Answer:
50 N
Explanation:
Let the natural length of the spring = L
so
100 = k(40 - L)       (1)
200 = k(60 - L)       (2)
(2)/(1):   2 = (60 - L)/(40 - L)
60 - L = 2(40 - L)
60 - L = 80 - 2L
2L - L = 80 - 60
L = 20
Sub it into (1):
100 = k(40 - 20) = 20k
k = 100/20 = 5 N/in
Now
X = k(30 - L) = 5(30 - 20) = 50 N
 
        
             
        
        
        
Answer:
d = 68.5 x 10⁻⁶ m = 68.5 μm
Explanation:
The complete question is as follows:
An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is  1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?
The answer can be given by using the formula derived from Young's Double Slit Experiment:

where,
d = slit separation = ?
λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m
L = distance from screen (detector) = 1.7 m
y = distance between bright fringes = 15.7 mm = 0.0157 m
Therefore,

<u>d = 68.5 x 10⁻⁶ m = 68.5 μm</u>
 
        
             
        
        
        
Answer:
The tank is losing  
 

Explanation:
According to the Bernoulli’s equation:
 
We are being informed that both the tank and the hole is being exposed to air : 
∴ P₁ = P₂
Also as the tank is voluminous ; we take the initial volume   ≅ 0 ;
 ≅ 0 ;
 then  can be determined as:
 can be determined as:![\sqrt{[2g (h_1- h_2)]](https://tex.z-dn.net/?f=%5Csqrt%7B%5B2g%20%28h_1-%20h_2%29%5D)
h₁ = 5 + 15 = 20 m; 
h₂ = 15 m
![v_2 = \sqrt{[2*9.81*(20 - 15)]](https://tex.z-dn.net/?f=v_2%20%3D%20%5Csqrt%7B%5B2%2A9.81%2A%2820%20-%2015%29%5D)
![v_2 = \sqrt{[2*9.81*(5)]](https://tex.z-dn.net/?f=v_2%20%3D%20%5Csqrt%7B%5B2%2A9.81%2A%285%29%5D)
 as it leaves the hole at the base.
  as it leaves the hole at the base.
radius r = d/2  = 4/2 = 2.0 mm
(a) From the law of continuity; its equation can be expressed as:
J = 
J = πr² 
    
J =
J =
b) 
 How fast is the water from the hole moving just as it reaches the ground?
In order to determine that; we use the relation of the velocity from the equation of motion which says:
v² = u² + 2gh
₂
v² = 9.9² + 2×9.81×15 
v² = 392.31
The velocity of how fast the water from the hole is moving just as it reaches the ground is : 

 
        
             
        
        
        
Answer: Use Question cove you can get it faster you can get the answer faster! ;) hope this helps ;) but yeah use that and answer is done right away
Explanation: HOPE THIS HELPSS!! ;))
 
        
                    
             
        
        
        
Answer:
D. Dylan is incorrect because a 90-degree launch angle results in the largest vertical range
Explanation:
Projectile is the motion of an object thrown into space. When an object is thrown into space, the only force which acts on it is the acceleration due to gravity.
An object thrown into space would reach maximum height (vertical range) if it is launched at an angle of 90 degrees. For maximum horizontal range, the object needs to be launched at an angle of 45 degrees. 
Therefore Dylan is incorrect because a 90-degree launch angle results in the largest vertical range