4. With a diameter that's 11 times larger than Earth's, ______

Which of the following is a form of pollution created when vehicle exhaust interacts with sunlight?

SVEN [57.7K]

Photochemical smog is formed when primary air pollutants interact with sunlight.

Photochemical smog is the result of the reaction between pollutants like nitrogen oxides (NO), sunlight and volatile organic compound (VOC) in the atmosphere. The sources of NO are car exhaust, coal power plants, factory emissions, etc. This type of smog is also known by the name Los Angeles smog.

Air pollutants are the particles present dissolved in the air, which when inhaled by the organisms can cause serious health issues. These pollutants are :ozone, particulate matter, gaseous oxides, etc. These pollutants majorly affect the respiratory system of the humans.

Therefore, photochemical smog is a form of pollution created when vehicle exhaust interacts with sunlight.

To know more about photochemical smog, here: brainly.com/question/15728274

#SPJ4

8 0

1 year ago

An 85,000 kg stunt plane performs a loop-the-loop, flying in a 260-m-diameter vertical circle. at the point where the plane is f

konstantin123 [22]

A) When the plane is flying straight down, there are three forces acting on it:
- the centripetal force $F=m \frac{v^2}{r}$ , directed toward the center of the circle (so, horizontally)
- the weight of the plane: $W=mg$ , downward, so vertically
- a third force, given by the propulsion of the plane, which is accelerating it towards the ground (because the problem says that the plane has an acceleration of a=12 m/s^2 towards the ground)

The radius of the circle is $r= \frac{260 m}{2} = 130 m$ , so the centripetal force acting on the plane is
$F_c=m \frac{v^2}{r} = \frac{(85000 kg)(55 m/s)^2}{130 m}=1.98 \cdot 10^6 N$
On the vertical axis, we have two forces: the weight
$W=mg=(85000 kg)(9.81 m/s^2)=8.34 \cdot 10^5 N$
and the other force F given by the propulsion. Since we know that their sum should generate an acceleration equal to $a=12 m/s^2$ , we can find the magnitude of this other force F by using Newton's second law:
$F+mg=ma$
$F=m(a-g)=(85000kg)(12 m/s^2-9.81 m/s^2)=1.86 \cdot 10^5 N$

So, the net force acting on the plane will be the resultant of the centripetal force (acting in the horizontal direction) and the two forces W and F (acting in the vertical direction):
$R= \sqrt{(F_c^2+(W+F)^2}=$
$= \sqrt{(1.98\cdot 10^6N)^2+(8.34 \cdot 10^5N+1.86 \cdot 10^5 N)^2} =2.23 \cdot 10^6 N$

(b) The tangent of the angle with respect to the horizontal is the ratio between the sum of the forces in the vertical direction (taken with negative sign, since they are directed downward) and the forces acting in the horizontal direction, so:
$\tan \theta = \frac{-(W+F)}{F_c}= -0.5$
And so, the angle is
$\theta = \arctan (-0.5)=-26.8 ^{\circ}$

7 0

3 years ago

A capacitor with a very large capacitance is in series with a capacitor that has a very small capacitance. what can we say about

Misha Larkins [42]

<span>A capacitor with a very large capacitance is in series with a capacitor
that has a very small capacitance.

The capacitance of the series combination is slightly smaller than the
capacitance of the small capacitor. (choice-C)

The capacitance of a series combination is

1 / (1/A + 1/B + 1/C + 1/D + .....) .

If you wisk, fold, knead, and mash that expression for a while,
you find that for only two capacitors in series, (or 2 resistors or
two inductors in parallel), the combination is

(product of the 2 individuals) / (sum of the individuals) .

In this problem, we have a humongous one and a tiny one.
Let's call them 1000 and 1 .
Then the series combination is

(1000 x 1) / (1000 + 1)

= (1000) / (1001)

= 0.999 000 999 . . .

which is smaller than the smaller individual.

It'll always be that way. </span>

5 0

3 years ago

The Hubble Space Telescope in orbit above the Earth has a 2.4 m circular aperture. The telescope has equipment for detecting ult

strojnjashka [21]

Answer:

Option d

The minimum angular separation between two objects that the Hubble Space Telescope can resolve is $4.8x10^{-8}rad$ .

Explanation:

The resulting image in a telescope that will be gotten from an object is a diffraction pattern instead of a perfect point (point spread function (PSF)).

That diffraction pattern is gotten because the light encounters different obstacles on its path inside the telescope (interacts with the walls and edges of the instrument).

The diffraction pattern is composed by a central disk, called Airy disk, and diffraction rings.

The angular resolution is defined as the minimal separation at which two sources can be resolved one for another, or in other words, when the distance between the two diffraction pattern maxima is greater than the radius of the Airy disk.

The angular resolution can be determined in analytical way by means of the Rayleigh criterion.

$\theta = 1.22\frac{\lambda}{D}$ (1)

Where $\lambda$ is the wavelength and D is the diameter of the telescope.

Notice that it is necessary to express the wavelength in the same units than the diameter.

$\lambda = 95nm \cdot \frac{1x10^{-9}m}{1nm}$ ⇒ $9.5x10^{-8}m$

Finally, equation 1 can be used.

$\theta = 1.22(\frac{9.5x10^{-8}m}{2.4m})$

$\theta = 4.8x10^{-8}rad$

Hence, the minimum angular separation between two objects that the Hubble Space Telescope can resolve is $4.8x10^{-8}rad$ .

5 0

4 years ago

When we apply the energy conversation principle to a cylinder rolling down an incline without sliding, we exclude the work done

NikAS [45]

Answer:

D. the linear velocity of the point of contact (relative to the inclined surface) is zero

Explanation:

The force of friction emerges only when there is relative velocity between two objects . In case of perfect rolling , there is no sliding so relative velocity between the surface and the point of contact is zero . In other words the velocity of point of contact becomes zero , even though , the whole body is in linear motion . It happens due point of contact having two velocities which are equal and opposite . One of the velocity is in forward direction and the other velocity which is due to rotation is in backward direction . So net velocity of point of contact becomes zero . Due to absence of sliding , displacement due to friction becomes zero . Hence work done by friction becomes zero.

5 0

3 years ago

4. With a diameter that's 11 times larger than Earth's, _______ is the largest planet.