C. Members of the same species work together for survival
A sphere is charged with electrons to −9 × 10−6 C. The value given is the total charge of all the electrons present in the sphere. To calculate the number of electrons in the sphere, we divide the the total charge with the charge of one electron.
N = 9 × 10−6 C / 1.6 × 10−19 C
N = 5.6 x 10^13
<h2>
Density of the unknown liquid is 771.93 kg/m³</h2>
Explanation:
An empty graduated cylinder weighs 55.26 g
Weight of empty cylinder = 55.26 g = 0.05526 kg
Volume of liquid filled = 48.1 mL = 48.1 x 10⁻⁶ m³
Weight of cylinder plus liquid = 92.39 g = 0.09239 kg
Weight of liquid = 0.09239 - 0.05526
Weight of liquid = 0.03713 kg
We have
Mass = Volume x Density
0.03713 = 48.1 x 10⁻⁶ x Density
Density = 771.93 kg/m³
Density of the unknown liquid is 771.93 kg/m³
<h3>
Answer:</h3>
225 meters
<h3>
Explanation:</h3>
Acceleration is the rate of change in velocity of an object in motion.
In our case we are given;
Acceleration, a = 2.0 m/s²
Time, t = 15 s
We are required to find the length of the slope;
Assuming the student started at rest, then the initial velocity, V₀ is Zero.
<h3>Step 1: Calculate the final velocity, Vf</h3>
Using the equation of linear motion;
Vf = V₀ + at
Therefore;
Vf = 0 + (2 × 15)
= 30 m/s
Thus, the final velocity of the student is 30 m/s
<h3>Step 2: Calculate the length (displacement) of the slope </h3>
Using the other equation of linear motion;
S = 0.5 at + V₀t
We can calculate the length, S of the slope
That is;
S = (0.5 × 2 × 15² ) - (0 × 15)
= 225 m
Therefore, the length of the slope is 225 m
Answer:
![a_{cm}=\frac{F}{2M}\\\\F_{fr}=\frac{F}{2}](https://tex.z-dn.net/?f=a_%7Bcm%7D%3D%5Cfrac%7BF%7D%7B2M%7D%5C%5C%5C%5CF_%7Bfr%7D%3D%5Cfrac%7BF%7D%7B2%7D)
Explanation:
Given the mass as M, the rotational inertia of the mower is;
![I_{cm}=MR^2](https://tex.z-dn.net/?f=I_%7Bcm%7D%3DMR%5E2)
-The roller doesn't slip while rolling;
![v_{cm}=wR, a_{cm}=\alpha R](https://tex.z-dn.net/?f=v_%7Bcm%7D%3DwR%2C%20a_%7Bcm%7D%3D%5Calpha%20R)
![\sum F_x=Ma_x, -F+F_{fr}=-Ma_{cm}\\\\a_{cm}=\frac{F-Fr}{M} \ \ \ \ \ \ \ \ eqtn1\\\\\sum \tau_{cm}=I_{cm}\alpha, F_{fr}(R)}=(MR^2)(\frac{a_{cm}}{R}), ->F_{fr}=Ma_c_m\ \ \ \ \ \ \ eqtn2\\\\a_{cm}=\frac{F-Ma_{cm}}{M}, ->F=2Ma_{cm}\\\\a_{cm}=\frac{F}{2M}\\\\\\\therefore F_{fr}=M(\frac{F}{2M})\\\\F_{fr}=\frac{F}{2}](https://tex.z-dn.net/?f=%5Csum%20F_x%3DMa_x%2C%20-F%2BF_%7Bfr%7D%3D-Ma_%7Bcm%7D%5C%5C%5C%5Ca_%7Bcm%7D%3D%5Cfrac%7BF-Fr%7D%7BM%7D%20%20%20%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20eqtn1%5C%5C%5C%5C%5Csum%20%5Ctau_%7Bcm%7D%3DI_%7Bcm%7D%5Calpha%2C%20F_%7Bfr%7D%28R%29%7D%3D%28MR%5E2%29%28%5Cfrac%7Ba_%7Bcm%7D%7D%7BR%7D%29%2C%20-%3EF_%7Bfr%7D%3DMa_c_m%5C%20%5C%20%5C%20%20%5C%20%5C%20%5C%20%5C%20eqtn2%5C%5C%5C%5Ca_%7Bcm%7D%3D%5Cfrac%7BF-Ma_%7Bcm%7D%7D%7BM%7D%2C%20-%3EF%3D2Ma_%7Bcm%7D%5C%5C%5C%5Ca_%7Bcm%7D%3D%5Cfrac%7BF%7D%7B2M%7D%5C%5C%5C%5C%5C%5C%5Ctherefore%20F_%7Bfr%7D%3DM%28%5Cfrac%7BF%7D%7B2M%7D%29%5C%5C%5C%5CF_%7Bfr%7D%3D%5Cfrac%7BF%7D%7B2%7D)