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Snezhnost [94]
2 years ago
8

Use the information and model to answer the following question.

Physics
2 answers:
olga_2 [115]2 years ago
7 0

Answer:

D

Explanation:

cestrela7 [59]2 years ago
4 0

Answer:

D

Explanation:

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During a rainy day, as a result of colliding clouds an observer saw lighting and a heard thunder sound. The time between seeing
hodyreva [135]

Answer:

d = 1700 meters

Explanation:

During a rainy day, as a result of colliding clouds an observer saw lighting and a heard thunder sound.  The time between seeing the lighting and hearing the sound was 5 second, t = 5 seconds

Speed of sound, v = 340 m/s (say)

Let d is the distance of the colliding cloud from the observer. The distance covered by the object. It is given by :

d=v\times t

d=340\ m/s\times 5\ s  

d = 1700 meters

So, the distance of the colliding cloud from the observer is 1700 meters. Hence, this is the required solution.    

8 0
2 years ago
What evidence is there that the Universe began in a big bang?<br><br>​
NeTakaya

Answer: For me personally i have a Biblical faith and first of all there is evidence that looks convincing that the big bang made the universe, but man wants answers and God in the Bible says he created the universe for us it is hard to believe it but i have faith.

5 0
2 years ago
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How much power will it take to move a 50Kg box 10m across a floor that has -50N of Frictional forces in 3 seconds?
likoan [24]

Answer: 83.3 W

Explanation: I think, I’m not sure. If I’m wrong correct me ;)

8 0
3 years ago
What effect would decreasing the distance between objects have on their gravitational attraction to each other?
Lady_Fox [76]
Decreasing the distance between two objects having a considerable mass would increase the attraction on gravitation. The reverse is true that if you separate or inrease the objects distance would substantially decrease their gravitational attraction. Most object in our planet is held by its gravitational force towards it's center.
5 0
3 years ago
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a bubble of air of volume 1cm^3 is released by a deep sea diver at a depth where the pressure is 4.0 atmospheres. assuming its t
lys-0071 [83]

Answer:

hope this helps!

Explanation:

Volume of the air bubble, V1=1.0cm3=1.0×10−6m3

Bubble rises to height, d=40m

Temperature at a depth of 40 m, T1=12oC=285K

Temperature at the surface of the lake, T2=35oC=308K

The pressure on the surface of the lake: P2=1atm=1×1.103×105Pa 

The pressure at the depth of 40 m: P1=1atm+dρg

Where,

ρ is the density of water =103kg/m3

g is the acceleration due to gravity =9.8m/s2

∴P1=1.103×105+40×103×9.8=493300Pa

We have T1P1V1=T2P2V2

Where, V2 is the volume of the air bubble when it reaches the surface.

V2=

8 0
2 years ago
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