Answer:
16.03m(2dp)
Explanation:
Ep=m x g x h
1100=7.0x 9.8( gravitational field strength) x h
Height= 1100/7.0 x 9.8
=16.03498542
= 16.03m (2dp)
Answer: a) 456.66 s ; b) 564.3 m
Explanation: The time spend to cover any distance a constant velocity is given by:
v= distance/time so t=distance/v
The slower student time is: t=780m/0.9 m/s= 866.66 s
For the faster students t=780 m/1,9 m/s= 410.52 s
Therefore the time difference is 866.66-410.52= 456.14 s
In order to calculate the distance that faster student should walk
to arrive 5,5 m before that slower student, we consider the follow expressions:
distance =vslower*time1
distance= vfaster*time 2
The time difference is 5.5 m that is equal to 330 s
replacing in the above expression we have
time 1= 627 s
time2 = 297 s
The distance traveled is 564,3 m
<span>Answer:
KE = (11/2)mω²r²,
particle B must have mass of 2m, while A has mass m.
Then the moment of inertia of the system is
I = Σ md² = m*(3r)² + 2m*r² = 11mr²
and then
KE = ½Iω² = ½ * 11mr² * ω² = 11mr²ω² / 2
So I'll proceed under that assumption.
For particle A, translational KEa = ½mv²
but v = ω*d = ω*3r, so KEa = ½m(3ωr)² = (9/2)mω²r²
For particld B, translational KEb = ½(2m)v²
but v = ω*r, so KEb = ½(2m)ω²r²
so total translational KE = (9/2 + 2/2)mω²r² = 11mω²r² / 2
which is equal to our rotational KE.</span>
Answer:
A
The current through the solution is
B
The current is moving from B to A
Explanation:
From the question we are told that
The number of that move from A to B is
The time taken to move from A to B
Since the value of 1 charge is
The quantity of charge Q that flow from A to B is mathematically given as
The number of that move from A to B is
Since time taken to move from A to B is equal to time taken to move from B to A
The quantity of charge Q that flow from B to A is mathematically given as
The total quantity of charge is
The current flowing through the solution is
The flow is from B to A cause current flow from the positive terminal to negative terminal