Answers are:
(1) KE = 1 kg m^2/s^2
(2) KE = 2 kg m^2/s^2
(3) KE = 3 kg m^2/s^2
(4) KE = 4 kg m^2/s^2
Explanation:
(1) Given mass = 0.125 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.125 * (16)
KE = 1 kg m^2/s^2
(2) Given mass = 0.250 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.250 * (16)
KE = 2 kg m^2/s^2
(3) Given mass = 0.375 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.375 * (16)
KE = 3 kg m^2/s^2
(4) Given mass = 0.500 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.5 * (16)
KE = 4 kg m^2/s^2
Answer:
184 feets
Explanation:
Given the data:
time (sec) __ velocity (ft/sec)
0 __________30
1 __________ 54
2 __________56
3 __________34
4 __________ 8
5 __________ 2
6 __________22
Using left end approximation:
(0,1) ___ f(0) = 30
(1,2) ___ f(1) = 54
(2,3) ___f(2) = 56
(3,4) ___f(3) = 34
(4,5) ___f(4) = 8
(5,6) __ f(5) = 2
Hence, the Total distance traveled during the 6 second interval is:
Change ; dT = 1
1 * (30 + 54 + 56 + 34 + 8 + 2) = 184
Answer:
distance r from the uranium atom is 18.27 nm
Explanation:
given data
uranium and iron atom distance R = 44.10 nm
uranium atom = singly ionized
iron atom = doubly ionized
to find out
distance r from the uranium atom
solution
we consider here that uranium electron at distance = r
and electron between uranium and iron so here
so we can say electron and iron distance = ( 44.10 - r ) nm
and we know single ionized uranium charge q2= 1.602 × 
C
and charge on iron will be q3 = 2 × 1.602 × C
so charge on electron is q1 = - 1.602 × C
and we know F = 
so now by equilibrium
Fu = Fi
=
put here k = 
and find r
= 
r = 18.27 nm
distance r from the uranium atom is 18.27 nm
Answer:
Displacement is 565.69 m at 45° west of north
Explanation:
Let north represent positive y axis and east represent positive x axis.
We have journey started from 600 N. Michigan Avenue, and walked 3 blocks toward north, 4 blocks toward west, and 1 block toward north to a train station.
3 blocks toward north = 300 j m
4 blocks toward west = -400 i m
1 blocks toward north = 100 j m
Total displacement = -400 i + 400 j m
Magnitude
Direction,
Direction is 45° west of north.
Displacement is 565.69 m at 45° west of north
The sprinter experiences muscle cramps and a stray dog on the track from A to B, followed immediately by acute nausea and a gaggle of news photographers from B to C.
If none of these selections appears on the list of choices, then perhaps we can do better if you'll permit us a glance at the picture that accompanies the question, even if you still don't let us see the choices.
