For number one the answer is Iodine because it is in the same group as fluorine. For number two the answer is Germanium for the same reason. For number three the answer is Aluminum for the same reason.
Answer:
Second option 6.3 N at 162° counterclockwise from
F1->
Explanation:
Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.
For the address x we have:
![-F_3sin(b) + F_1 = 0](https://tex.z-dn.net/?f=-F_3sin%28b%29%20%2B%20F_1%20%3D%200)
For the address and we have:
![-F_3cos(b) + F_2 = 0](https://tex.z-dn.net/?f=-F_3cos%28b%29%20%2B%20F_2%20%3D%200)
The forces
and
are known
![F_1 = 5.7\ N\\\\F_2 = 1.9\ N](https://tex.z-dn.net/?f=F_1%20%3D%205.7%5C%20N%5C%5C%5C%5CF_2%20%3D%201.9%5C%20N)
We have 2 unknowns (
and b) and we have 2 equations.
Now we clear
from the second equation and introduce it into the first equation.
![F_3 = \frac{F_2}{cos (b)}](https://tex.z-dn.net/?f=F_3%20%3D%20%5Cfrac%7BF_2%7D%7Bcos%20%28b%29%7D)
Then
![-\frac{F_2}{cos (b)}sin(b)+F_1 = 0\\\\F_1 = \frac{F_2}{cos (b)}sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = \frac{F_1}{F_2}\\\\tan(b) = \frac{5.7}{1.9}\\\\tan^{-1}(\frac{5.7}{1.9}) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°](https://tex.z-dn.net/?f=-%5Cfrac%7BF_2%7D%7Bcos%20%28b%29%7Dsin%28b%29%2BF_1%20%3D%200%5C%5C%5C%5CF_1%20%3D%20%5Cfrac%7BF_2%7D%7Bcos%20%28b%29%7Dsin%28b%29%5C%5C%5C%5CF_1%20%3D%20F_2tan%28b%29%5C%5C%5C%5Ctan%28b%29%20%3D%20%5Cfrac%7BF_1%7D%7BF_2%7D%5C%5C%5C%5Ctan%28b%29%20%3D%20%5Cfrac%7B5.7%7D%7B1.9%7D%5C%5C%5C%5Ctan%5E%7B-1%7D%28%5Cfrac%7B5.7%7D%7B1.9%7D%29%20%3D%20b%5C%5C%5C%5Cb%3D%2072%5C%C2%B0%5C%5C%5C%5Cm%20%3D%20b%20%2B90%5C%5C%5C%5C%5Cm%3D%20162%5C%C2%B0)
Then we find the value of ![F_3](https://tex.z-dn.net/?f=F_3)
![F_3 = \frac{F_1}{sin(b)}\\\\F_3 =\frac{5.7}{sin(72\°)}\\\\F_3 = 6.01 N](https://tex.z-dn.net/?f=F_3%20%3D%20%5Cfrac%7BF_1%7D%7Bsin%28b%29%7D%5C%5C%5C%5CF_3%20%3D%5Cfrac%7B5.7%7D%7Bsin%2872%5C%C2%B0%29%7D%5C%5C%5C%5CF_3%20%3D%206.01%20N)
Finally the answer is 6.3 N at 162° counterclockwise from
F1->
Answer:
Standard deviation = 3
Explanation:
Given
![N = 1201](https://tex.z-dn.net/?f=N%20%3D%201201)
![SS = 10800](https://tex.z-dn.net/?f=SS%20%3D%2010800)
Required
Determine the standard deviation
First, we need to determine the variance;
![Variance = \frac{SS}{N - 1}](https://tex.z-dn.net/?f=Variance%20%3D%20%5Cfrac%7BSS%7D%7BN%20-%201%7D)
This gives:
![Variance = \frac{10800}{1201 - 1}](https://tex.z-dn.net/?f=Variance%20%3D%20%5Cfrac%7B10800%7D%7B1201%20-%201%7D)
![Variance = \frac{10800}{1200}](https://tex.z-dn.net/?f=Variance%20%3D%20%5Cfrac%7B10800%7D%7B1200%7D)
![Variance = 9](https://tex.z-dn.net/?f=Variance%20%3D%209)
Know that:
![Variance = SD^2](https://tex.z-dn.net/?f=Variance%20%3D%20SD%5E2)
Where SD represents standard deviation
This gives
![9 = SD^2](https://tex.z-dn.net/?f=9%20%3D%20SD%5E2)
Take square root
![SD = \sqrt 9](https://tex.z-dn.net/?f=SD%20%3D%20%5Csqrt%209)
![SD = 3](https://tex.z-dn.net/?f=SD%20%3D%203)
Answer:
![\boxed {\boxed {\sf v_i= 4 \ m/s}}](https://tex.z-dn.net/?f=%5Cboxed%20%7B%5Cboxed%20%7B%5Csf%20v_i%3D%204%20%5C%20m%2Fs%7D%7D)
Explanation:
We are asked to find the cyclist's initial velocity. We are given the acceleration, final velocity, and time, so we will use the following kinematic equation.
![v_f= v_i + at](https://tex.z-dn.net/?f=v_f%3D%20v_i%20%2B%20at)
The cyclist is acceleration at 1.2 meters per second squared. After 10 seconds, the velocity is 16 meters per second.
= 16 m/s - a= 1.2 m/s²
- t= 10 s
Substitute the values into the formula.
![16 \ m/s = v_i + (1.2 \ m/s^2)(10 \ s)](https://tex.z-dn.net/?f=16%20%5C%20m%2Fs%20%3D%20v_i%20%2B%20%281.2%20%5C%20m%2Fs%5E2%29%2810%20%5C%20s%29)
Multiply.
![16 \ m/s = v_i + (1.2 \ m/s^2 * 10 \ s)](https://tex.z-dn.net/?f=16%20%5C%20m%2Fs%20%3D%20v_i%20%2B%20%281.2%20%5C%20m%2Fs%5E2%20%2A%2010%20%5C%20s%29)
![16 \ m/s = v_i + 12 \ m/s](https://tex.z-dn.net/?f=16%20%5C%20m%2Fs%20%3D%20v_i%20%2B%2012%20%5C%20m%2Fs)
We are solving for the initial velocity, so we must isolate the variable
. Subtract 12 meters per second from both sides of the equation.
![16 \ m/s - 12 \ m/s = v_i + 12 \ m/s -12 \ m/s](https://tex.z-dn.net/?f=16%20%5C%20m%2Fs%20-%2012%20%5C%20m%2Fs%20%3D%20v_i%20%2B%2012%20%5C%20m%2Fs%20-12%20%5C%20m%2Fs)
![4 \ m/s = v_i](https://tex.z-dn.net/?f=4%20%5C%20m%2Fs%20%3D%20v_i)
The cyclist's initial velocity is <u>4 meters per second.</u>
To solve this problem we will apply the concepts of the Magnetic Force. This expression will be expressed in both the vector and the scalar ways. Through this second we can directly use the presented values and replace them to obtain the value of the magnitude. Mathematically this can be described as,
![\vec{F_B} = q(\vec{V}\times \vec{B})](https://tex.z-dn.net/?f=%5Cvec%7BF_B%7D%20%3D%20q%28%5Cvec%7BV%7D%5Ctimes%20%5Cvec%7BB%7D%29)
![F_B = q|v||B| sin\theta](https://tex.z-dn.net/?f=F_B%20%3D%20q%7Cv%7C%7CB%7C%20sin%5Ctheta)
Here,
q = Charge
v = Velocity
B = Magnetic field
![\theta = \text{Angle between } \vec{B} \text{ and } \vec{V}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Ctext%7BAngle%20between%20%7D%20%5Cvec%7BB%7D%20%5Ctext%7B%20and%20%7D%20%5Cvec%7BV%7D)
Our values are given as,
![\theta = 35.7\°](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2035.7%5C%C2%B0)
![q = 22.5*10^{-9}C](https://tex.z-dn.net/?f=q%20%3D%2022.5%2A10%5E%7B-9%7DC)
![B = 1.05*10^{-5}T](https://tex.z-dn.net/?f=B%20%3D%201.05%2A10%5E%7B-5%7DT)
![v = 3.11m/s](https://tex.z-dn.net/?f=v%20%3D%203.11m%2Fs)
Replacing,
![F_B = (22.5*10^{-9}C)(3.11 \times 1.05*10^{-5}) sin(35.1\°)](https://tex.z-dn.net/?f=F_B%20%3D%20%2822.5%2A10%5E%7B-9%7DC%29%283.11%20%5Ctimes%201.05%2A10%5E%7B-5%7D%29%20sin%2835.1%5C%C2%B0%29)
![F_B = 4.224*10^{-13}N](https://tex.z-dn.net/?f=F_B%20%3D%204.224%2A10%5E%7B-13%7DN)
Therefore the size of the magnetic force acting on the bumble bee is ![4.22*10^{-13}N](https://tex.z-dn.net/?f=4.22%2A10%5E%7B-13%7DN)