Answer:
F=94.32*10⁻⁹N , The force F is repusilve because both charges have the same sign (+)
Explanation:
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(m)
Equivalence
1nC= 10⁻⁹C
Data
K=8.99x10⁹N*m²/C²
q₁ = 7.94-nC= 7.94*10⁻⁹C
q₂= 4.14-nC= 4.14 *10⁻⁹C
d= 1.77 m
Magnitude of the electrostatic force that one charge exerts on the other
We apply formula (1):
F=94.32*10⁻⁹N , The force F is repusilve because both charges have the same sign (+)
Answer:
(a) 32.5 kgm/s
(b) 32.5 Ns
(c) 10.8 N
Explanation:
The change in momentum can be calculated from the definition of linear momentum:
Then, the change in momentum of the body is of 32.5 kgm/s (a).
Now, from the impulse-momentum theorem, we know that the change in momentum of a body 
is equal to the impulse 
exerted to it. So, the impulse produced by the force equals 32.5 kgm/s (or 32.5 Ns) (b).
Finally, since we know the value of the impulse and the interval of time, we can easily solve for the magnitude of the force:
It means that the magnitude of the force is of 10.8N (c).
Answer:
Explanation:
The x- and y- components of the velocity vector can be written as following:
Since the angle θ and the magnitude of the velocity is given, the vector representation can be written as follows:
Answer:
135°.
Explanation:
R = 75 ohm, L = 0.01 H, C = 4 micro F = 4 x 10^-6 F
Frequency is equal to the half of resonant frequency.
Let f0 be the resonant frequency.
f0 = 796.2 Hz
f = f0 / 2 = 398.1 Hz
So, XL = 2 x 3.14 x f x L = 2 x 3.14 x 398.1 x 0.01 = 25 ohm
Xc = 100 ohm
tan Ф = (25 - 100) / 75 = - 1
Ф = 135°
Thus, the phase difference is 135°.
Answer: The electric field is given in three regions well defined; 0<r<2; 2<r<4; 4<r<5 and r>5
Explanation: In order to solve this problem we have to use the gaussian law in the mentioned regions.
Region 1; 0<r<2
∫E.ds=Qinside the gaussian surface/ε0
inside of the solid conducting sphere the elevctric field is zero because the charge is located at the surface on this sphere.
Region 2; 2<r<4;
E.4*π*r^2=8,84/ε0
E=8,84/(4*π*ε0*r^2)
Region 3; 4<r<5
E=0 because is inside the conductor.
Finally
Region 4; r>5
E.4*π*r^2=(8,84-2.02)/ε0
