Answer:
Distance from the moon= dm= 384,403km
A quarter diameter= dq= 2.3 cm= 0.000023km
No of quarters= 16,713,173, 913
To convert cm into km divide cm by 100 and then by 1000
as
1m= 100cm
1km= 1000m
Therefore
2.3/100= 0.023 m
And
0.023/1000= 0.000023 km
Dividing the distance from the moon by the diameter of the quarter laid end to end would give the number of the quarters needed.
No. of quarters= Distance from the moon/ Diameter of the quarter
= 384,403km/0.000023 km
= 16,713,173,913.043
Rounding 16,713,173,913.043 gives 16,713,173,913 quarters
Explanation:
Brainliest, please!
-ZERO- No 17.6 pound Earth-bound cat is going to be able to jump to the top of a 6 feet 7 inch bookshelf unassisted! Someone should call the SPCA on the writer of this textbook question ;-)
Assuming that this overweight cat is content being placed in such a lofty position, your professor probably wants the answer of <u>156.8 Joules</u>
Mass x Acceleration of Gravity (on Earth) x elevation = Potential Energy
8 Kg x 9.8 m/s2 x 2 m = 156.8 J
How did you take a picture?
Answer: Proximity to active volcanoes
Explanation:
When volcanoes erupt, they can generate tsunamis in the following ways:
-Parts of the volcano (some of them may be underwater) could collapse, generating a big displacement of water and resulting in a tsunami.
-The magma chamber could also collapse as it empties, also generating water displacement.
-If the volcanoes are near to the shore, debris resulting from the eruption could hit the water at fast speeds, producing big waves and eventually a tsunami.
In this case, the proximity of Palu city to active volcanoes, especially ones that are so close to the shore, increases the risk of a tsunami.
The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C
R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m
Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:
Substitute numerical values:
The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.
As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).
Learn more about Gaussian sphere here:
brainly.com/question/2004529
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