Answer:
None
Explanation:
An scale is the factor by which actual features on ground are enlarged or reduced for representing on a plane. There are different kinds of scales:
- Verbal scale use of words to represent scale information on the map. The distance or linear units are used for depicting this scale on the map. For example: 1 inch = 1 Kilo meter.
- Fractional scale uses the numbers or values for showing the scale instead of words. As the name says, it is represented using a fraction or ratio. Example: 1: 10,000 or 1/10,000
- In large scale more details are shown in a map, however, less area coverage will be shown in a single map as the scale is large and more details are given. Example: 1:500
- Small scale is exactly opposite to the large scale, less details are shown as magnification is not enough, however a large amount of area can be shown in a single map. Example: 1:25,000
- A graphic scale is a bar that has been calibrated to show map distances. On maps that have been reduced or enlarged the original ratio and written scales are incorrect, since the relationship between map distance and real world distance has been altered, graphic scale is enlarged or reduced to the same extent as the map, this makes it the right option.
I hope you find this information useful and interesting! Good luck!
Answer:
a) 5.851× 10¹⁰m/s²
b) 2.411×10⁻¹¹s
c) 1.70×10⁻¹¹m
d) 1.661×10⁻²⁷KJ
Explanation:
A proton in the field experience a downward force of magnitude,
F = eE. The force of gravity on the proton will be negligible compared to the electric force
F = eE
a= eE/m
= 1.602×10⁻¹⁹ × 610/1.67×10⁻²⁷
= 5.851× 10¹⁰m/s²
b)
V = u + at
u= 0
v= 1.4106m/s
v= (0)t + at
t= v/a
= 1.4106m/s/5.851 ×10¹⁰
= 2.411×10⁻¹¹s
c)
S = ut + at²
= (o)t + 5.851×10¹⁰×(2.411×10⁻¹¹)²
= 1.70×10⁻¹¹m
d)
Ke = 1/2mv²
= (1.67×10⁻²⁷×)(1.4106)²/2
= 1.661×10⁻²⁷KJ
Since it was stated that it must move at constant
velocity, so the only force it must overpower is the frictional force.
So the equation is:
F cos θ = Ff
F cos 36 = 65 N
F = 80.34 N
<span>So the nurse must exert 80.34 N of force</span>
Answer:
The Moon's distance from the Earth varies during its orbit
Explanation:
The correct statement is ,The Moon's distance from the Earth varies during its orbit.
Important point regarding moon:
1 .Moon is a natural satellite of the earth.
2. Moon is the fifth largest satellite in solar system.
3.Second densest satellite in solar system.
4.Moon rotates about earth.
5.Moon is an astronomical body .
Answer:
The distance will be x = 41.7 [m]
Explanation:
We must first find the components in the x & y axes of the initial velocity.
![(v_{o})_{x} = 15*cos(20)= 14.09[m/s]\\(v_{o})_{y} = 15*sin(20)= 5.13[m/s]](https://tex.z-dn.net/?f=%28v_%7Bo%7D%29_%7Bx%7D%20%3D%2015%2Acos%2820%29%3D%2014.09%5Bm%2Fs%5D%5C%5C%28v_%7Bo%7D%29_%7By%7D%20%3D%2015%2Asin%2820%29%3D%205.13%5Bm%2Fs%5D)
The acceleration is the gravity acceleration therefore.
g = 9.81 [m/s^2]
Now we can calculate how long it takes to fall.
![y=(v_{o})_{y}*t-0.5*g*t^2\\-28 = 5.13*t-0.5*9.81*t^2\\-28=-4.905*t^2+5.13*t\\4.905*t^2-5.13*t=28\\t = 2.96[s]](https://tex.z-dn.net/?f=y%3D%28v_%7Bo%7D%29_%7By%7D%2At-0.5%2Ag%2At%5E2%5C%5C-28%20%3D%205.13%2At-0.5%2A9.81%2At%5E2%5C%5C-28%3D-4.905%2At%5E2%2B5.13%2At%5C%5C4.905%2At%5E2-5.13%2At%3D28%5C%5Ct%20%3D%202.96%5Bs%5D)
With this time we can find the horizontal distance that runs the projectile.
![x=(v_{o})_{x}*t\\x=14.09*2.96\\x=41.7[m]](https://tex.z-dn.net/?f=x%3D%28v_%7Bo%7D%29_%7Bx%7D%2At%5C%5Cx%3D14.09%2A2.96%5C%5Cx%3D41.7%5Bm%5D)
