Consider the uniform electric field E = (2.5 j + 3.5 k) × 103 N/C. (a) Calculate the electric flux through a circular area of ra
1 answer:
You might be interested in
Answer:
See the explanation below.
Explanation:
12) When an object is falling, how does the objects velocity change? what formula do you use?
The speed of a falling object is increased by a value of 9.81 meters per second per second. That is if we throw any body regardless of mass from a considerable height, its speed in the first second will be 9.81[ m/ s] , in the next second will be equal to 19.62 [m/s] in the next will be equal to 29.43 [m/ s].
The formula is:
where:
vo = initial velocity = 0
g = gravity = 9.81[m/s^2]
t = time [s]
13)
what is a falling speed at 6s, 9s, 112s?
v = 0 + (9.81*6) = 58.86[m/s]
v = 0 + (9.81*9) = 88.29 [m/s]
v = 0 + (9*112) = 1098.72 [m/s]
14)
If an object is falling at 65 [m/s]. How long has it been falling ? what is the formula that you use?
The formula is the same:
65 = 0 + 9.81*t
t = 65/9.81
t = 6.62[s]
15)
What formula is used to determine the distance an object is falling ?
where:
y = distance [m]
yo = initial distance, in most of the cases and depending of the reference point it will be eqaul to zero
vo = initial velocity, if it is free fall, then = 0
t = time [s]
g = gravity = 9.81[m/s^2]
This equation will be reduce to:
16)
using the times given in problem 13. Determine the distance fallen for each.
y = 0.5*9,81*(6)^2 = 176.58 [m]
y = 0.5*9,81*(9)^2 = 397.3 [m]
y = 0.5*9,81*(112)^2 = 61528.3 [m]
17)
If an object has fallen a distance of 87.3 [m]. How long was it falling?
87.3 = 0.5*9.81*t^2
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