The answer should be A weighlifter lifting 100lb. bell bar from his chest to a certain height.
Answer:
8.46E+1
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = 39 C
Charge 2 (q₂) = –53 C
Force (F) of attraction = 26×10⁸ N
Electrical constant K) = 9×10⁹ Nm²/C²
Distance apart (r) =?
The distance between the two charges can be obtained as follow:
F = Kq₁q₂ / r²
26×10⁸ = 9×10⁹ × 39 × 53 / r²
26×10⁸ = 1.8603×10¹³ / r²
Cross multiply
26×10⁸ × r² = 1.8603×10¹³
Divide both side by 26×10⁸
r² = 1.8603×10¹³ / 26×10⁸
r² = 7155
Take the square root of both side
r = √7155
r = 84.6 m
r = 8.46E+1 m
1) Current
2) Atoms
3) Wire
4) Negative
5) Neutron
6) Shock
7) Switch
8) Static
9) Volt
10) Battery
11) Dam
12) Thomas Edison
13) Benjamin Franklin
14) Alessandro Volta
15) Michael Faraday
I would say that these would be your correct answers, btw I'm doing something that is close to the same right now
Hope this helps :)
Answer:
imma need a few answer choices so i can do my research
Explanation:
Please any thank you
The Professor's centripetal acceleration is 0.044 m/s²
Centripetal acceleration is the acceleration of an object moving in circular motion. It is usually directed towards the center of the rotation.
It is given by:
a = v²/r
where v is the velocity and r is the radius.
Given that the radius (r) = 4 m, velocity (v) = 0.419 m/s, hence:
a = v²/r = 0.419²/4 = 0.044 m/s²
The Professor's centripetal acceleration is 0.044 m/s²
Find out more at: brainly.com/question/6082363
