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2 years ago

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An airplane pilot wishes to fly directly westward. According to the weather bureau, a wind of 75.0 km/hour is blowing southward.

The speed of the plane relative to the air (called the air speed) as measured by instruments aboard the plane is 310 km/hour . In which direction should the pilot head?

1 answer:

Alex17521 [72]2 years ago

6 0

Answer:

The speed of the plane relative to the ground is 300.79 km/h.

Explanation:

Given that,

Speed of wind = 75.0 km/hr

Speed of plane relative to the air = 310 km/hr

Suppose, determine the speed of the plane relative to the ground

We need to calculate the angle

Using formula of angle

$\sin\theta=\dfrac{v'}{v}$

Where, v'=speed of wind

v= speed of plane

Put the value into the formula

$\sin\theta=\dfrac{75}{310}$

$\theta=\sin^{-1}(\dfrac{75}{310})$

$\theta=14.0^{\circ}$

We need to calculate the resultant speed

Using formula of resultant speed

$\cos\theta=\dfrac{v''}{v}$

Put the value into the formula

$\cos14=\dfrac{v''}{310}$

$v''=\cos14\times310$

$v''=300.79\ km/h$

Hence, The speed of the plane relative to the ground is 300.79 km/h.

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The free-body diagram of a crate is shown. What is the net force acting on the crate? 352 N to the left 176 N to the left 528 N

Umnica [9.8K]

As per given conditions there are two directions along which forces are acting

1. Net force along left direction is given as

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2. Net force towards right direction is given as

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now since the two forces here in opposite direction so here we will have net force given as

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$F_{net} = 968 - 528$

$F_{net} = 440 N$

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2 years ago

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Given three different locations on Earth's surface, where will the weight of a person be greatest? in New York City, which is ab

zhenek [66]

Answer:

B. South Pole.

Explanation:

In order to answer this question, we simply have to refer to the laws of the equations of gravitational mechanics.

The equation given by Newton tells us that

$F = \frac {Gm_1m_2} {r^2}$

In the case where we compare a specific place where the Force of Gravity is greater or lesser, we focus on the term assigned to the Planet's Radius.

In the case of $G, m_1, m_2$ , we understand that they are constant.

We can easily notice that the more the Radius (Height seen from a viewer on the ground), the lower the force will be.

<em>In other words, the smaller the radius in which the measurement is made with respect to the center of the earth, the greater the gravitational force.</em>

In that order of ideas the smallest radio has South Pole, which is about 6356 km from the center of the Earth on the Equator line

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2 years ago

A carbon-dioxide laser emits infrared light with a wavelength of 10.6 μm. What is the length of a tube that will oscillate in th

alex41 [277]

Answer:

The length of a tube and number of rounds are 0.848 m and $1.77\times10^{8}\ trip\ per\ second$ .

Explanation:

Given that,

Wavelength $\lambda= 10.6\mu m$

m = 160000

We need to calculate the length

Using formula of wavelength

Laser tube behave like closed pipe

$m\dfrac{\lambda}{2}=L$

$L=160000\times\dfrac{10.6\times10^{-6}}{2}$

$L=0.848\ m$

Distance traveled by pulse of light in one back and fourth trip

$d=2L$

$d=2\times0.848$

$d=1.696\ m$

We need to calculate the time

Using formula for time

$t = \dfrac{d}{c}$

$t=\dfrac{1.696}{3\times10^{8}}$

$t=5.653\times10^{-9}\ s$

We need to calculate the number of round

Using formula of number of round

$N=\dfrac{1}{t}$

$N= \dfrac{1}{5.653\times10^{-9}}$

$N=1.77\times10^{8}\ trip\ per\ second$

Hence, The length of a tube and number of rounds are 0.848 m and $1.77\times10^{8}\ trip\ per\ second$ .

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2 years ago

pitot tube on an airplane flying at a standard sea level reads 1.07 x 105 N/m2. What is the velocity of the airplane?

Allushta [10]

Answer:

V_infinty=98.772 m/s

Explanation:

complete question is:

The following problem assume an inviscid, incompressible flow. Also, standard sea level density and pressure are 1.23kg/m3(0.002377slug/ft3) and 1.01imes105N/m2(2116lb/ft2), respectively. A Pitot tube on an airplane flying at standard sea level reads 1.07imes105N/m2. What is the velocity of the airplane?

<u>solution:</u>

<u>given:</u>

<em>p_o=1.07*10^5 N/m^2</em>

<em>ρ_infinity=1.23 kg/m^2</em>

<em>p_infinity=1.01*10^5 N/m^2</em>

p_o=p_infinity+(1/2)*(ρ_infinity)*V_infinty^2

V_infinty^2=9756.097

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Answer:

The electric field strength is $4.5\times 10^{4} N/C$

Solution:

As per the question:

Area of the electrode, $A_{e} = 25\times 25\times 10^{- 4} m^{2} = 0.0625 m^{2}$

Charge, q = 50 nC = $50\times 10^{- 9} C[/etx]Distance, x = 2 mm = [tex]2\times 10^{- 3} m$

Now,

To calculate the electric field strength, we first calculate the surface charge density which is given by:

$\sigma = \frac{q}{A_{e}} = \frac{50\times 10^{- 9}}{0.0625} = 8\times 10^{- 7}C/m^{2}$

Now, the electric field strength of the electrode is:

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where

$\epsilon_{o} = 8.85\times 10^{- 12} F/m$

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$\vec{E} = 4.5\times 10^{4} N/C$

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2 years ago