Answer:
1 Frequency
2 Wavelength
3 Amplitude
4 Crest
Hope it helps pls mark brainliest
Answer:
C = 1.01
Explanation:
Given that,
Mass, m = 75 kg
The terminal velocity of the mass, ![v_t=60\ m/s](https://tex.z-dn.net/?f=v_t%3D60%5C%20m%2Fs)
Area of cross section, ![A=0.33\ m^2](https://tex.z-dn.net/?f=A%3D0.33%5C%20m%5E2)
We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,
R = W
or
![\dfrac{1}{2}\rho CAv_t^2=mg](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7D%5Crho%20CAv_t%5E2%3Dmg)
Where
is the density of air = 1.225 kg/m³
C is drag coefficient
So,
![C=\dfrac{2mg}{\rho Av_t^2}\\\\C=\dfrac{2\times 75\times 9.8}{1.225\times 0.33\times (60)^2}\\\\C=1.01](https://tex.z-dn.net/?f=C%3D%5Cdfrac%7B2mg%7D%7B%5Crho%20Av_t%5E2%7D%5C%5C%5C%5CC%3D%5Cdfrac%7B2%5Ctimes%2075%5Ctimes%209.8%7D%7B1.225%5Ctimes%200.33%5Ctimes%20%2860%29%5E2%7D%5C%5C%5C%5CC%3D1.01)
So, the drag coefficient is 1.01.
Answer:
Letter b is wavelength. Letter a is amplitude.
Explanation:
Let's imagine a simple experiment. Imagine you have a long thick rope which one end is at your hands, and you start an oscillatory motion in it, moving your hand up and down. Then a friend of you take a picture of the rope in motion, looking at the rope laterally. Now let's find the wavelength and amplitude. Amplitude is "The distance from the center of the oscillation of the rope (when the rope was not in motion) to its high or low point", or the vertical displacement, in our experiment. On the other hand, wavelength is "The distance between one high point /low point and the next high point /low point". Take a look at a photo of a wave in your textbook and you will find the answer as well. ; )
Answer:
![-4.941*10^8J.](https://tex.z-dn.net/?f=-4.941%2A10%5E8J.)
Explanation:
To solve this exercise it is necessary to take into account the concepts related to gravitational potential energy, as well as the concept of perigee and apogee of a celestial body.
By conservation of energy we know that,
![\Delta U = \Delta_{perogee}-\Delta_{Apogee}](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%20%5CDelta_%7Bperogee%7D-%5CDelta_%7BApogee%7D)
Where,
![U= \frac{-GmM_e}{r}](https://tex.z-dn.net/?f=U%3D%20%5Cfrac%7B-GmM_e%7D%7Br%7D)
Replacing
![\Delta U = \frac{-GmM_e}{r_p}- \frac{-GmM_e}{r_a}](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%20%5Cfrac%7B-GmM_e%7D%7Br_p%7D-%20%5Cfrac%7B-GmM_e%7D%7Br_a%7D)
![\Delta U = GmM_e (\frac{1}{r_A}-\frac{1}{r_p})](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%20GmM_e%20%28%5Cfrac%7B1%7D%7Br_A%7D-%5Cfrac%7B1%7D%7Br_p%7D%29)
Our values are given by,
![m = 85.5Kg](https://tex.z-dn.net/?f=m%20%3D%2085.5Kg)
![M_e = 5.97*10^{24}Kg](https://tex.z-dn.net/?f=M_e%20%3D%205.97%2A10%5E%7B24%7DKg)
![r_A = 7330Km](https://tex.z-dn.net/?f=r_A%20%3D%207330Km)
![r_p = 6610Km](https://tex.z-dn.net/?f=r_p%20%3D%206610Km)
![G = 6.67*10^{-11}Nm^2/Kg^2](https://tex.z-dn.net/?f=G%20%3D%206.67%2A10%5E%7B-11%7DNm%5E2%2FKg%5E2)
Replacing at the equation,
![\Delta U = (6.67*10^{-11})(85.5)(5.97*10^{24}) (\frac{1}{7330}-\frac{1}{6610})](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%20%286.67%2A10%5E%7B-11%7D%29%2885.5%29%285.97%2A10%5E%7B24%7D%29%20%28%5Cfrac%7B1%7D%7B7330%7D-%5Cfrac%7B1%7D%7B6610%7D%29)
![\Delta U = -4.941*10^8J](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%20-4.941%2A10%5E8J)
Therefore the Energy necessary for Sputnik I as it moved from apogee to perigee was ![-4.941*10^8J.](https://tex.z-dn.net/?f=-4.941%2A10%5E8J.)
Answer:
The term mass is used to refer to the amount of matter in any given object For instance, a person or object may be weightless on the moon because of the lack of gravity but that same person or object maintains the same mass regardless of location
Explanation: