<h3>Question -:</h3>
The Earth orbits around the sun because the gravitational force that the sun
exerts on the Earth:
O A. causes Earth's acceleration toward the sun.
O B. is very small because the sun is so far from the Earth.
O c. is smaller than the force the Earth exerts on the sun.
O D. pushes the Earth away from the sun.
<h3>Answer -:</h3>
O A. causes Earth's acceleration toward the sun.
<em>I </em><em>hope </em><em>this</em><em> </em><em>helps</em><em>,</em><em> </em><em>have </em><em>a </em><em>nice </em><em>time </em><em>ahead!</em>
Answer:
Explanation:
The acceleration of a circular motion is given by
where 
is the angular velocity and 
is the radius.
Angular velocity is related to the period, T, by
Substitute into the previous formula.
This acceleration does not depend on the linear or angular displacement. Hence, the amount of rotation does not change it.
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
Answer:
w = 11.211 KN/m
Explanation:
Given:
diameter, d = 50 mm
F.S = 2
L = 3
Due to symmetry, we have:
To find the maximum intensity, w, let's take the Pcr formula, we have:
Let's take k = 1
Substituting figures, we have:
Solving for w, we have:
w = 11211.14 N/m = 11.211 KN/m
Since Area, A= pi * (0.05)²
. This means it is safe
The maximum intensity w = 11.211KN/m
