Answer:
Part a)
![t = 1.65 s](https://tex.z-dn.net/?f=t%20%3D%201.65%20s)
Part b)
![x = 40.4 m](https://tex.z-dn.net/?f=x%20%3D%2040.4%20m)
Since the distance of other building is 15 m so YES it can make it to other building
Part c)
![v = 27.3 m/s](https://tex.z-dn.net/?f=v%20%3D%2027.3%20m%2Fs)
direction of velocity is given as
![[tex]\theta = 26.35 degree](https://tex.z-dn.net/?f=%5Btex%5D%5Ctheta%20%3D%2026.35%20degree)
Explanation:
Part a)
acceleration due to gravity on this planet is 3/4 times the gravity on earth
So the acceleration due to gravity on this new planet is given as
![a = \frac{3}{4}(9.81)](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B3%7D%7B4%7D%289.81%29)
![a = 7.36 m/s^2](https://tex.z-dn.net/?f=a%20%3D%207.36%20m%2Fs%5E2)
now the vertical displacement covered by the canister is given as
![y = 10 m](https://tex.z-dn.net/?f=y%20%3D%2010%20m)
now by kinematics we have
![y = \frac{1}{2}gt^2](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
![10 = \frac{1}{2}(7.36)t^2](https://tex.z-dn.net/?f=10%20%3D%20%5Cfrac%7B1%7D%7B2%7D%287.36%29t%5E2)
![t = 1.65 s](https://tex.z-dn.net/?f=t%20%3D%201.65%20s)
Part b)
Horizontal speed of the canister is given as
![v_x = 24.5 m/s](https://tex.z-dn.net/?f=v_x%20%3D%2024.5%20m%2Fs)
now the distance moved by it
![x = v_x t](https://tex.z-dn.net/?f=x%20%3D%20v_x%20t)
![x = 24.5 (1.65)](https://tex.z-dn.net/?f=x%20%3D%2024.5%20%281.65%29)
![x = 40.4 m](https://tex.z-dn.net/?f=x%20%3D%2040.4%20m)
Since the distance of other building is 15 m so YES it can make it to other building
Part c)
Final velocity in X direction will remains the same
![v_x = 24.5 m/s](https://tex.z-dn.net/?f=v_x%20%3D%2024.5%20m%2Fs)
final velocity in Y direction
![v_y = v_i + at](https://tex.z-dn.net/?f=v_y%20%3D%20v_i%20%2B%20at)
![v_y = 0 + (7.36)(1.65)](https://tex.z-dn.net/?f=v_y%20%3D%200%20%2B%20%287.36%29%281.65%29)
![v_y = 12.14 m/s](https://tex.z-dn.net/?f=v_y%20%3D%2012.14%20m%2Fs)
now magnitude of velocity is given as
![v = \sqrt{v_x^2 + v_y^2}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7Bv_x%5E2%20%2B%20v_y%5E2%7D)
![v = \sqrt{24.5^2 + 12.14^2}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B24.5%5E2%20%2B%2012.14%5E2%7D)
![v = 27.3 m/s](https://tex.z-dn.net/?f=v%20%3D%2027.3%20m%2Fs)
direction of velocity is given as
![\theta = tan^{-1}\frac{v_y}{v_x}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%5E%7B-1%7D%5Cfrac%7Bv_y%7D%7Bv_x%7D)
![\theta = tan^{-1}\frac{12.14}{24.5}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%5E%7B-1%7D%5Cfrac%7B12.14%7D%7B24.5%7D)
![[tex]\theta = 26.35 degree](https://tex.z-dn.net/?f=%5Btex%5D%5Ctheta%20%3D%2026.35%20degree)
Answer:
![\frac{F}{W} = 9.37 \times 10^{-4}](https://tex.z-dn.net/?f=%5Cfrac%7BF%7D%7BW%7D%20%3D%209.37%20%5Ctimes%2010%5E%7B-4%7D)
Explanation:
Radius of the pollen is given as
![r = 12.0 \mu m](https://tex.z-dn.net/?f=r%20%3D%2012.0%20%5Cmu%20m)
Volume of the pollen is given as
![V = \frac{4}{3}\pi r^3](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B4%7D%7B3%7D%5Cpi%20r%5E3)
![V = \frac{4}{3}\pi (12\mu m)^3](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B4%7D%7B3%7D%5Cpi%20%2812%5Cmu%20m%29%5E3)
![V = 7.24 \times 10^{-15} m^3](https://tex.z-dn.net/?f=V%20%3D%207.24%20%5Ctimes%2010%5E%7B-15%7D%20m%5E3)
mass of the pollen is given as
![m = \rho V](https://tex.z-dn.net/?f=m%20%3D%20%5Crho%20V)
![m = 7.24 \times 10^{-12}](https://tex.z-dn.net/?f=m%20%3D%207.24%20%5Ctimes%2010%5E%7B-12%7D)
so weight of the pollen is given as
![W = mg](https://tex.z-dn.net/?f=W%20%3D%20mg)
![W = (7.24 \times 10^{-12})(9.81)](https://tex.z-dn.net/?f=W%20%3D%20%287.24%20%5Ctimes%2010%5E%7B-12%7D%29%289.81%29)
![W = 7.1 \times 10^{-11}](https://tex.z-dn.net/?f=W%20%3D%207.1%20%5Ctimes%2010%5E%7B-11%7D)
Now electric force on the pollen is given
![F = qE](https://tex.z-dn.net/?f=F%20%3D%20qE)
![F = (-0.700\times 10^{-15})(95)](https://tex.z-dn.net/?f=F%20%3D%20%28-0.700%5Ctimes%2010%5E%7B-15%7D%29%2895%29)
![F = 6.65 \times 10^{-14} N](https://tex.z-dn.net/?f=F%20%3D%206.65%20%5Ctimes%2010%5E%7B-14%7D%20N)
now ratio of electric force and weight is given as
![\frac{F}{W} = \frac{6.65 \times 10^{-14}}{7.1 \times 10^{-11}}](https://tex.z-dn.net/?f=%5Cfrac%7BF%7D%7BW%7D%20%3D%20%5Cfrac%7B6.65%20%5Ctimes%2010%5E%7B-14%7D%7D%7B7.1%20%5Ctimes%2010%5E%7B-11%7D%7D)
![\frac{F}{W} = 9.37 \times 10^{-4}](https://tex.z-dn.net/?f=%5Cfrac%7BF%7D%7BW%7D%20%3D%209.37%20%5Ctimes%2010%5E%7B-4%7D)
Answer:
well theirs health, idustry, transport, comunication, information tech, and plenty more
Explanation:
can i plz have brainlyest it helps me continew helping people like you
As it stands now, that statement is false.
There are two ways to make the statement true:
#1). Exchange the places of the words "Opposite" and "like".
#2). Exchange the places of the words "repel" and "attract".
Either ONE of these changes will make the statement true.
Doing BOTH of these changes will leave it false.