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harina [27]
3 years ago
12

If you could see the electric field coming off a proton you would notice that electric field lines are what?

Physics
1 answer:
Readme [11.4K]3 years ago
3 0
The electric field generated by a proton is a radial field, so we would see the field lines going radially from the proton toward every direction. I've attached a picture representing the electric field generated by a positive charge, such as a proton, and we can see the field lines going radially from the charge toward every direction. 

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PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
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A. To express large and small numbers


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The difference in the composition of the terrestrial planets and the gas giants can be explained by the different condensation
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3 years ago
A golf club rotates 215 degrees and has a length (radius) equal to 29 inches. The time it took to swing the club was 0.8 seconds
vichka [17]

Answer:

The average linear velocity (inches/second) of the golf club is 136.01 inches/second

Explanation:

Given;

length of the club, L = 29 inches

rotation angle, θ = 215⁰

time of motion, t = 0.8 s

The angular speed of the club is calculated as follows;

\omega = (\frac{\theta}{360} \times 2\pi, \ rad) \times \frac{1}{t} \\\\\omega =  (\frac{215}{360} \times 2\pi, \ rad) \times \frac{1}{0.8 \ s} \\\\\omega = 4.69 \ rad/s

The average linear velocity (inches/second) of the golf club is calculated as;

v = ωr

v = 4.69 rad/s  x  29 inches

v = 136.01 inches/second

Therefore, the average linear velocity (inches/second) of the golf club is 136.01 inches/second

8 0
2 years ago
Why is it important for you to understand the basics of organ systems?
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3 years ago
Water initially at 200 kPa and 300°C is contained in a piston–cylinder device fitted with stops. The water is allowed to cool at
netineya [11]

Answer:

Δu=1300kJ/kg  

Explanation:

Energy at the initial state

p_{1}=200kPa\\t_{1}=300^{o}\\u_{1}=2808.8kJ/kg(tableA-5)

Is saturated vapor at initial pressure we have

p_{2}=200kPa\\x_{2}=1(stat.vapor)\\v_{2}=0.8858m^3/kg(tableA-5)

Process 2-3 is a constant volume process

p_{3}=100kPa\\v_{3}=v_{2}=0.8858m^{3}/kg\\u_{3}=1508.6kJ/kg(tableA-5)

The overall in internal energy

Δu=u₁-u₃

We replace the values in equation

Δu=u₁-u₃

=2808.8kJ/kg-1508.6kJ/kg\\=1300kJ/kg

Δu=1300kJ/kg  

3 0
3 years ago
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