I think true. I'm pretty sure, but check w/ others too.
A metallic bond would be formed
Answer:
(A) 667.5 N/m
(B)
Explanation:
(A) Let the spring constant be k.
Using the formula F = kx
k = 251 / 0.376
K = 667.5 N/m
(B)
Work done
W = 0.5 × kx^2
W = 0.5 × 667.5 × 0.376 × 0.376
W = 47.2 J
Answer:
The Sun and planets are shown to the same scale. The small terrestrial planets and tiny Pluto are in the box---the Earth is the blue dot near the center of the box (montage created by Nick Strobel using NASA images).
Size
The Sun is by far the biggest thing in the solar system. From its angular size of about 0.5° and its distance of almost 150 million kilometers, its diameter is determined to be 1,392,000 kilometers. This is equal to 109 Earth diameters and almost 10 times the size of the largest planet, Jupiter. All of the planets orbit the Sun because of its enormous gravity. It has about 333,000 times the Earth's mass and is over 1,000 times as massive as Jupiter. It has so much mass that it is able to produce its own light. This feature is what distinguishes stars from planets.
Composition
What is the Sun made of? Spectroscopy shows that hydrogen makes up about 94% of the solar material, helium makes up about 6% of the Sun, and all the other elements make up just 0.13% (with oxygen, carbon, and nitrogen the three most abundant ``metals''---they make up 0.11%). In astronomy, any atom heavier than helium is called a ``metal'' atom. The Sun also has traces of neon, sodium, magnesium, aluminum, silicon, phosphorus, sulfur, potassium, and iron. The percentages quoted here are by the relative number of atoms. If you use the percentage by mass, you find that hydrogen makes up 78.5% of the Sun's mass, helium 19.7%, oxygen 0.86%, carbon 0.4%, iron 0.14%, and the other elements are 0.54%.
Explanation:
Answer:
Electric field, 
Explanation:
It is given that,
Mass of sphere, m = 2.1 g = 0.0021 kg
Charge, 
We need to find the magnitude of electric field that balances the weight of a plastic spheres. So,
a = g
or
Hence, the magnitude of electric field that balances its weight is 
. Hence, this is the required solution.
