The force on the clove will have the same magnitude but a negative sign with it that is F = - 14 N
<h3>What is force and how do we calculate it ?</h3>
The definition of force is: The pushing or pulling on a massed object changes its velocity. Force is an external agent that has the power to alter a body's resting or moving condition.
It has a size and a movement.
Newton's second law of motion defines the force formula as follows:
A force is equal to an object's mass times its acceleration, or F = m a. You must use SI units when applying this formula: kilograms for mass, meters per second squared for acceleration, and newtons for force.
To know more about force you may visit the link which is :
brainly.com/question/26115859
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To solve this exercise it is necessary to apply the equations related to the magnetic moment, that is, the amount of force that an image can exert on the electric currents and the torque that a magnetic field exerts on them.
The diple moment associated with an iron bar is given by,

Where,
Dipole momento associated with an Atom
N = Number of atoms
y previously given in the problem and its value is 2.8*10^{-23}J/T


The number of the atoms N, can be calculated as,

Where
Density
Molar Mass
A = Area
L = Length
Avogadro number


Then applying the equation about the dipole moment associated with an iron bar we have,



PART B) With the dipole moment we can now calculate the Torque in the system, which is



<em>Note: The angle generated is perpendicular, so it takes 90 ° for the calculation made.</em>
Answer:
<u>For M84:</u>
M = 590.7 * 10³⁶ kg
<u>For M87:</u>
M = 2307.46 * 10³⁶ kg
Explanation:
1 parsec, pc = 3.08 * 10¹⁶ m
The equation of the orbit speed can be used to calculate the doppler velocity:

making m the subject of the formula in the equation above to calculate the mass of the black hole:
.............(1)
<u>For M84:</u>
r = 8 pc = 8 * 3.08 * 10¹⁶
r = 24.64 * 10¹⁶ m
v = 400 km/s = 4 * 10⁵ m/s
G = 6.674 * 10⁻¹¹ m³/kgs²
Substituting these values into equation (1)

M = 590.7 * 10³⁶ kg
<u>For M87:</u>
r = 20 pc = 20 * 3.08 * 10¹⁶
r = 61.6* 10¹⁶ m
v = 500 km/s = 5 * 10⁵ m/s
G = 6.674 * 10⁻¹¹ m³/kgs²
Substituting these values into equation (1)

M = 2307.46 * 10³⁶ kg
The mass of the black hole in the galaxies is measured using the doppler shift.
The assumption made is that the intrinsic velocity dispersion is needed to match the line widths that are observed.
A jet fighter flies from the airbase A 300 km East to the point M. Then 350 km at 30° West of North.
It means : at 60° North of West. So the distance from the final point to the line AM is :
350 · cos 60° = 350 · 0.866 = 303.1 km
Let`s assume that there is a line N on AM.
AN = 125 km and NM = 175 km.
And finally jet fighter flies 150 km North to arrive at airbase B.
NB = 303.1 + 150 = 453.1 km
Then we can use the Pythagorean theorem.
d ( AB ) = √(453.1² + 125²) = √(205,299.61 + 15,625) = 470 km
Also foe a direction: cos α = 125 / 470 = 0.266
α = cos^(-1) 0.266 = 74.6°
90° - 74.6° = 15.4°
Answer: The distance between the airbase A and B is 470 km.
Direction is : 15.4° East from the North.
Explanation:
<em>Hello</em><em> </em><em>there</em><em>!</em><em>!</em><em>!</em>
<em>You</em><em> </em><em>just</em><em> </em><em>need</em><em> </em><em>to</em><em> </em><em>use</em><em> </em><em>simple</em><em> </em><em>formula</em><em> </em><em>for</em><em> </em><em>force</em><em> </em><em>and</em><em> </em><em>momentum</em><em>, </em>
<em>F</em><em>=</em><em> </em><em>m.a</em>
<em>and</em><em> </em><em>momentum</em><em> </em><em>(</em><em>p</em><em>)</em><em>=</em><em> </em><em>m.v</em>
<em>where</em><em> </em><em>m</em><em>=</em><em> </em><em>mass</em>
<em>v</em><em>=</em><em> </em><em>velocity</em><em>.</em>
<em>a</em><em>=</em><em> </em><em>acceleration</em><em> </em><em>.</em>
<em>And</em><em> </em><em>the</em><em> </em><em>solutions</em><em> </em><em>are</em><em> </em><em>in</em><em> </em><em>pictures</em><em>. </em>
<em><u>Hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em>