I don’t know sorry ;khbadkhb didhwbck( khwdicdwbihwd
A person's weight will change if they move from the earth to the moon. This does not however, change the person's mass. Mass is the amount of matter that makes up an object, and volume is how much space it takes up. On the moon, there is a lighter gravitational pull on said person, so they will not weigh as much if they stepped on a scale.
Answer:
5.71428571422 m/s
Explanation:
u = Initial velocity = 20 m/s
v = Final velocity
s = Displacement
a = Acceleration
Time taken = 15-1 = 14 s
Distance traveled in 1 second = ![20\times 1=20\ m](https://tex.z-dn.net/?f=20%5Ctimes%201%3D20%5C%20m)
![s=ut+\frac{1}{2}at^2\\\Rightarrow 200-20=20\times 14+\frac{1}{2}\times a\times 14^2\\\Rightarrow a=\frac{2(180-20\times14)}{14^2}\\\Rightarrow a=-1.02040816327\ m/s^2](https://tex.z-dn.net/?f=s%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2%5C%5C%5CRightarrow%20200-20%3D20%5Ctimes%2014%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20a%5Ctimes%2014%5E2%5C%5C%5CRightarrow%20a%3D%5Cfrac%7B2%28180-20%5Ctimes14%29%7D%7B14%5E2%7D%5C%5C%5CRightarrow%20a%3D-1.02040816327%5C%20m%2Fs%5E2)
![v=u+at\\\Rightarrow v=20-1.02040816327\times 14\\\Rightarrow v=5.71428571422\ m/s](https://tex.z-dn.net/?f=v%3Du%2Bat%5C%5C%5CRightarrow%20v%3D20-1.02040816327%5Ctimes%2014%5C%5C%5CRightarrow%20v%3D5.71428571422%5C%20m%2Fs)
The speed as she reaches the light at the instant it turns green is 5.71428571422 m/s
Answer:
Take-off velocity = v = 81.39[m/s]
Explanation:
We can calculate the takeoff speed easily, using the following kinematic equation.
![v_{f}^{2}=v_{i}^{2} +2*a*x](https://tex.z-dn.net/?f=v_%7Bf%7D%5E%7B2%7D%3Dv_%7Bi%7D%5E%7B2%7D%20%2B2%2Aa%2Ax)
where:
a = acceleration = 4[m/s^2]
x = distance = 750[m]
vi = initial velocity = 25 [m/s]
vf = final velocity
![v_{f}=\sqrt{(25)^{2}+(2*4*750) } \\v_{f}=81.39[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3D%5Csqrt%7B%2825%29%5E%7B2%7D%2B%282%2A4%2A750%29%20%7D%20%5C%5Cv_%7Bf%7D%3D81.39%5Bm%2Fs%5D)
Explanation:
The charge on the electron is, ![q=-1.6\times 10^{-19}\ C](https://tex.z-dn.net/?f=q%3D-1.6%5Ctimes%2010%5E%7B-19%7D%5C%20C)
The electric field at a distance r from the electron is :
![E=k\dfrac{q}{r^2}](https://tex.z-dn.net/?f=E%3Dk%5Cdfrac%7Bq%7D%7Br%5E2%7D)
Where
k is the electrostatic constant, ![k=\dfrac{1}{4\pi \epsilon_o}](https://tex.z-dn.net/?f=k%3D%5Cdfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_o%7D)
We know that the electric field lines starts from positive charge and ends at the negative charge. Also, for a positive charge the field lines are outwards while for a negative charge the field lines are inwards.
So, the correct option is " the electric field is directed toward the electron and has a magnitude of
. Hence, this is the required solution.