Which best describes the electric field created by a positive charge?
2 answers:
Answer:
Its rays point away from the charge.
Explanation:
The electric field due to a point charge is given by :
Where
k is the electrostatic constant
q is the point charge
r is the distance from charge particle
It is clear that the electric field is directly proportional to the charge and inversely proportional to the distance from charge particle.
There are two types of charged particle i.e. negative and positive charge. For a positive charge, the direction of electric field points away from it and for a negative charge, the direction of electric field points towards it.
Hence, the correct option is (c).
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Answer:
Magnitude of the net force on q₁-
Fn₁=1403 N
Magnitude of the net force on q₂+
Fn₂= 810 N
Magnitude of the net force on q₃+
Fn₃= 810 N
Explanation:
Look at the attached graphic:
The charges of the same sign exert forces of repulsion and the charges of opposite sign exert forces of attraction.
Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:
F= (k*q*q)/(d)²
F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N
Magnitude of the net force on q₁-
Fn₁x= 0
Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N
Fn₁=1403 N
Magnitude of the net force on q₃+
Fn₃x= 810- 810 cos 60° = 405 N
Fn₃y= 810*sin 60° = 701.5 N
Fn₃ = 810 N
Magnitude of the net force on q₂+
Fn₂ = Fn₃ = 810 N
Answer:
The punnet chart indicate the cross between a black cat and a spotted cat.
Explanation:
As the punnet chart was not given with the question, the question is searched and found online which is as attached herewith.
As seen from the punned table the column head has two capital H which indicate that the cat is a black cat.
White the row heads indicate a Capital and a small h thus this is a spotted cat
Thus the punnet chart indicate the cross between a black cat and a spotted cat.
Answer
given,
D = 50 mm = 0.05 m
d = 10 mm = 0.01 m
Force to compress the spring
F = 3160 N
stress correction factor from stress correction curve is equal to 1.1
now, calculation of corrected stress
= 442.6 Mpa
The tensile strength of the steel material of ASTM A229 is equal to 1300 Mpa
now,
since corrected stress is less than the
hence, spring will return to its original shape.