Answer:
If there is no damping, the amount of transmitted vibration that the microscope experienced is   = 
Explanation:
The motion of the ceiling is y = Y sinωt
y = 0.05 sin (2 π × 2) t
y = 0.05 sin 4 π t
K = 25 lb/ft  × 4  sorings
K = 100 lb/ft 
Amplitude of the microscope  ![\frac{X}{Y}= [\frac{1+2 \epsilon (\omega/ W_n)^2}{(1-(\frac{\omega}{W_n})^2)^2+(2 \epsilon  \frac{\omega}{W_n})^2}]](https://tex.z-dn.net/?f=%5Cfrac%7BX%7D%7BY%7D%3D%20%5B%5Cfrac%7B1%2B2%20%5Cepsilon%20%28%5Comega%2F%20W_n%29%5E2%7D%7B%281-%28%5Cfrac%7B%5Comega%7D%7BW_n%7D%29%5E2%29%5E2%2B%282%20%5Cepsilon%20%20%5Cfrac%7B%5Comega%7D%7BW_n%7D%29%5E2%7D%5D)
where;


 = 
= 4.0124
replacing them into the above equation and making X the subject of the formula:
 
 
 
 
 
 
Therefore; If there is no damping, the amount of transmitted vibration that the microscope experienced is   = 
 
        
             
        
        
        
Answer:
1.86 m
Explanation:
First, find the time it takes to travel the horizontal distance.  Given:
Δx = 52 m
v₀ = 26 m/s cos 31.5° ≈ 22.2 m/s
a = 0 m/s²
Find: t
Δx = v₀ t + ½ at²
52 m = (22.2 m/s) t + ½ (0 m/s²) t²
t = 2.35 s
Next, find the vertical displacement.  Given:
v₀ = 26 m/s sin 31.5° ≈ 13.6 m/s
a = -9.8 m/s²
t = 2.35 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (13.6 m/s) (2.35 s) + ½ (-9.8 m/s²) (2.35 s)²
Δy = 4.91 m
The distance between the ball and the crossbar is:
4.91 m − 3.05 m = 1.86 m
 
        
             
        
        
        
Answer:
yes  ( true)
Explanation:
positive  effects on all  the body systems.
 
        
             
        
        
        
The specific heat of the unknown substance with a mass of 0.158kg is 0.5478 J/g°C
HOW TO CALCULATE SPECIFIC HEAT CAPACITY:
The specific heat capacity of a substance can be calculated using the following formula:
Q = m × c × ∆T
Where;
- Q = quantity of heat absorbed (J)
- c = specific heat capacity (4.18 J/g°C)
- m = mass of substance
- ∆T = change in temperature (°C)
According to this question, 2,510.0 J of heat is required to heat the 0.158kg substance from 32.0°C to 61.0°C. The specific heat capacity can be calculated:
2510 = 158 × c × (61°C - 32°C)
2510 = 4582c
c = 2510 ÷ 4582
c = 0.5478 J/g°C
Therefore, the specific heat capacity of the unknown substance that has a mass of 0.158 kg is 0.5478 J/g°C.
Learn more about specific heat capacity at: brainly.com/question/2530523
 
        
             
        
        
        
Answer: The atomic number is the number of protons in an atom, and isotopes have the same atomic number but differ in the number of neutrons. The number of protons in an atom is called its atomic number. This number is very important because it is unique for atoms of a given element. All atoms of an element have the same number of protons, and every element has a different number of protons in its atoms.
Explanation: