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3 years ago

14

The Hubble space telescope has been maintained by which of these?

1 answer:

ra1l [238]3 years ago

6 0

The answer to your question is NASA's Hubble Space Telescope<span> is the first major infrared-optical-ultraviolet</span>telescope<span> to be placed into orbit around the Earth.And it Located high above Earth's obscuring atmosphere, the </span>telescope has<span> provided the clearest views of the universe yet obtained in optical astronomy. Its in the International Space Station
Hope this helps</span>

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(science)<br>pls help, due today and I have alot more to do

Effectus [21]

Answer:

I can help

Explanation:

hold on

8 0

3 years ago

You push a heavy crate out of a carpeted room and down a hallway with a waxed linoleum floor. While pushing the crate 2.5 m out

Alinara [238K]

Answer:

W = 1562.5 J

Explanation:

Path 1:

W₁ = F₁*d₁ = 385 N * 2.5 m = 962.5 J

Path 2:

W₂ = F₂*d₂ = 130 N * 10 m = 1300 J

Path 3:

W₃ = F₃*d₃ = (-350 N) * 2 m = - 700 J (opposite to the motion)

We get

W = W₁ + W₂ + W₃ = 962.5 J + 1300 J + (- 700 J) = 1562.5 J

3 0

3 years ago

What kind of water that most used in industrial

ValentinkaMS [17]

<em>Hard </em><em>water </em><em>is </em><em>mostly </em><em>used </em><em>in </em><em>Industries </em>

3 0

3 years ago

A 591 μF capacitor is discharged through a resistor, whereby its potential difference decreases from its initial value of 88.5 V

Stels [109]

Answer:

2.6 kilo Ohm

Explanation:

Capacitance, C = 591 μF = 591 x 10^-6 F

Vo = 88.5 V

V = 11.9 V

t = 3.09 s

Let the resistance is R.

$V = V_{0}e^{\frac{-t}{RC}}$

$\frac{11.9}{88.5} = e^{\frac{-t}{RC}}$

$0.135 = e^{\frac{-t}{RC}}$

Take natural log on oth the sides

ln 0.135 = - 3.09 / RC

RC = 1.545

R = 1.545 / ( 591 x 10^-6)

R = 2614.2 ohm

R = 2.6 kilo Ohm

Thus the resistance is 2.6 kilo Ohm.

8 0

3 years ago

Ablok slides on ahorizonted sur fors which has been lubricated with heavy oil such that the blok soffers a viscous resistance th

svetoff [14.1K]

Answer:

$v = (v_0^{\frac{3}{2}}-\frac{3cx}{2m})^\frac{2}{3}$

$x = \frac{2m}{c}*v_0^{\frac{3}{2}}$

Explanation:

Given

$f(v) =- cv^\frac{1}{2}$

To start with, we begin with

$F = ma$

Substitute the expression for F

$-cv^\frac{1}{2} = ma$

$-ma = cv^\frac{1}{2}$

Acceleration (a) is:

$a = \frac{dv}{dt}$

So, the expression becomes:

$m\frac{dv}{dt} = -cv^\frac{1}{2}$

-----------------------------------------------------------------------------------------

Velocity (v) is:

$v = \frac{dx}{dt}$ --- distance/time

$v * \frac{dv}{dx}= \frac{dx}{dt}* \frac{dv}{dx}$

$v * \frac{dv}{dx}= \frac{dv}{dt}$

$\frac{dv}{dt} = v * \frac{dv}{dx}$

--------------------------------------------------------------------------------------------

So, we have:

$m\frac{dv}{dt} = -cv^\frac{1}{2}$

$mv * \frac{dv}{dx} = -cv^\frac{1}{2}$

Divide both sides by $v^\frac{1}{2}$

$mv^{1-\frac{1}{2}} * \frac{dv}{dx} = -c$

$mv^{\frac{1}{2}} * \frac{dv}{dx} = -c$

Divide both sides by m

$v^{\frac{1}{2}} * \frac{dv}{dx} = -\frac{c}{m}$

$v^{\frac{1}{2}} * dv = -\frac{c}{m} * dx$

Integrate:

$\int\limits^v_{v_0} {v^{\frac{1}{2}}} \, dv = -\frac{c}{m}\int\limits^x_0 {}} \, dx$

$\frac{2}{3}v^{\frac{3}{2}}|\limits^v_{v_0} = -\frac{c}{m}x|\limits^x_0$

$\frac{2}{3}(v^{\frac{3}{2}} - v_0^{\frac{3}{2}} ) = -\frac{cx}{m}$

$v^{\frac{3}{2}} - v_0^{\frac{3}{2}} = -\frac{3cx}{2m}$

$v^{\frac{3}{2}} = v_0^{\frac{3}{2}}-\frac{3cx}{2m}$

$v = (v_0^{\frac{3}{2}}-\frac{3cx}{2m})^\frac{2}{3}$

Next, is to get the maximum velocity by distance x.

To do this, we find the derivation by x

$\frac{dv}{dx} = 0$

$\frac{2}{3}(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{2}{3} - 1} * -\frac{3c}{2m} = 0$

$\frac{2}{3}(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{2-3}{3}} * -\frac{3c}{2m} = 0$

$\frac{2}{3}(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{-1}{3}} * -\frac{3c}{2m} = 0$

$(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{-1}{3}} * -\frac{c}{m} = 0$

Divide both sides by $-\frac{c}{m}$

$(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{-1}{3}} = 0$

Take cube roots of both sides

$(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{-1} = 0$

$v_0^{\frac{3}{2}}-\frac{3cx}{2m} = 0$

$\frac{3cx}{2m} = v_0^{\frac{3}{2}}$

$x = \frac{2m}{c}*v_0^{\frac{3}{2}}$

7 0

3 years ago