Answer:
The answer is....
Explanation:
How hard the surfaces push together and the types of surfaces involved.
Since Jim's speed is constant and he is moving in a straight line, he is not accelerating, and we know the net force on him is zero. There is no Force anywhere doing any work. So no power is being added to him or dissipated by him.
- The wavelength of the red light in "nanometer" is 7×

- Wavelength is given as : 7×
meter
- 1 nanometer = (
meter)
- Let X= value of the wavelength in nanometer.
1 nanometer =
meter
X nanometer = 7×
meter
- <em>If we Cross multiply</em>
X nanometer = (
)
X= 7×
nanometer
Therefore, the wavelength in "nanometer" is 7×
Learn more at :brainly.com/question/12924624?referrer=searchResults
The relationships can best be described as follows:
As frequency increases, wavelength decreases. <span>The greater the </span>energy<span>, the larger the frequency </span>and<span> the shorter (smaller) the </span>wavelength<span>. </span>
<span>a) wavelength vs. frequency = inversely proportional
b) wavelength vs. energy = inversely proportional
c) frequency vs. energy = directly proportional
Hope this answers the questions. Have a nice day. Feel free to ask more questions.</span>
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