Answer:
On Edg 2021, the answer was "B = The moon's orbital period and distance from Earth"
Explanation:
I hope this is correct for you, if it isn't I am very sorry. Good luck! :)
Answer:
0.49m
Explanation:
So you need to change the original equation for finding fields to find distance, and then just plug in the numbers
Which equals 0.49meters
Also it was right on Acellus :)
Hope this helps
Answer:
y₀ = 1020.3 m
Explanation:
This is a projectile launching exercise, in this case as the package is released its initial vertical velocity is zero.
y = y₀ +
t - ½ g t²
when it reaches the ground its height is zero
0 = y₀ + 0 - ½ g t²
y₀ = ½ g t²
let's calculate
y₀ = ½ 9.8 14.43²
y₀ = 1020.3 m
Answer:0.478 c
Explanation:
Given
mass of lighter Particle![(m_1)=3\times 10^{-28} kg](https://tex.z-dn.net/?f=%28m_1%29%3D3%5Ctimes%2010%5E%7B-28%7D%20kg)
mass of heavier Particle![(m_2)=1.51\times 10^{-27} kg](https://tex.z-dn.net/?f=%28m_2%29%3D1.51%5Ctimes%2010%5E%7B-27%7D%20kg)
speed of lighter particle![(v_1)=0.834 c](https://tex.z-dn.net/?f=%28v_1%29%3D0.834%20c)
Let speed of heavier particle![=v_2](https://tex.z-dn.net/?f=%3Dv_2)
and Momentum of the particle is given by
![P=\frac{mv}{\sqrt{1-(\frac{v}{c})^2}}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7Bmv%7D%7B%5Csqrt%7B1-%28%5Cfrac%7Bv%7D%7Bc%7D%29%5E2%7D%7D)
![P_1=\frac{m_1v_1}{\sqrt{1-(\frac{v_1}{c})^2}}](https://tex.z-dn.net/?f=P_1%3D%5Cfrac%7Bm_1v_1%7D%7B%5Csqrt%7B1-%28%5Cfrac%7Bv_1%7D%7Bc%7D%29%5E2%7D%7D)
![P_1=\frac{3\times 10^{-28}\times 0.834 c}{\sqrt{1-(\frac{0.834 c}{c})^2}}](https://tex.z-dn.net/?f=P_1%3D%5Cfrac%7B3%5Ctimes%2010%5E%7B-28%7D%5Ctimes%200.834%20c%7D%7B%5Csqrt%7B1-%28%5Cfrac%7B0.834%20c%7D%7Bc%7D%29%5E2%7D%7D)
![P_1=8.219\times 10^{-28} kg c](https://tex.z-dn.net/?f=P_1%3D8.219%5Ctimes%2010%5E%7B-28%7D%20kg%20c)
![P_2=\frac{m_2v_2}{\sqrt{1-(\frac{v_2}{c})^2}}](https://tex.z-dn.net/?f=P_2%3D%5Cfrac%7Bm_2v_2%7D%7B%5Csqrt%7B1-%28%5Cfrac%7Bv_2%7D%7Bc%7D%29%5E2%7D%7D)
as momentum is conserved therefore ![P_1=P_2](https://tex.z-dn.net/?f=P_1%3DP_2)
![8.219\times 10^{-28} kg c=\frac{1.51\times 10^{-27}\times v_2}{\sqrt{1-(\frac{v_2}{c})^2}}](https://tex.z-dn.net/?f=8.219%5Ctimes%2010%5E%7B-28%7D%20kg%20c%3D%5Cfrac%7B1.51%5Ctimes%2010%5E%7B-27%7D%5Ctimes%20v_2%7D%7B%5Csqrt%7B1-%28%5Cfrac%7Bv_2%7D%7Bc%7D%29%5E2%7D%7D)
![v_2=0.478 c](https://tex.z-dn.net/?f=v_2%3D0.478%20c)
Answer:
0.25hr
Explanation:
Given parameters:
Distance = 200km
Time = 2hrs
Now, let us find the rate which is the speed;
Speed =
Insert parameters:
Speed =
= 100km/hr
Since the speed of Imamu = 2 x speed of Mosi = 200km/hr
Time taken =
=
= 0.25hr