Your weight on the moon given the data from the question is 110.5 N
<h3>Definition of mass and weight </h3>
Mass is simply defined as the quantity of matter present in an object. The mass of an object is constant irrespective of the location of the object.
Weight is simply defined as the gravitational pull on an object. The weight of an object varies from place to place due to gravity.
<h3>Relationship between mass and weight </h3>
Mass and weight are related according to the following equation
Weight (W) = mass (m) × Acceleration due to gravity (g)
<h3>How to determine the weight on the moon</h3>
- Mass (m) = 65 Kg
- Acceleration due to gravity on the moon (g) = 1.7 m/s²
- Weight (W) =?
W = mg
W = 65 × 1.7
W = 110.5 N
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Answer:the voltmeter measures the potential difference of the circuit,. Voltmeter is a device used to measure potential difference.
Explanation:
Answer:
Explanation:
Density = Mass / Volume = 850 / 40*10*5 = 0.425 g /cm^3
Answer:
If there is no damping, the amount of transmitted vibration that the microscope experienced is = ![5.676*10^{-3} \ mm](https://tex.z-dn.net/?f=5.676%2A10%5E%7B-3%7D%20%5C%20mm)
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;
![\epsilon = 0](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%200)
![W_n = \sqrt { \frac{k}{m}}](https://tex.z-dn.net/?f=W_n%20%3D%20%5Csqrt%20%7B%20%5Cfrac%7Bk%7D%7Bm%7D%7D)
= ![\sqrt { \frac{100*32.2}{200}}](https://tex.z-dn.net/?f=%5Csqrt%20%7B%20%5Cfrac%7B100%2A32.2%7D%7B200%7D%7D)
= 4.0124
replacing them into the above equation and making X the subject of the formula:
![Y * \frac{1}{\sqrt{(1-(\frac{\omega}{W_n})^2)^2})}}](https://tex.z-dn.net/?f=Y%20%2A%20%5Cfrac%7B1%7D%7B%5Csqrt%7B%281-%28%5Cfrac%7B%5Comega%7D%7BW_n%7D%29%5E2%29%5E2%7D%29%7D%7D)
![0.05 * \frac{1}{\sqrt{(1-(\frac{4 \pi}{4.0124})^2)^2})}}](https://tex.z-dn.net/?f=0.05%20%2A%20%5Cfrac%7B1%7D%7B%5Csqrt%7B%281-%28%5Cfrac%7B4%20%5Cpi%7D%7B4.0124%7D%29%5E2%29%5E2%7D%29%7D%7D)
![5.676*10^{-3} \ mm](https://tex.z-dn.net/?f=5.676%2A10%5E%7B-3%7D%20%5C%20mm)
Therefore; If there is no damping, the amount of transmitted vibration that the microscope experienced is = ![5.676*10^{-3} \ mm](https://tex.z-dn.net/?f=5.676%2A10%5E%7B-3%7D%20%5C%20mm)