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3 years ago

Which of the following describes a protective put?

Physics

1 answer:

11111nata11111 [884]3 years ago

7 0

A long put option on a stock plus a long position in the stock describes a protective put.

Option a

<h3>Explanation:</h3>

A protective put position can be defined by buying or owning stock and buying put options on a share-for-share basis. It is a "risk-management strategy" that uses the options contracts which investors employ to guard themselves against the loss of owning a stock or asset. In this strategy, traders believe that the price of the asset may decline in the future.

For example: Suppose 50 shares are purchased (or owned) and one put is purchased. So, when the stock price declines, the purchased put protects the strike price.

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The plane of a rectangular coil, 7.2 cm by 3.7 cm, is perpendicular to the direction of a uniform magnetic field B. If the coil

netineya [11]

Answer:

The rate of change of magnetic field is 2.23 T/s.

Explanation:

Given that,

Dimension of rectangular coil is 7.2 cm by 3.7 cm.

Number of turns in the coil, N = 104

Resistance of the coil, R = 12.4 ohms

Current, I = 0.05 A

We need to find the rate of change of magnetic field in the coil. The induced emf is given by the rate of change of magnetic flux. So,

$\epsilon=-\dfrac{d\phi}{dt}$

Ohm's law is :

$\epsilon=IR$

So,

$IR=-\dfrac{d\phi}{dt}\\\\IR=-\dfrac{d(NBA)}{dt}\\\\IR=-NA\dfrac{dB}{dt}\\\\\dfrac{dB}{dt}=\dfrac{IR}{NA}\\\\\dfrac{dB}{dt}=\dfrac{0.05\times 12.4}{104\times 7.2\times 10^{-2}\times 3.7\times 10^{-2}}\\\\\dfrac{dB}{dt}=2.23\ T/s$

So, the rate of change of magnetic field is 2.23 T/s.

4 0

3 years ago

A cannon is shot downwards with an initial speed of 10m/s at an angle of 20 degrees South of East off a 60m cliff. Find the init

Sliva [168]

Answer:hmm

Explanation:

7 0

3 years ago

a ball was projected in such a way that it attained the maximum horizontal distance with a velocity of 20m/s calculate the verti

Eduardwww [97]

Answer:

80m

Explanation:

u=20,R=?,sin theta=1,g=10

R=u²sin2theta/g

R=20²x2/10

R=400x2=800/10

R=80m

7 0

3 years ago

A cubic metal box with sides of 17 cm contains air at a pressure of 1 atm and a temperature of 278 K. The box is sealed so that

const2013 [10]

Answer:

F = 3.98 kN

Explanation:

GIVEN DATA:

sides of box = 17 cm

pressure = 1 atm = 101325 N/m2

T2 = 378K

T1 = 278 K

final pressure can be calculate by using below relation

$\frac{P_{1}}{P_{2}}=\frac{T_{1}}{T_{2}}$

we know that

force = pressure * area

therefore force is

$F =(\frac{T_{1}}{T_{2}}*P_{1})A$

$F =(\frac{378}{278}*101325)(17*10^{-2})^{2}$

F = 3.98 kN

5 0

3 years ago

Fiber-optic cables are used widely for internet wiring, data transmission, and surgeries. When light passes through a fiber-opti

Gwar [14]

After one meter, 3.4% of the light is gone ... either soaked up in the fiber
material or escaped from it. So only (100 - 3.4) = 96.6% of the light
remains, to go on to the next meter.

After the second meter, 96.6% of what entered it emerges from it, and
that's 96.6% of 96.6% of the original signal that entered the beginning
of the fiber.

==> After 2 meters, the intensity has dwindled to (0.966)² of its original level.
It's that exponent of ' 2 ' that corresponds to the number of meters that the light
has traveled through.

==> After 'x' meters of fiber, the remaininglight intensity is (0.966) ^x-power
of its original value.

If you shine 1,500 lumens into the front of the fiber, then after 'x' meters of
cable, you'll have
(1,500) · (0.966)^x
lumens of light remaining.

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

The genius engineers in the fiber design industry would not handle it this way.
When they look up the 'attenuation' of the cable in the fiber manufacturer's
catalog, it would say "15dB per 100 meters".

What does that mean ? Break it down: 15dB in 100 meters is 0.15dB per meter.
Now, watch this:

Up at the top, the problem told us that the loss in 1 meter is 3.4% . We applied
super high mathematics to that and calculated that 96.6% remains, or 0.966.

Look at this ==> 10 log(0.966) = -0.15  <== loss per meter, in dB .

Armed with this information, the engineer ... calculating the loss in 'x' meters of
fiber cable, doesn't have to mess with raising numbers to powers. All he has to
do is say ...

-- 0.15 dB loss per meter

-- 'x' meters of cable

-- 0.15x dB of loss.

If 'x' happens to be, say, 72 meters, then the loss is (72) (0.15) = 10.8 dB .

and 10 ^ (-10.8/10) = 10 ^ -1.08 = 0.083 = 8.3% <== That's how much light
he'll have left after 72 meters, and all he had to do was a simple multiplication.

Sorry. Didn't mean to ramble on. But I do stuff like this every day.

5 0

3 years ago