Work = (force) x (distance)
Each time she lifts the weight, she does
(550 N) x (0.5 m) = 275 joules of work against gravity.
Each time she lets the bar down gently, gravity does
(550 N) x (0.5 m) = 275 joules of work against her muscles.
If the human physical muscular system were 100% efficient, and capable
of absorbing work as well as spending it, then the bodybuilder would do
exactly zero work in the process of 1-up followed by 1-down.
Answer:
no
Explanation:
it is faster at the equator
<span>a. KE in electron volts is 1020 eV.
b. KE in Joules is e(1020) = (1.6022E-19)(1020) = 1.634E-16
c. KE = (1/2)mv^2, so v = sqrt[2*KE/m] = 18.94E6 m/s
note: m is the mass of an electron = 9.109e-31 kg
I hope my answer has come to your help. Thank you for posting your question here in Brainly.
</span>
The time for half the atoms in a radioactive substance to disintegrate
https://www.vocabulary.com/dictionary/half-life
Complete question:
Point charges q1=- 4.10nC and q2=+ 4.10nC are separated by a distance of 3.60mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×10−9 N⋅m ? Express your answer in newtons per coulomb to three significant figures.
Answer:
The magnitude of this field is 826 N/C
Explanation:
Given;
The torque exerted on the dipole, T = 7.3 x 10⁻⁹ N.m
PEsinθ = T
where;
E is the magnitude of the electric field
P is the dipole moment
First, we determine the magnitude dipole moment;
Magnitude of dipole moment = q*r
P = 4.1 x 10⁻⁹ x 3.6 x 10⁻³ = 1.476 x 10⁻¹¹ C.m
Finally, we determine the magnitude of this field;

E = 826 N/C (in three significant figures)
Therefore, the magnitude of this field is 826 N/C