer:

The crash rate is 22 vehicles per 1 million vehicles

Explanation:

In this question, we are asked to determine the crash rate per million vehicles.

Crash rate is calculated using average crash frequency.

The crash rates are calculated based on the number of crashes per million vehicle miles travelled in a year.

Mathematically;

crash rate = (48 * 1,000,000)/ [(980 + 1560 + 1230 + 900 + 1435)* 365]

= 48,000,000/(6105*365)

= 21.54 approximately 22

4 0
4 years ago
Consider a 1.80-m-tall man standing vertically in water and completely submerged in a pool. Determine the difference between the
defon

Answer:

17.658 kPa

Explanation:

The hydrostatic pressure of a fluid is the weight of a column of that fluid divided by the base of that column.

P = \frac{weight}{base}

Also, the weight of a column is its volume multiplied by it's density and the acceleration of gravity:

weight = \delta * v * g

Meanwhile, the volume of a column is the area of the base multiplied by the height:

V = base * h

Replacing:

P = \frac{\delta * base * h * g}{base}

The base cancels out, so:

P = \delta * h * g

The pressure depends only on the height of the fluid column, the density of the fluid and the gravity.

If you have two point at different heights (or depths in the case of objects submerged in water) each point will have its own column of fluid exerting pressure on it. Since the density of the fluid and the acceleration of gravity are the same for both points (in the case of hydrostatics density is about constant for all points, it is not the case in the atmosphere), we can write:

\Delta P = \rho * g * (h1 - h2)

We do not know at what depth the man of this problem is, but it doesn't matter, because we know the difference in height of the two points of interes (h1 - h2) = 1.8 m. So:

\Delta P = 9.81 \frac{m}{s^{2} } * 1000 \frac{kg}{m^3} * 1.8 m = 17658 Pa = 17.658 kPa

4 0
3 years ago
Who knows about welding ??
shtirl [24]
They joint together melting metal
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4 years ago
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