10dm/s=...m/phút=......km/h
1 answer:
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Answer:
Density of 127 I =
Also,
Explanation:
Given, the radius of a nucleus is given as
.
where,
- A is the mass number of the nucleus.
The density of the nucleus is defined as the mass of the nucleus M per unit volume V.
For the nucleus 127 I,
Mass, M =
Mass number, A = 127.
Therefore, the density of the 127 I nucleus is given by
On comparing with the density of the solid iodine,
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