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Nastasia [14]
3 years ago
15

The fact that quasars can be detected from distances where even the biggest and most luminous galaxies cannot be seen means that

Select one: A. they must be in directions where intergalactic absorption by dark matter is minimum, allowing us to see them. B. they must be intrinsically far more luminous than the brightest galaxies. C. they have not been as redshifted by their motion as have galaxies, and hence they can still be seen. D. they must be in directions where gravitational focusing by the masses of nearer galaxies makes them visible from Earth.
Physics
1 answer:
kirill115 [55]3 years ago
6 0

Answer:

B. they must be intrinsically far more luminous than the brightest galaxies.

Explanation:

Quasar is famous for being an intergalactic object which is billions of years away from the earth yet can still be seen, unlike the other star body, unlike giant galaxies.

Hence, the fact that quasars can be detected from distances where even the biggest and most luminous galaxies cannot be seen means that "they must be intrinsically far more luminous than the brightest galaxies."

This condition, including other related evidence gotten in recent years concerning our galaxy, has shown that quasars are probably the central nuclei of very distant, very active galaxies.

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An electric motor can drive grinding wheel at two different speeds. When set to high the angular speed is 2000 rpm. The wheel tu
shutvik [7]

a) The initial angular speed is 209.3 m/s

b) The angular acceleration is -1.74 rad/s^2

c) The angular speed after 40 s is 139.7 rad/s

d) The wheel makes 1501 revolutions

Explanation:

a)

The initial angular speed of the wheel is

\omega_i = 2000 rpm

which means 2000 revolutions per minute.

We have to convert it into rad/s. Keeping in mind that:

1 rev = 2\pi rad

1 min = 60 s

We find:

\omega_i = 2000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=209.3 rad/s

b)

To find the angular acceleration, we have to convert the final angular speed also from rev/min to rad/s.

Using the same procedure used in part a),

\omega_f = 1000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=104.7 rad/s

Now we can find the angular acceleration, given by

\alpha = \frac{\omega_f - \omega_i}{t}

where

\omega_i = 209.3 rad/s is the initial angular speed

\omega_f = 104.7 rad/s is the final angular speed

t = 60 s is the time interval

Substituting,

\alpha = \frac{104.7-209.3}{60}=-1.74  rad/s^2

c)

To find the angular speed 40 seconds after the initial moment, we use the equivalent of the suvat equations for circular motion:

\omega' = \omega_i + \alpha t

where we have

\omega_i = 209.3 rad/s

\alpha = -1.74 rad/s^2

And substituting t = 40 s, we find

\omega' = 209.3 + (-1.74)(40)=139.7 rad/s

d)

The angular displacement of the wheel in a certain time interval t is given by

\theta=\omega_i t + \frac{1}{2}\alpha t^2

where

\omega_i = 209.3 rad/s

\alpha = -1.74 rad/s^2

And substituting t = 60 s, we find:

\theta=(209.3)(60) + \frac{1}{2}(-1.74)(60)^2=9426 rad

So, the wheel turns 9426 radians in the 60 seconds of slowing down. Converting this value into revolutions,

\theta = \frac{9426 rad}{2\pi rad/rev}=1501 rev

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

8 0
3 years ago
A battery produces less electrical energy than the total amount of chemical energy contained in its reactive substances. Which s
labwork [276]

Answer:

The energy conversion does not fully occur due to various energy losses.

Explanation:

There are many different types of batteries, but all have three basic components: positive electrode (cathode, or "positive terminal"), negative electrode (anode or "negative terminal"), and electrolyte.

Charging a battery forces ions from  cathode to  anode;  the battery reverses the flow. Over a period of time, this process wears out  cathode, which results in reduced capacity.

Also there can be heat loss inside the system as well which is produced by the conversion from chemical energy.

Hence the chemical energy does not fully convert to electrical energy.

3 0
4 years ago
A car is being pulled by a tow truck. The net force on the car is 3000 N and it is being pulled with an
vaieri [72.5K]

Answer:

1500 kg

Explanation:

m=F/a

m=3000/2=1500 kg

3 0
4 years ago
What is the change in the cell voltage when the ion concentrations in the cathode half-cell are increased by a factor of 10?
VMariaS [17]
When the ion concentrations in the cathode half-cell are increased by a factor of 10, the change in the cell voltage is ln10 times greater. This comes from the Nernst equation. Use variation to solve for the cell voltage in terms of the initial cell voltage before increasing the ion concentrations by 10.
8 0
4 years ago
Read 2 more answers
A tank is is half full of oil that has a density of 900 kg/m3. Find the work W required to pump the oil out of the spout. (Use 9
rusak2 [61]

Answer:

3.9 × 10^7 J

Explanation:

Given that a tank is is half full of oil that has a density of 900 kg/m3. Find the work W required to pump the oil out of the spout. (Use 9.8 m/s2 for g. Assume r = 15 m and h = 5 m.) W = 1.59 J

Solution

Since the tank is half full, the height = 2.5m

Pressure = density × gravity × height

Pressure = 900 × 9.8 × 2.5

Pressure = 22050 Pascal

The cross sectional area of the pump will be area of a circle.

A = πr^2

A = π × 15^2

A = 706.858 m^2

Using the formula

Density = mass/volume

Mass = density × volume

Mass = 900 × 706.86 × 2.5

Mass = 1590.435

Energy = mgh

Energy = 1590.435 × 9.8 × 2.5

Energy = 38965657.8 J

Since the work done = energy

Therefore, the work done = 3.9 × 10^7 J

7 0
3 years ago
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