Answer:
The velocity of the man from the frame of reference of a stationary observer is, V₂ = 5 m/s
Explanation:
Given,
Your velocity, V₁ = 2 m/
The velocity of the person, V₂ =?
The velocity of the person relative to you, V₂₁ = 3 m/s
According to the relative velocity of two
V₂₁ = V₂ -V₁
∴ V₂ = V₂₁ + V₁
On substitution
V₂ = 3 + 2
= 5 m/s
Hence, the velocity of the man from the frame of reference of a stationary observe is, V₂ = 5 m/s
Answer:
A body travels 10 meters during the first 5 seconds of its travel,and a total of 30 meters over the first 10 seconds of its travel
20miles / 5sec = 4miles /sec would be the average speed for the last 20 m
Explanation:
The answer is 4 m/s.
In the first 5 seconds, a body travelled 10 meters. In the first 10 seconds of the travel, the body travelled a total of 30 meters, which means that in the last 5 seconds, it travelled 20 meters (30m + 10m).
The relation of speed (v), distance (d), and time (t) can be expressed as:
v = d/t
We need to calculate the speed of the second 5 seconds of the travel:
d = 20 m (total 30 meters - first 10 meters)
t = 5 s (time from t = 5 seconds to t = 10 seconds)
Thus:
v = 20m / 5s = 4 m/s
PLEASE GIVE BRAINIEST!! HOPE THIS HELPS
Answer:
Electric switch is commonly known as the key of an electric circuit.
Answer:
a) C.M ![=(\bar x, \bar y)=(0.767,0.7)m](https://tex.z-dn.net/?f=%3D%28%5Cbar%20x%2C%20%5Cbar%20y%29%3D%280.767%2C0.7%29m)
b) ![(x_4,y_4)=(-1.917,-1.75)m](https://tex.z-dn.net/?f=%28x_4%2Cy_4%29%3D%28-1.917%2C-1.75%29m)
Explanation:
The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"
The center of mass on a two dimensional plane is defined with the following formulas:
![\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5EN%20m_i%20x_i%7D%7BM%7D)
![\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}](https://tex.z-dn.net/?f=%5Cbar%20y%20%3D%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5EN%20m_i%20y_i%7D%7BM%7D)
Where M represent the sum of all the masses on the system.
And the center of mass C.M ![=(\bar x, \bar y)](https://tex.z-dn.net/?f=%3D%28%5Cbar%20x%2C%20%5Cbar%20y%29)
Part a
represent the masses.
represent the coordinates for the masses with the units on meters.
So we have everything in order to find the center of mass, if we begin with the x coordinate we have:
![\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%5Cfrac%7B%283kg%2A0m%29%2B%285kg%2A2.3m%29%2B%287kg%2A0m%29%7D%7B3kg%2B5kg%2B7kg%7D%3D0.767m)
![\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m](https://tex.z-dn.net/?f=%5Cbar%20y%20%3D%5Cfrac%7B%283kg%2A0m%29%2B%285kg%2A0m%29%2B%287kg%2A1.5m%29%7D%7B3kg%2B5kg%2B7kg%7D%3D0.7m)
C.M ![=(\bar x, \bar y)=(0.767,0.7)m](https://tex.z-dn.net/?f=%3D%28%5Cbar%20x%2C%20%5Cbar%20y%29%3D%280.767%2C0.7%29m)
Part b
For this case we have an additional mass
and we know that the resulting new center of mass it at the origin C.M
and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)
![\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%5Cfrac%7B%283kg%2A0m%29%2B%285kg%2A2.3m%29%2B%287kg%2A0m%29%2B%286kg%2Aa%29%7D%7B3kg%2B5kg%2B7kg%2B6kg%7D%3D0m)
If we solve for a we got:
![(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0](https://tex.z-dn.net/?f=%283kg%2A0m%29%2B%285kg%2A2.3m%29%2B%287kg%2A0m%29%2B%286kg%2Aa%29%3D0)
![a=-\frac{(5kg*2.3m)}{6kg}=-1.917m](https://tex.z-dn.net/?f=a%3D-%5Cfrac%7B%285kg%2A2.3m%29%7D%7B6kg%7D%3D-1.917m)
![\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m](https://tex.z-dn.net/?f=%5Cbar%20y%20%3D%5Cfrac%7B%283kg%2A0m%29%2B%285kg%2A0m%29%2B%287kg%2A1.5m%29%2B%286kg%2Ab%29%7D%7B3kg%2B5kg%2B7kg%2B6kg%7D%3D0m)
![(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0](https://tex.z-dn.net/?f=%283kg%2A0m%29%2B%285kg%2A0m%29%2B%287kg%2A1.5m%29%2B%286kg%2Ab%29%3D0)
And solving for b we got:
![b=-\frac{(7kg*1.5m)}{6kg}=-1.75m](https://tex.z-dn.net/?f=b%3D-%5Cfrac%7B%287kg%2A1.5m%29%7D%7B6kg%7D%3D-1.75m)
So the coordinates for this new particle are:
![(x_4,y_4)=(-1.917,-1.75)m](https://tex.z-dn.net/?f=%28x_4%2Cy_4%29%3D%28-1.917%2C-1.75%29m)
Answer: He has only move 0.2 yards
Explanation: When you subtract 18.3 from 18.5 you get 0.2 and that is how much he's moved