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WITCHER [35]
3 years ago
14

The amount of electrons that an atom loses, shares or gains is the ________________.

Physics
1 answer:
mr Goodwill [35]3 years ago
4 0
It’s known as valency
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How would the atmosphere of earth differ if the earth was much larger
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If the earth just doubles its size in an instant, the oxygen present in the atmosphere could be much larger. Due to these, we could theorize that insects could grow much larger because the size of their body is heavily dependent on the maximum amount of oxygen present in the atmosphere. Air pressure also increases thus those aerial enthusiasts would enjoy their professions.
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4 years ago
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A solenoid with 3,000.0 turns is 70.0 cm long. If its self-inductance is 25.0 mH, what is its radius? (The value of μ0 is 4π x 1
nevsk [136]

Answer:

A. 2.2*10^-2m

Explanation:

Using

Area = length x L/ uo xN²

So A = 0.7m * 25 x 10^-3H /( 4π x10^-7*

3000²)

A = 17.5*10^-3/ 1.13*10^-5

= 15.5*10^-2m²

Area= π r ²

15.5E-2/3.142 = r²

2.2*10^2m

Explanation:

5 0
3 years ago
Resolve the vector shown below into its components.
yan [13]

Answer:

D. s=3x+4y

Explanation:

The line is at the 3rd column in the 4th row.

7 0
2 years ago
A heavy object is dropped From a vertical height of 8.0 m. What is it’s spend when it hits the ground ?
otez555 [7]

Answer:

Final speed of the ball = Vf= 12.52 m/s when it hits the ground

Explanation:

Height = S= 8.0 m

Initial Velocity= Vi = 0 m/s

Final Velocity= Vf =?

Acceleration due to gravity= a=9.8m/s²

Using Newtons third equation of motion

2as= Vf²- Vi²

2(9.8) 8= Vf²- 0

156.8= Vf²

Vf=√156.8= 12.52m/s

Final Speed of the ball is 12.52 m/s

4 0
3 years ago
The power rating on a light bulb indicates how much power it would dissipate when it is hooked up to the standard household volt
erastova [34]

Answer:

A. P = 18.75 watts

B. P = 75 watts

Explanation:

V = 120 Volts

P = VI

I = P/V = 75/120 = 0.625 Amps

V = IR

R = V/I

R = 120/0.625 = 192 Ω

So the resistance of the bulb is 192 Ω and it does not change as it is given in the question.  

A. How much power is dissipated in a light bulb that is normally rated at 75 W, if instead we hook it up to a potential difference of 60 V?

As P = VI and I = V/R

P = V*(V/R)

P = V²/R

P = (60)/192

P = 18.75 watts

As expected, it will dissipate less power (18.75 watts) than rated power due to not having rated voltage of 120 Volts.

I = V/R = 60/192 = 0.3125 Amps

or I = P/V = 18.75/60 = 0.3125 Amps

Since the resistance is being held constant, decreasing voltage will also decrease current as V = IR voltage is directly proportional to the current.

B. How much power is dissipated in a light bulb that is normally rated at 75 W, if instead we hook it up to a potential difference of 120 V?

P = V*(V/R)

P = V²/R

P = 120/192 = 75 watts

I = P/V = 75/120 = 0.625 Amps

As expected, it will dissipate rated power of 75 watts at rated voltage of 120 Volts.

5 0
4 years ago
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