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uranmaximum [27]
3 years ago
13

Rutherford's atomic theory included which idea?

Physics
1 answer:
Murljashka [212]3 years ago
4 0
Your answer would be C Atoms are composed of a positively charged sphere that contains electrons. 
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the mass of an object is 90 kg calculate weight of the same object on the surface of the moon.(acceleration due to gravity on th
aalyn [17]

Explanation:

the weight of an object is its force exerted by gravity.

on earth we take gravitational acceleration to be 9.8 m/s^2

so the weight of an object on earth is given by F = mg

hence,

F = 90 × 9.8

F = 882 Newtons (weight on earth)

gravitational acceleration on the moon is equal to 9.8/6. so, we can just use the same formula to find the weight on the moon.

F = 90 × 9.8/6

F = 147 Newtons (weight on the moon)

make sure to ask if you need any further guidance.

5 0
3 years ago
Read 2 more answers
Your friend wants to join the school track team, and has asked for your help to determine how fast she can run. 4. What kind of
DENIUS [597]

Answer:

I think you would need too time your friend and know how too divide that time by other racers to see how much faster she needs to get. I THINK.

7 0
2 years ago
The process by which hot and cold air are transferred in the atmosphere is
Nuetrik [128]
That process is convection.
6 0
3 years ago
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Positive ions from a base and negative ions from an acid form a .
Trava [24]

Positive ions from a base and negative ion from an acid form salt.

 

<span>To add, table salt or common salt is a mineral composed primarily of sodium chloride, a chemical compound belonging to the larger class of salts. Rock salt or halite is also the common term for salt in its natural form.</span>

3 0
3 years ago
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Calculate the mass 9f the earth, assuring that uts is sphere with radius 6.67×10^6m.​
MA_775_DIABLO [31]

Answer:

6.86 × 10²⁴ kg

Explanation:

The mass of the earth m = density of earth, ρ × volume of earth, V

m = ρV

The density of the earth, ρ = 5515 kg/m³ and since the earth is a sphere, its volume is the volume of a sphere V = 4πr³/3 where r = radius of the earth = 6.67 × 10⁶ m

Since m = ρV

m = ρ4πr³/3

So, substituting the values of the variables into the equation for the mass of the earth, m, we have

m = 5515 kg/m³ × 4π(6.67 × 10⁶ m)³/3

m = 5515 kg/m³ × 4π × 296.741 × 10¹⁸ m³/3

m = 5515 kg/m³ × 1189.9639π × 10¹⁸ m³/3

m = 6546105.64378π × 10¹⁸ kg/3

m = 20565197.400122 × 10¹⁸ kg/3

m = 6855065.8 × 10¹⁸ kg

m = 6.8550658 × 10²⁴ kg

m ≅ 6.86 × 10²⁴ kg

8 0
3 years ago
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