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Amanda [17]
4 years ago
5

What condition must exist if a bond’s coupon rate is to equal both the bond’s current yield and its yield to maturity? Assume th

e market rate of interest for this bond is positive.
Physics
1 answer:
insens350 [35]4 years ago
6 0

The condition that must exist if a bond’s coupon rate is to equal both the bond’s current yield and its yield to maturity is that THE BOND MUST BE PRICED AT PAR.

Explanation:

  • The par value is also the amount upon which the entity calculates the interest that it owes to investors. Thus, if the stated interest rate on a bond is 10% and the bond par value is $1,000, then the issuing entity must pay $100 every year until it redeems the bond.
  • Par value is the face value of a bond. Par value is important for a bond or fixed-income instrument because it determines its maturity value as well as the dollar value of coupon payments.
  • When a bond is issued at par value it is sold for the face value amount. This generally means that the bond's market and contract rates are equal to each other, meaning that there is no bond premium or discount.

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Only question 7. I keep getting it wrong and i’m not sure
Akimi4 [234]

Answer:

I think you just have to input the horizontal velocity component.

Explanation:

Assuming no air resistance

The horizontal velocity is constant at

vx = 17.5cos71 = 5.6974... = 5.70 i m/s

The vertical velocity varies with gravity

vy = 17.5sin71 + -9.81(22) = -199.273... = -199 j m/s

v = 5.70i - 199j m/s

arguably one should round to only two significant digits

3 0
3 years ago
Premium
Luda [366]

Explanation:

the formula of speed is distance traveled by time it work

5 0
3 years ago
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Atoms of what two elements are in every organic compound?
Georgia [21]
Carbon atoms and hydrogen I hope this helps
8 0
3 years ago
You are designing a new piece of fishing equipment that will allow you to catch a fish from the surface. You know that water ref
OleMash [197]

Answer:

Law of refraction

Explanation:

An experiment to analyze the refraction of light in water can easily be performed with a laser pointer and protractor.

We throw the fishing rod line into the water, place the protractor at the point where the line touches the water and use the direction of the line for the direction of the laser pointer (on), the laser is visible by the reflection on the particles in the air.

The vertical line is called Normal and all angles must be measured with respect to this reference in optics.

Having these angles and the refractive index of water we can use the law of refraction

         n₁ sin θ₁  = n₂ sin θ₂

         θ₂ = sin^{-1} ( \frac{n_1}{n_2} \ sin \ \theta_1 )

we can repeat several times to analyze several different input points (different angles) and to decrease the errors in the measurements.

the refractive index of air is n1 = 1 and n2= 1.33  (water)

4 0
3 years ago
A 1.10 μF capacitor is connected in series with a 1.92 μF capacitor. The 1.10 μF capacitor carries a charge of +10.1 μC on one p
miss Akunina [59]

Answer:

(a). The potential on the negative plate is 42.32 V.

(b). The equivalent capacitance of the two capacitors is 0.69 μF.

Explanation:

Given that,

Charge = 10.1 μC

Capacitor C₁ = 1.10 μF

Capacitor C₂ = 1.92 μF

Capacitor C₃ = 1.10 μF

Potential V₁ = 51.5 V

Let V₁ and V₂ be the potentials on the two plates of the capacitor.

(a). We need to calculate the potential on the negative plate of the 1.10 μF capacitor

Using formula of potential difference

V_{1}=\dfrac{Q}{C_{1}}

Put the value into the formula

V_{1}=\dfrac{10.1 \times10^{-6}}{1.10\times10^{-6}}

V_{1}=9.18\ V

The potential on the second plate

V_{2}=V-V_{1}

V_{2}=51.5 -9.18

V_{2}=42.32\ v

(b). We need to calculate the equivalent capacitance of the two capacitors

Using formula of equivalent capacitance

C=\dfrac{C_{1}\timesC_{2}}{C_{1}+C_{2}}

Put the value into the formula

C=\dfrac{1.10\times10^{-6}\times1.92\times10^{-6}}{(1.10+1.92)\times10^{-6}}

C=6.99\times10^{-7}\ F

C=0.69\ \mu F

Hence, (a). The potential on the negative plate is 42.32 V.

(b). The equivalent capacitance of the two capacitors is 0.69 μF.

7 0
3 years ago
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