Answer:
Option (2)
Explanation:
From the figure attached,
Horizontal component, 
![A_x=12[\text{Sin}(37)]](https://tex.z-dn.net/?f=A_x%3D12%5B%5Ctext%7BSin%7D%2837%29%5D)
= 7.22 m
Vertical component, ![A_y=A[\text{Cos}(37)]](https://tex.z-dn.net/?f=A_y%3DA%5B%5Ctext%7BCos%7D%2837%29%5D)
= 9.58 m
Similarly, Horizontal component of vector C,
= C[Cos(60)]
= 6[Cos(60)]
= 
= 3 m
![C_y=6[\text{Sin}(60)]](https://tex.z-dn.net/?f=C_y%3D6%5B%5Ctext%7BSin%7D%2860%29%5D)
= 5.20 m
Resultant Horizontal component of the vectors A + C,
m
= 4.38 m
Now magnitude of the resultant will be,
From ΔOBC,

= 
= 
= 6.1 m
Direction of the resultant will be towards vector A.
tan(∠COB) = 
= 
= 
m∠COB = 
= 46°
Therefore, magnitude of the resultant vector will be 6.1 m and direction will be 46°.
Option (2) will be the answer.
<u>Answer:</u> The Young's modulus for the wire is 
<u>Explanation:</u>
Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.
The equation representing Young's Modulus is:

where,
Y = Young's Modulus
F = force exerted by the weight = 
m = mass of the ball = 10 kg
g = acceleration due to gravity = 
l = length of wire = 2.6 m
A = area of cross section = 
r = radius of the wire =
(Conversion factor: 1 m = 1000 mm)
= change in length = 1.99 mm = 
Putting values in above equation, we get:

Hence, the Young's modulus for the wire is 
It evaporates into a vapor
~~~hope this helps~~~
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Answer:
B. 175 N
Explanation:
Net force can be defined as the vector sum of all the forces acting on a body or an object i.e the sum of all forces acting simultaneously on a body or an object.
Mathematically, net force is given by the formula;
Where;
Fnet is the net force
Fapp is the applied force
Fg is the force due to gravitation
In this scenario, we observed that both forces are acting in the same direction.
Therefore:
Net force = 100 N + 75 N
Net force = 175 Newton
Answer:
The answer to this question can be defined as follows:
Explanation:
Therefore the 4th harmonicas its node is right and over the pickup so, can not be captured from 16.25, which is 1:4 out of 65. Normally, it's only conceptual for the certain harmonic, this will be low, would still be heard by the catcher.
Instead, every harmonic node has maximum fractions along its string; the very first node is the complete string length and the second node is half a mile to the third node, which is one-third up and so on.