Answer:
with the 2 north poles it will repulse with the north and south it will retract
Explanation:
north and north wont come together. same for south and south. opposite sides always attract one another. hope it helped!!!
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
When the core temperature rises, stars start burning helium to carbon. Hydrogen and helium are fused in a shell surrounding the core, raising the temperature.
The most prevalent method of generating energy in the cosmos is hydrogen fusion. The gravitational force is utilised to fuse hydrogen at the centres of our Sun and other stars, where it has a density more than 70 times greater than that of steel. Thus, all of the heat and light on earth comes from fusion. Helium is a chemical element with the atomic number 2 and the symbol He. The first member of the noble gas group in the periodic table, it is a colourless, odourless, tasteless, inert, monatomic gas that is not harmful.
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The minimum average speed it must have in the second half of the event in order to qualify is 414.7 km/h.
<h3>
What is average speed?</h3>
The average speed of an object is the ratio of total distance traveled by the object to the total time of motion of the object.
<h3>Total time taken by the car during the entire race</h3>
time = distance/average speed
time = (1.41 km) / (278 km/h)
time = 0.0051 hr
The car travels the first half of the race, d (¹/₂ x 1410 m) at 210 km/h;
d = 705 m = 0.705 km
t1 = 0.705/210
t1 = 0.0034 hr
<h3>time for the second half</h3>
t2 = 0.0051 - 0.0034 hr
t2 = 0.0017 hr
<h3>minimum average speed of the second half</h3>
v = d/t
v = 0.705 km / 0.0017 hr
v = 414.7 km/hr
Thus, the minimum average speed it must have in the second half of the event in order to qualify is 414.7 km/h.
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Answer:
666000000000 J
391764.71 N
Explanation:
u = Initial velocity
v = Final velocity
s = Displacement
Change in energy
Work done be the force is 666000000000 J
The magnitude of force is 391764.71 N
