Answer:
Fn: magnitude of the net force.
Fn=30.11N , oriented 75.3 ° clockwise from the -x axis
Explanation:
Components on the x-y axes of the 17 N force(F₁)
F₁x=17*cos48°= 11.38N
F₁y=17*sin48° = 12.63 N
Components on the x-y axes of the the second force(F₂)
F₂x= −19.0 N
F₂y= 16.5 N
Components on the x-y axes of the net force (Fn)
Fnx= F₁x +F₂x= 11.38N−19.0 N= -7.62 N
Fny= F₁y +F₂y= 12.63 N +16.5 N = 29.13 N
Magnitude of the net force.



Direction of the net force (β)

β=75.3°
Magnitude and direction of the net force
Fn= 30.11N , oriented 75.3 ° clockwise from the -x axis
In the attached graph we can observe the magnitude and direction of the net force
1 ft =12 in
4 in = 0.333 ft
volume = (п/4)*(0.333)² = 0.087 ft²
vol. flow = spead *volume
=3 ft/s * 0.087 ft²
vol flow = 0.261 ft³/s
<h2>
a)Acceleration due to gravity on the surface of the Sun is 274.21 m/s²</h2><h2>b)
Factor of increase in weight is 27.95</h2>
Explanation:
a) Acceleration due to gravity

Here we need to find acceleration due to gravity of Sun,
G = 6.67259 x 10⁻¹¹ N m²/kg²
Mass of sun, M = 1.989 × 10³⁰ kg
Radius of sun, r = 6.957 x 10⁸ m
Substituting,

Acceleration due to gravity on the surface of the Sun = 274.21 m/s²
b) Acceleration due to gravity in earth = 9.81 m/s²
Ratio of gravity = 274.21/9.81 = 27.95
Weight = mg
Factor of increase in weight = 27.95
To solve this problem we will use the Ampere-Maxwell law, which describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

Where,
B= Magnetic Field
l = length
= Vacuum permeability
= Vacuum permittivity
Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

Recall that the speed of light is equivalent to

Then replacing,


Our values are given as




Replacing we have,



Therefore the magnetic field around this circular area is 