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Explanation: HOPE THIS HELPSS!! ;))
 
        
                    
             
        
        
        
Answer:
Net force: 20 N to the right
mass of the bag: 20.489 kg
acceleration:  0.976  m/s^2
Explanation:
Since the normal force and the weight are equal in magnitude but opposite in direction, they add up to zero in the vertical direction. In the horizontal direction, the 195 N tension to the right minus the 175 force of friction to the left render a net force towards the right of magnitude:
195 N - 175 N = 20 N
So net force on the bag is 20 N to the right. 
The mass of the bag can be found using the value of the weight force: 201 N:
mass = Weight/g = 201 / 9.81 = 20.489 kg
and the acceleration of the bag can be found as the net force divided by the mass we just found:
acceleration = 20 N / 20.489 kg = 0.976  m/s^2
 
        
             
        
        
        
Answer:
   λ = 162 10⁻⁷ m
Explanation:
Bohr's model for the hydrogen atom gives energy by the equation
           = - k²e² / 2m (1 / n²)
 = - k²e² / 2m (1 / n²)
Where k is the Coulomb constant, e and m the charge and mass of the electron respectively and n is an integer
The Planck equation
            E = h f
The speed of light is
           c = λ f
           E = h c /λ
For a transition between two states we have
            -
 -  = - k²e² / 2m (1 /
 = - k²e² / 2m (1 /  ² -1 /
² -1 /  ²)
²)
            h c / λ = -k² e² / 2m (1 /  ² - 1/
² - 1/  ²)
²)
            1 / λ = (- k² e² / 2m h c) (1 /  ² - 1/
² - 1/ ²)
²)
 The Rydberg constant with a value of 1,097 107 m-1 is the result of the constant in parentheses
Let's calculate the emission of the transition
             1 /λ = 1.097 10⁷ (1/10² - 1/8²)
             1 / λ = 1.097 10⁷ (0.01 - 0.015625)
             1 /λ = 0.006170625 10⁷
             λ = 162 10⁻⁷ m
 
        
             
        
        
        
Answer:
a)   x = 4.33 m
,   b)  w = 2 rad / s
,  f = 0.318 Hz
,  c) a = - 17.31 cm / s²,  
d) T =  3.15 s,  e)  A = 5.0 cm
Explanation:
In this exercise on simple harmonic motion we are given the expression for motion
           x = 5 cos (2t + π / 6)
they ask us for t = 0
a) the position of the particle
       x = 5 cos (π / 6)
       x = 4.33 m
remember angles are in radians
  
b) The general form of the equation is
           x = A cos (w t + Ф)
when comparing the two equations
          w = 2 rad / s
angular velocity and frequency are related
           w = 2π f
            f = w / 2π
            f = 2 / 2pi
            f = 0.318 Hz
c) the acceleration is defined by
       a == d²x / dt²
       a = - A w² cos (wt + Ф)
for t = 0
,  we substitute
       a = - 5,0 2² cos (π / 6)
       a = - 17.31 cm / s²
d) El period is
           T = 1/f
          T= 1/0.318
          T =  3.15 s
e) the amplitude
         A = 5.0 cm