first object gors highest
second object goes lowest.
Yes. Your answer is correct. I hope you do well on the quiz or whatever it is.
For this problem, we use the Coulomb's law written in equation as:
F = kQ₁Q₂/d²
where
F is the electrical force
k is a constant equal to 9×10⁹
Q₁ and Q₂ are the charge of the two objects
d is the distance between the two objects
Substituting the values:
F = (9×10⁹)(-22×10⁻⁹ C)(-22×10⁻⁹ C)/(0.10 m)²
F = 0.0004356 N
Answer:
Explanation:
A ) When gymnast is motionless , he is in equilibrium
T = mg
= 63 x 9.81
= 618.03 N
B )
When gymnast climbs up at a constant rate , he is still in equilibrium ie net force acting on it is zero as acceleration is zero.
T = mg
= 618.03 N
C ) If the gymnast climbs up the rope with an upward acceleration of magnitude 0.600 m/s2
Net force on it = T - mg , acting in upward direction
T - mg = m a
T = mg + m a
= m ( g + a )
= 63 ( 9.81 + .6)
= 655.83 N
D ) If the gymnast slides down the rope with a downward acceleration of magnitude 0.600 m/s2
Net force acting in downward direction
mg - T = ma
T = m ( g - a )
= 63 x ( 9.81 - .6 )
= 580.23 N
<span>The correct frequency when you tune a guitar is
when you hear the right tune in your own hearing and standard. The measure
frequency of a guitar string is when you measure the tune of the string
correctly. This is not the same because manual tuning is affected by many
factors.</span>
