Answer:
A) If one travels around a closed path adding the voltages for which one enters the negative reference and subtracting the voltages for which one enters the positive reference, the total is zero.
Explanation:
Kirchhoff's voltage law deals with the conservation of energy when the current flows in a closed-loop path.
It states that the algebraic sum of the voltages around any closed loop in a circuit is always zero.
In other words, the algebraic sum of all the potential differences through a loop must be equal to zero.
As per the question Bob drops the bag full with feathers from the top of the building.
The mass of the bag(m)= 1.0 lb
Let the air resistance is neglected.As the bag is under free fall ,hence the only force that acts on the bag is the force of gravity which is in vertical downward direction.
Here the acceleration produced on bag due to the free fall will be nothing else except the acceleration due to gravity i.e g =9.8 m/s^2
Here we are asked to calculate the distance travelled by the bag at the instant 1.5 s
Hence time t= 1.5 s
From equation of kinematics we know that -
S=ut + 0.5at^2 [ here S is the distance travelled]
For motion under free fall initial velocity (u)=0.
Hence S= 0×1.5+{0.5×(-9.8)×(1.5)^2}
⇒ -S =0-11.025 m
⇒ S= 11.025 m
=11 m
Here the negative sign is taken only due to the vertical downward motion of the body .we may take is positive depending on our frame of reference .
Hence the correct option is B.
<span>It's pretty easy problem once you set it up.
Earth------------P--------------Moon
"P" is where the gravitational forces from both bodies are acting equally on a mass m
Let's define a few distances.
Rep = distance from center of earth to P
Rpm = distance from P to center of moon
Rem = distance from center of earth to center of moon
You are correct to use that equation. If the gravitational forces are equal then
GMearth*m/Rep² = Gm*Mmoon/Rpm²
Mearth/Mmoon = Rep² / Rpm²
Since Rep is what you're looking for we can't touch that. We can however rewrite Rpm to be
Rpm = Rem - Rep
Mearth / Mmoon = Rep² / (Rem - Rep)²
Since Mmoon = 1/81 * Mearth
81 = Rep² / (Rem - Rep)²
Everything is done now. The most complicated part now is the algebra,
so bear with me as we solve for Rep. I may skip some obvious or
too-long-to-type steps.
81*(Rem - Rep)² = Rep²
81*Rep² - 162*Rem*Rep + 81*Rem² = Rep²
80*Rep² - 162*Rem*Rep + 81*Rem² = 0
We use the quadratic formula to solve for Rep:
Rep = (81/80)*Rem ± (9/80)*Rem
Rep = (9/8)*Rem and (9/10)*Rem
Obviously, point P cannot be 9/8 of the way to the moon because it'll be
beyond the moon. Therefore, the logical answer would be 9/10 the way
to the moon or B.
Edit: The great thing about this idealized 2-body problem, James, is
that it is disguised as a problem where you need to know a lot of values
but in reality, a lot of them cancel out once you do the math. Funny
thing is, I never saw this problem in physics during Freshman year. I
saw it orbital mechanics in my junior year in Aerospace Engineering. </span>
sylent_reality
· 8 years ago
Answer:
a) factor 
b) factor 
c) factor 
d) factor 
Explanation:
Time period of oscillating spring-mass system is given as:


where:
frequency of oscillation
mass of the object attached to the spring
stiffness constant of the spring
a) <u>On doubling the mass:</u>
- New mass,

<u>Then the new time period:</u>




where the factor
as asked in the question.
b) On quadrupling the stiffness constant while other factors are constant:
New stiffness constant, 
<u>Then the new time period:</u>

where the factor
as asked in the question.
c) On quadrupling the stiffness constant as well as mass:
New stiffness constant, 
New mas, 
<u>Then the new time period:</u>

where factor
as asked in the question.
d) On quadrupling the amplitude there will be no effect on the time period because T is independent of amplitude as we can observe in the equation.
so, factor 