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BlackZzzverrR [31]
3 years ago
9

We normally write our address as a street number in a city or town, which is in a state or province, followed by a country. How

would you state our cosmic address? Place its elements in sequence, starting with our planet and moving outward.- Universe- Milky Way Galaxy- Solar System- Supercluster- Earth- Local Group
Physics
1 answer:
enot [183]3 years ago
7 0

Answer:

Earth, Solar system, Milky way galaxy,Local group, super cluster, Universe.

Explanation:

We usually write our address starting from house no then street no then the town or city and after that comes the province and country. So we write address starting from smaller to larger. Let's do it in this case.First comes the planet earth which lies in the solar system. So our solar system is a smaller part of the milky way galaxy. Milky way galaxy is just a one galaxy in the cluster of galaxies which is called the local group.Local group is further a part of a super cluster. Then comes our universe which consists of 10 million super clusters.

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A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7
aleksandrvk [35]

(a) 1440.5 Hz

The general formula for the Doppler effect is

f'=(\frac{v+v_r}{v+v_s})f

where

f is the original frequency

f is the apparent frequency

v is the velocity of the wave

v_r is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

v_s is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

v_s = v_A = -28.7 m/s (surface A is the source, which is moving towards the receiver)

v_r = +62.2 m/s (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz

(b) 0.232 m

The wavelength of a wave is given by

\lambda=\frac{v}{f}

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m

(c) 1481.2 Hz

Again, we can use the same formula

f'=(\frac{v+v_r}{v+v_s})f

In the source frame (= on surface A), we have

v_s = v_B = -62.2 m/s (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

v_r = +28.7 m/s (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz

(d) 0.225 m

The wavelength of the wave is given by

\lambda=\frac{v}{f}

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m

8 0
3 years ago
The brick wall (of thermal conductivity 0.35 W/m ·◦ C) of a building has dimensions of 2.7 m by 6 m and is 16 cm thick. How much
Gwar [14]

Answer:

Q = 54.577\,MJ

Explanation:

The heat transfer through brick wall is:

\dot Q = \frac{k\cdot A}{L}\cdot \Delta T

\dot Q = \frac{\left(0.35\,\frac{W}{m\cdot ^{\circ}C} \right)\cdot (2.7\,m)\cdot (6\,m)}{0.16\,m} \cdot (31^{\circ}C - 8^{\circ}C)

\dot Q = 815.063\,W

The heat flow in a 18.6-h period is:

Q = \dot Q \cdot \Delta t

Q = (815.063\,W)\cdot (18.6\,h)\cdot \left(\frac{3600\,s}{1\,h} \right)

Q = 54576618.48\,J

Q = 54.577\,MJ

8 0
4 years ago
Read 2 more answers
two tugboats pull a disabled supertanker. each tug exerts a constant force of 2.20×106 n , one at an angle 19.0 ∘ west of north,
asambeis [7]

The amount of work the two tugboats completed on the supertanker W = 3.12 × 10^9 joules .

Work is a physics term used to describe the energy transfer that takes place when an object is moved over a distance by an external force, at least some . The component of the force acting along the path multiplied by the length of the path can be used to calculate work if the force is constant. Mathematically, this idea is expressed as W = fd, where W is the work and f is the force multiplied by d, the distance. Work is completed when the force is applied at an angle of with respect to the displacement. Performing work on a body involves moving it in its entirety from one location to another as well as other methods.  However, the work is thought to be negative if the applied force is in opposition to the item's motion, indicating that energy is being pulled away from the object

Learn more about work here:

brainly.com/question/1374468

#SPJ4

7 0
1 year ago
At 2 P.M., ship A is 150 km west of ship B. Ship A is sailing east at 35 km/h and ship B is sailing north at 25 km/h. How fast i
labwork [276]

Answer:

The distance between the ships changing at 6PM is 21.29Km/h

Explanation:

Ship A is sailing east at 35Km/h and ship B is sailing West at 25Km/h

Given

dx/dt= 35

dy/dt= 25

dv/dt= ???? at t= 6PM - 2PM= 4

Therefore t=4

We know ship A travels at 150km in the x-direction and Ship A at t=4 travels at 4.35 Which is 140 also in x-direction

So, we use:

D^2 = (150 - x)^2 + y^2;

D^2 = (150 - 140)^2 + y^2

But ship B travels at t=4, at 4.25 =100 in the y-direction

so, let's use the equation:

D^2 = 10^2 + 100^2

= D= sqrt*(10 + 100)

Lets use 2DD' = 2xx' + 2yy'

Differentiating with respect to t we have:

D•d(D)/dt = -(10)•dx/dt + 100•dy/dt

=100.5 d(D)/dt = (-10)•35 + (100)•25

When t=4, we have x=(140-150) =10 and y=100

= D = sqrt*(10^2 + 100^2)

=100.5

= 100.5 dD/dt = 10.35 +100.25

= dD/dt = 21.29km/h

7 0
3 years ago
 15cm3 block of gold weighs  2.8N it is carefully submerged in a tank of mercury. one cm3 of mercury weight 0.!3n a will the mer
NNADVOKAT [17]
You have said that 15cm³ of gold weighs 2.8N.  So I may infer that each cm³
of gold weighs about 0.19N.  When I compare that figure with the 0.13N per cm³
of mercury, it becomes immediately apparent that the gold is more dense than
mercury. Therefore, the sample of gold, no matter what its size or weight, will
displace its total volume of mercury, and will go on to sink entirely beneath the
waves in the mercury.
3 0
3 years ago
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