Answer:
-611.32 N/C
0.43723 m
Explanation:
k = Coulomb constant = 
q = Charge = -4.25 nC
r = Distance from particle = 0.25 m
Electric field is given by
The magnitude is 611.32 N/C
The electric field will point straight down as the sign is negative towards the particle.
The distance from the electric field is 1.71436 m
First we write the corresponding kinematics equations:
a = -g
v = -g * t + vo
y = -g * ((t ^ 2) / 2) + vo * t + yo
Substituting the values:
y = - (9.81) * (((0.50) ^ 2) / 2) + (19) * (0.50) + (0) = 8.27m
answer:
the displacement at the time of 0.50s is 8.27m
Thw question is not complete. The complete question is;
Charge of uniform linear density (6.7 nCim) is distributed along the entire x axis. Determine the magnitude of the electric field on the y axis at y = 1.6 m. a. 32 N/C b. 150 NC c 75 N/C d. 49 N/C e. 63 NC
Answer:
Option C: E = 75 N/C
Explanation:
We are given;
Uniform linear density; λ = 6.7 nC/m = 6.7 × 10^(-9) C/m
Distance on the y-axis; d = 1.6 m
Now, the formula for electric field with uniform linear density is given as;
E = λ/(2•π•r•ε_o)
Where;
E is electric field
λ is uniform linear density = 6.7 × 10^(-9) C/m
r is distance = 1.6m
ε_o is a constant = 8.85 × 10^(-12) C²/N.m²
Thus;
E = (6.7 × 10^(-9))/(2π × 1.6 × 8.85 × 10^(-12))
E = 75.31 N/C ≈ 75 N/C
I think this answer would be D. It was seen as a refueling station for naval vessels
Answer:
g ’= ¼ g correct answer is 2
Explanation:
The expression for the angular velocity of a simple pendulum is
w = √ g / L
The angular velocity and the period are related
w = 2π / T
T = 2π √L / g
Let's look for the length of the pendulum with the data on Earth
L = T² g / 4π²
L = 1² g / 4π²
L = 2.533 10⁻² 9.8
L = 0.2482 m
Now let's look for the gravity of the planet
g’= 4π² L / T’²
g’= 4π² 0.2483 / 2²
g’= 2.45 m / s²
The relationship between this value and the earth's gravity is
g ’/ g = 2.45 / 9.80
g ’= 0.25 g
g ’= ¼ g
correct anwers is 2
