I would say its a positive cgarge
To solve this problem it is necessary to consider two concepts. The first of these is the flow rate that can be defined as the volumetric quantity that a channel travels in a given time. The flow rate can also be calculated from the Area and speed, that is,
Q = V*A
Where,
A= Cross-sectional Area
V = Velocity
The second concept related to the calculation of this problem is continuity, which is defined as the proportion that exists between the input channel and the output channel. It is understood as well as the geometric section of entry and exit, defined as,


Our values are given as,


Re-arrange the equation to find the first ratio of rates we have:



The second ratio of rates is



Answer:
y = 10.44cos(2t - 0.291) cm
Explanation:
y = Acos(2πt/T + φ) = Acos(2πt/π + φ) = Acos(2t + φ)
v = y' = -2Αsin(2t + φ)
10 = Acos(2(0) + φ) = Acosφ
6 = -2Αsin(2(0) + φ) = -2Asinφ
6/10 = -2Asinφ/Acosφ = -2tanφ
tanφ = -0.3
φ = -0.291 radians
10 = Acos(-0.291)
A = 10/cos(-0.291) = 10.44
Answer:
The smallest distance the student that the student could be possibly be from the starting point is 6.5 meters.
Explanation:
For 2 quantities A and B represented as
and 
The sum is represented as
For the the values given to us the sum is calculated as

Now the since the uncertainity inthe sum is 
The closest possible distance at which the student can be is obtained by taking the negative sign in the uncertainity
Thus closest distance equals
meters
Answer:
Explanation:
just use the gravational force equation which is G x m of earth x m of object divided by r squared (which is radius of earth)