A hot piece of iron is thrown into the ocean and its temperature eventually stabilizes. Which of the following statements concer
1 answer:
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Answer:
q₁= +0.5nC
Explanation:
Theory of electrical forces
Because the particle q3 is close to three other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.
To solve this problem we apply Coulomb's law:
Two point charges (q1, q2) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:
o solve this problem we apply Coulomb's law:
Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:
F=K*q₁*q₂/d² Formula (1)
F: Electric force in Newtons (N)
K : Coulomb constant in N*m²/C²
q₁,q₂:Charges in Coulombs (C)
d: distance between the charges in meters
Data:
Equivalences
1nC= 10⁻⁹ C
1cm= 10⁻² m
Data
q₃=+5.00 nC =+5* 10⁻⁹ C
q₂= -2.00 nC =-2* 10⁻⁹ C
d₂= 5.00 cm= 5*10⁻² m
d₁= 2.50 cm= 2.5*10⁻² m
k = 8.99*10⁹ N*m²/C²
Calculation of magnitude and sign of q1
Fn₃=0 : net force on q3 equals zero
F₂₃:The force F₂₃ that exerts q₂ on q₃ is attractive because the charges have opposite signs,in direction +x.
F₁₃:The force F₂₃ that exerts q₂ on q₃ must go in the -x direction so that Fn₃ is zero, therefore q₁ must be positive and F₂₃ is repulsive.
We propose the algebraic sum of the forces on q₃
F₂₃ - F₁₃=0
We eliminate k*q₃ of the equation
q₁= +0.5*10⁻⁹ C
q₁= +0.5nC
Answer:
2023857702.507m
Explanation:
recall from newton's law of gravitation
G=gravitational constant
mshew=50g
melephant=5*10^3kg
rearth=radius of the earth 6400km or 6400000m
mearth= masss of the earth
Gm(shrew)m(earth)/r(earth)^2 = Gm(elephant)m(earth)/r^2
strike out the left hand side and right hand side variables
m(shrew)/r(earth)^2 = m(elephant)/r^2
r^2 = m(elephant).r(earth)^2 / m(shrew) .........make r^2 the subject of the equation
r^2=
r^2=40960000000000
r=2023857702.507m