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egoroff_w [7]
3 years ago
13

A risk-free, zero-coupon bond with a $5000 face value has ten years to maturity. The bond currently trades at $3650. What is the

yield to maturity of this bond
Physics
1 answer:
PtichkaEL [24]3 years ago
5 0

Answer:

The Yield to Maturity of the bond is YTM = 3.20%

Explanation:

Mathematically the Yield to Maturity of the bond  YTM is as follows

           YTM = \frac{C + \frac{F-P}{n} }{\frac{F+P}{2} }

Where C is the amount of payment to be made = $0

           P is the price i.e the present value =$3650

           F is the face value of the bond=$5000

          n is the year of maturity of the bond = 10 years

            YTM = \frac{0+\frac{5000-3650}{10} }{\frac{5000+3650}{2} } * \frac{100}{1}

                      =3.20%

                     

                     

     

   

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Answer:

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Answer:The correct options are:

1. A system is a group of objects analyzed as one unit.

2. Energy that moves across system boundaries is conserved.

Explanation:

A system is defined as group of interrelated or interacting items existing as a single unit or a whole to achieve a specific objective.Energy lost by the system is equal to the energy gained by the surroundings.

Two statements are true about a system:

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A series RLC circuit consists of a 52.0 Ω resistor, a 4.80 mH inductor, and a 330 nF capacitor. It is connected to an oscillator
Tju [1.3M]

Answer:

(G) 75.11 ohm

(H) 0.08 A

(I) 46.2 degree

Explanation:

R = 52 ohm

L = 4.8 m H = 4.8 x 106-3 H

C = 330 nF = 330 x 10^-9 F

Vo = 6 V

(G)

f = 5000 Hz

Let the impedance is Z.

X_{L}= 2 \pi fL = 2 \times 3.14\times 5000\times 4.8\times 10^{-3}=150.72 ohm

X_{c}= \frac{1}{2 \pi fC}=\frac{1}{2\times 3.14\times 5000\times 330\times 10^{-9}}=96.51 ohm

Z=\sqrt{R^{2}+\left ( X_{L}-X_{c} \right )^{2}}

Z=\sqrt{52^{2}+\left (150.72-96.51)^{2}}=75.11 ohm

(H) Let Io be the peak current

I_{0}=\frac{V_{0}}{Z}=\frac{6}{75.11}=0.0798 A = 0.08 A

(I) Let Ф be the phase angle

tan\phi = \frac{X_{L}-X_{C}}{R}

tan\phi =\frac{150.72-96.51}}{52}=1.0425

Ф = 46.2 degree

4 0
3 years ago
Find the required answer to the following by substituting the numbers. (Bonus points for knowing the units). PLEASE ANSWER
Alex73 [517]

Answer:

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F=13.2×10

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I=5

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3 years ago
Four identical masses of 2.5 kg each are located at the corners of a square with 1.0-m sides. What is the net force on any one o
Ghella [55]

Answer:

F=8.0*10^{-10}N

Explanation:

See the attached file for the masses distributions

The force between two masses at distance r is expressed as

F=\frac{Gm_{1}m_{2}  }{r^{2} }\\ G=Gravitional constant \\

since the masses are of the same value, the above formula can be reduce to

F=\frac{Gm^{2}}{r^{2} }\\

using vector notation,Let use consider the force on the lower left corner of the mass due to the upper left side of the mass is

F_{12} =\frac{Gm^{2}}{r^{2} }j\\

The force on the lower left corner of the mass due to the lower right side of the mass is

F_{14} =\frac{Gm^{2}}{r^{2} }i\\

The force on the lower left corner of the mass due to the upper right side of the mass is

F_{13} =\frac{Gm^{2}}{d^{2} }cos\alpha i +\frac{Gm^{2}}{d^{2} }sin\alpha j\\

The net force can be express as

F=\frac{Gm^{2}}{r^{2} }j +\frac{Gm^{2}}{r^{2} }i +\frac{Gm^{2}}{d^{2} }cos\alpha i +\frac{Gm^{2}}{d^{2} }sin\alpha j\\\\F=Gm^{2}[\frac{1}{r^{2}}+ \frac{1}{d^{2}cos\alpha }]i + Gm^{2}[\frac{1}{r^{2}}+ \frac{1}{d^{2}sin\alpha }]j\\\alpha=45^{0}, G=6.67*10^{-11}Nmkg^{-2}

if we insert values we arrive at

F=6.67*10^{-11}*2.5^{2}[\frac{1}{1^{2}}+ \frac{1}{\sqrt{2}^{2}cos45 }]i + 6.67*10^{-11}*2.5^{2}[\frac{1}{1^{2}}+ \frac{1}{\sqrt{2}^{2}sin45}]j\\F=5.643*10^{-10}i+5.643*10^{-10}j

if we solve for the magnitude, we arrive at

F=5.643*10^{-10}i+5.643*10^{-10}j \\F=\sqrt{(5.643*10^{-10})^{2} +(5.643*10^{-10})}^{2} \\F=8.0*10^{-10}

Hence the net force on one of the masses is

F=8.0*10^{-10}N

8 0
3 years ago
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