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Anika [276]
3 years ago
10

You purchase a ten-year 1,000 bond with semiannual coupons for 982. The bond has a 1,100 redemption payment at maturity, a nomin

al coupon rate of 7% for the first five years, and a nominal coupon rate of q% for the final five years. Christie calculated that her annual effective yield for the ten-year period was 7.35%. Find q
Physics
1 answer:
vladimir2022 [97]3 years ago
6 0

Answer:

ummmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

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What does the diagram show about phases and the phase of the substance as it is heated? Check all that apply.
Romashka [77]

Answer:

2 and 4

Explanation:

right on edge

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3 years ago
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A 1 pF capacitor is connected in parallel with a 2 pF capacitor, the parallel combination then being connected in series with a
asambeis [7]

Hi there!

We can approach this problem in many ways, but to show you how I arrive to the final conclusion, I will begin by solving the circuit with an assigned value for the power source.

Let's use a power source value of 6V (produces nice numbers).

Recall the following rules.

Capacitors in series:

  • Voltage ADDS up.
  • Charge is EQUAL across each.
  • Total capacitance uses the reciprocal rule.

Capacitors in parallel:

  • Voltage is EQUAL across each.
  • Charge ADDS up.
  • Total capacitance is simply the sum.

Solving for total capacitance:
C_P = 1 + 2 = 3 pF\\\\C_T = \frac{1}{\frac{1}{3} + \frac{1}{3}} = 1.5 pF

Using rules for capacitors in series and parallel, the total capacitance of the circuit is 1.5 pF.

Thus, the total charge is:
Q = C_TV\\Q = 1.5 pF * 6 V = 9 pC

9 pC will go through the parallel combination and the individual capacitor in series with the combination.

Since the voltage adds up, we can find the amount of voltage across the 3pF capacitor with the remaining going through the branches of the parallel combination.

V = \frac{Q}{C}\\\\V = \frac{9pC}{3pF} = 3V

Therefore, 3V goes through each branch since 6V - 3V = 3V.

Solving for the charge for each capacitor:
Q  = CV\\Q = (1 pF)(3V) = 3pC\\\\Q = (2pF)(3V) = 6pC\\\\Q = (3pF)(3V) = 9pC

<u>Thus, the capacitor with the greatest charge is the 3 pF capacitor. </u>

To explain without all of the work above, the equivalent capacitance of the parallel combination (1 pF + 2pF = 3pF) is equivalent to the capacitance of the capacitor in series (3pF). Thus, the voltage across the parallel capacitors (since voltage is the same across branches in parallel) and the series capacitor is equal. However, since charge SUMS UP for capacitors in parallel, they would have less charge than the single capacitor in series.

5 0
2 years ago
Place several E-Field Sensors at a few points on different equipotential lines, and look at the relationship between the electri
Vlada [557]

Answer:

2.

Explanation:

  • If a charge is moved along a equipotential line, no work is done on the charge.
  • If we remember that the work done by an external force, is just the product of the component of the force parallel to the displacement, if the force produces no work, this means that is perpendicular to the displacement.
  • So, as the electric field is just the force per unit charge, and has the same direction as the force (for a positive charge), it must be perpendicular to any equipotential line.
  • As the electric field (by convention) has the same direction as it would be taken by a positive test charge, and positive charges move from higher voltages to lower ones, the electric field is directed toward lines of lower voltages (like it happens between the plates of a capacitor).
3 0
3 years ago
Calculate the momentum, A bald eagle with a mass of 26.4 kg and velocity of 15.6 m/s​
AveGali [126]

Answer:

<u>411.84 kg m/s</u>

Explanation:

Formula :

<u>Momentum = mass × velocity</u>

<u />

=========================================================

Given :

⇒ mass = 26.4 kg

⇒ velocity = 15.6 m/s

=========================================================

Solving :

⇒ Momentum = 26.4 × 15.6

⇒ Momentum = <u>411.84 kg m/s</u>

3 0
2 years ago
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