In this case, you need the formula below where:
F = force
k = coulombs constant 8.99 x10^{9} N.m^{2} . C^{-2}
q1 = electric charge 1
q2 = electric charge 2
r = the distance between the charges
pls note: make sure your units are correct (in meters etc, not fm (<em>femto-meters</em>)).
Curiously, this question doesn't tell you what atom you are next to the nucleus of. Different numbers of protons in the nucleus of the atom will make for vastly different forces in your answer...
In a real system of levers, wheel or pulleys, the AMA (actual mechanical advantage) is less than the IMA (ideal mechanical advantage) because of the presence of friction.
In fact, the IMA and the AMA of a machine are defined as the ratio between the output force (the load) and the input force (the effort):
however, the difference is that the IMA does not take into account the presence of frictions, while the AMA does. As a result, the output force in the AMA is less than the output force in the IMA (because some energy is dissipated due to friction), and the AMA is less than the IMA.
Answer:
D) 4.9 m/s
Explanation:
Speed of Object in Free Fall
s = Rate of Acceleration, Due to Gravity × Length of Time
s = 9.8 × 2.0
s = m/s (Round Up to The Nearest Whole Number)
--
m/s = 9.8 ÷ 2.0 = 4.9 m/s
Answer:
after 6 second it will stop
he travel 36 m to stop
Explanation:
given data
speed = 12 m/s
distance = 100 m
decelerates rate = 2.00 m/s²
so acceleration a = - 2.00 m/s²
to find out
how long does it take to stop and how far does he travel
solution
we will apply here first equation of motion that is
v = u + at ......1
here u is speed 12 and v is 0 because we stop finally
put here all value in equation 1
0 = 12 + (-2) t
t = 6 s
so after 6 second it will stop
and
for distance we apply equation of motion
v²-u² = 2×a×s ..........2
here v is 0 u is 12 and a is -2 and find distance s
put all value in equation 2
0-12² = 2×(-2)×s
s = 36 m
so he travel 36 m to stop
Answer:
25.33 rpm
Explanation:
M = 100 kg
m1 = 22 kg
m2 = 28 kg
m3 = 33 kg
r = 1.60 m
f = 20 rpm
Let the new angular speed in rpm is f'.
According to the law of conservation of angular momentum, when no external torque is applied, then the angular momentum of the system remains constant.
Initial angular momentum = final angular momentum
(1/2 x M x r^2 + m1 x r^2 + m2 x r^2 + m3 x r^2) x ω =
(1/2 x M x r^2 + m1 x r^2 + m3 x r^2 ) x ω'
(1/2 M + m1 + m2 + m3) x 2 x π x f = (1/2 M + m1 + m3) x 2 x π x f'
( 1/2 x 100 + 22 + 28 + 33) x 20 = (1/2 x 100 + 22 + 33) x f'
2660 = 105 x f'
f' = 25.33 rpm
