Answer:
Explanation:
4.95s≈5s
Use equation :
h=G*t²/2
h=120m ----hight of cliff
G=9.81m/s²
t=?
-----------------------
h=G*t²/2
120m=9.81m/s²*t²/2
240m=9.81m/s²*t²
t²=240m/ 9.81m/s²
t²=24.46s
t=√24.46s²
t=4.95 s≈5s
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
The components that must be present for work to be considered is a force and a movement in the same direction as the force. In the basic definition of work, a magnitude and displacement that occurs in the same direction is what makes up work. Among the choices, the correct answer is the first one.
