A. 21 electrons
B. 44
C. 23 neutrons
Answer:
x = 4 m
Explanation:
For this exercise we must use the rotational equilibrium relationship, where we place zero at the turning point and counterclockwise rotations we will consider positive
as it indicates that the bar is in equilibrium, its center of mass coincides with the turning point, so the distance is zero and does not create torque on the system
∑τ = 0
W 3 - w x = 0
x = 3W / w
x = 3 Mg / mg
x = 3 M / m
let's calculate
x = 3 60/45
x = 4 m
Answer:
Diurnal motion, apparent daily motion of the heavens from east to west in which celestial objects seem to rise and set, a phenomenon that results from the Earth's rotation from west to east. The axis of this apparent motion coincides with the Earth's axis of rotation. Retrograde motion is an APPARENT change in the movement of the planet through the sky. It is not REAL in that the planet does not physically start moving backwards in its orbit. It just appears to do so because of the relative positions of the planet and Earth and how they are moving around the Sun.
Answer:
1.117935:1
Explanation:
Since the wires are of the same material, they will have the same resistivity
.
The cross-sectional area of the of a wire is given by;
![A=\pi\frac{d^2}{4}................(1)](https://tex.z-dn.net/?f=A%3D%5Cpi%5Cfrac%7Bd%5E2%7D%7B4%7D................%281%29)
where d is the diameter of the wire.
Also, the relationship between resistance R, cross-sectional area A and length l of a wire is given as follows;
![\rho=\frac{RA}{l}..................(2)](https://tex.z-dn.net/?f=%5Crho%3D%5Cfrac%7BRA%7D%7Bl%7D..................%282%29)
Since the resistivity same for both wires, say wire 1 and wire 2, we can wreite the following;
![\frac{R_1A_1}{l_1}=\frac{R_2A_2}{l_2}..................(3)](https://tex.z-dn.net/?f=%5Cfrac%7BR_1A_1%7D%7Bl_1%7D%3D%5Cfrac%7BR_2A_2%7D%7Bl_2%7D..................%283%29)
Hence from eqaution (3), the ration of wire 1 to 2 is expressed as;
![\frac{R_1}{R_2}=\frac{l_1A_2}{l_2A_1}..................(4)](https://tex.z-dn.net/?f=%5Cfrac%7BR_1%7D%7BR_2%7D%3D%5Cfrac%7Bl_1A_2%7D%7Bl_2A_1%7D..................%284%29)
Given;
![l_1=1.35l_2](https://tex.z-dn.net/?f=l_1%3D1.35l_2)
![\frac{R_1}{R_2}=\frac{1.35l_2A_2}{l_2A_1}\\\frac{R_1}{R_2}=\frac{1.35A_2}{A_1}.............(5)](https://tex.z-dn.net/?f=%5Cfrac%7BR_1%7D%7BR_2%7D%3D%5Cfrac%7B1.35l_2A_2%7D%7Bl_2A_1%7D%5C%5C%5Cfrac%7BR_1%7D%7BR_2%7D%3D%5Cfrac%7B1.35A_2%7D%7BA_1%7D.............%285%29)
We then use equation (1) to fine the ratio of the area
to ![A_2](https://tex.z-dn.net/?f=A_2)
bearing in mind that ![d_1=0.91d_2](https://tex.z-dn.net/?f=d_1%3D0.91d_2)
This ratio gives 0.8281. Substituting this into equation (5), we get the following;
![\frac{R_1}{R_2}= 1.35*0.8281=1.117935](https://tex.z-dn.net/?f=%5Cfrac%7BR_1%7D%7BR_2%7D%3D%201.35%2A0.8281%3D1.117935)
Answer:
46.8 kg
Explanation:
Mass = (density)(volume)
= (1.3)(36)
<u>M</u><u>a</u><u>s</u><u>s</u><u> </u><u>=</u><u> </u><u>4</u><u>6</u><u>.</u><u>8</u><u> </u><u>k</u><u>g</u>