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3 years ago

9

Quasars are thought to be the nuclei of active galaxies in the early stages of their formation. A typical quasar radiates energy

at the rate of 1040 W. At what rate is the mass of this quasar being reduced to supply this energy? Express your answer in solar mass units per year, where one solar mass unit (1 smu = 2.0 ✕ 1030 kg) is the mass of our Sun.

1 answer:

uranmaximum [27]3 years ago

6 0

Answer:

1.7531 smu/yn

Explanation:

check the attached file below for answer and explanation.

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When hydrogen peroxide antiseptic is tightly capped and left on the shelf for a long period of time, the bottle is observed to s

viktelen [127]

Answer:

The hydrogen peroxide decomposes as: $H_2O_2\rightarrow 2H_2O+O_2$

Explanation:

Hydrogen peroxide antiseptic although tightly capped but left on the shelf for a long time leads to swollen bottle and opens with a hissing sound because hydrogen peroxide reacts with air in presence of light to dissociate into water and oxygen molecules that leads to high volume of content in the tightly closed bottle and formation of water molecules from hydrogen peroxide.

Once the hydrogen peroxide has turned into water molecules it is no longer effective as an antiseptic.

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3 years ago

How does a distance-time graph show you which direction the runner is moving? Use your own words and science vocabulary.

Dmitry [639]

In a distance time graph, when the x and y values are positive (in first quadrant), the runner is moving forward.

While, if the distance value ( in the y axis) is negative, the runner is moving backwards ( towards the start) .

Hope it helps :)

8 0

3 years ago

A hiker walks 20.51 m at 33.16 degrees. What is the Y component of his displacement?

Serhud [2]

Answer:

<em>The y component of his displacement is 11.22 meters</em>

Explanation:

<u>Components of the displacement</u>

The displacement is a vector because it has a magnitude and a direction. Let's suppose a displacement has a magnitude r and a direction θ, measured with respect to the positive x-direction. The horizontal component of the displacement is calculated by:

$x=r\cos\theta$

The vertical component is calculated by:

$y=r\sin\theta$

The hiker has a displacement with magnitude r = 20.51 m at an angle of 33.16 degrees. Substituting in the above equation:

$y=20.51\sin(33.16^\circ)$

$y=11.22\ m$

The y component of his displacement is 11.22 meters

7 0

3 years ago

(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup

KengaRu [80]

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

$W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx$

Where

$x_{o}$ , $x_{f}$ - Initial and final position, respectively, measured in meters.

$F(x)$ - Force as a function of position, measured in newtons.

Given that $F = k\cdot x$ and the fact that $F = 25\,N$ when $x = 0.3\,m - 0.2\,m$ , the spring constant ( $k$ ), measured in newtons per meter, is:

$k = \frac{F}{x}$

$k = \frac{25\,N}{0.3\,m-0.2\,m}$

$k = 250\,\frac{N}{m}$

Now, the work function is obtained:

$W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx$

$W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]$

$W = 0.313\,J$

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be $r(\theta) = 2\cdot \sin 5\theta$ . The area of the region enclosed by one loop of the curve is given by the following integral:

$A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta$

$A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta$

By using trigonometrical identities, the integral is further simplified:

$A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta$

$A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta$

$A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta$

$A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)$

$A = 4\pi$

The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

5 0

3 years ago

Which statement best describes why an island's food web could be considered a closed system?

garik1379 [7]

I would say Option B) because Option C) is wrong since matter cannot be created. A closed system does not exchange matter so it's not Option D). Since an island is an isolated area, Option A) is wrong.

3 0

3 years ago

Read 2 more answers