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MariettaO [177]

1 year ago

9

An airplane flying parallel to the ground undergoes two consecutive dis- placements. The first is 75 km 30.0° west of north, and

the second is 155 km 60.0° east of north. What is the total displacement of the airplane?

1 answer:

torisob [31]1 year ago

5 0

Answer: displacement of airplane is 172 km in direction 34.2 degrees East of North

Explanation:

In constructing the two displacements it is noticed that the angle between the 75 km vector and the 155 km vector is a right angle (90 degrees).

Hence if the plane starts out at A, it travels to B, 75 km away, then turns 90 degrees to the right (clockwise) and travels to C, 155 km away from B. Angle ABC is 90 degrees, hence we can use Pythagoras theorem to solve for AC

AC2 = AB2 + BC2 ; AC^2 = 752 + 1552 ; from this we get AC = 172 km (3 significant figures)

Angle BAC = Tan-1(155/75) ; giving angle BAC = 64.2 degrees

Hence AC is in a direction (64.2 - 30) = 34.2 degrees East of North

Therefore the displacement of the airplane is 172 km in a direction 34.2 degrees East of North

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Estimate the change in the equilibrium melting point of copper caused by a change in pressure of 10 kbar. The molar volume of co

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Answer:

The change in the equilibrium melting point is 4.162 K.

Explanation:

Given that,

Pressure = 10 kbar

Molar volume of copper $V=8.0\times10^{-6}\ m^3$

Volume of liquid $V=7.6\times10^{-6}\ m^3$

Latent heat of fusion $L= 13.05 kJ$

Melting point =1085°C

We need to calculate the change temperature

Using Clapeyron equation

$\dfrac{\Delta P}{\Delta T}=\dfrac{\Delta H}{T\Delta V}$

Put the value into the formula

$\dfrac{1000\times10^{5}}{\Delta T}=\dfrac{13050}{(1085+273)\times(8.0-7.6)\times10^{-6}}$

$\Delta T=\dfrac{1000\times10^{-5}\times(1085+273)\times(8.0-7.6)\times10^{-6}}{13050}$

$\Delta T=4.162\ K$

Hence, The change in the equilibrium melting point is 4.162 K.

5 0

1 year ago

A diver explores a shallow reef off the coast of Belize. She initially swims d1 = 74.8 m north, makes a turn to the east and con

Nataly [62]

Answer:R=1607556m

θ=180degrees

Explanation:

d1=74.8m

d2=160.7km=160.7km*1000

d2=160700m

d3=80m

d4=198.1m

Using analytical method :

Rx=-(160700+75*cos(41.8))= -160755.9m

Ry= -(74.8+75sin(41.8))-198.1=73m

Magnitude, R:

R=√Rx+Ry

R=√160755.9^2+20^2=160755.916

R=160756m

Direction,θ:

θ=arctan(Rx/Ry)

θ=arctan(-73/160755.9)

θ=-7.9256*10^-6

Note that θ is in the second quadrant, so add 180

θ=180-7.9256*10^6=180degrees

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1 year ago

A container explodes and breaks into three fragments that fly off 120° apart from each other, with mass ratios 1: 4: 2. If the f

RSB [31]

Answer:

V₂ = 1.5 m/s

Explanation:

given,

speed of the first piece = 6 m/s

speed of the third piece = 3 m/s

speed of the second fragment = ?

mass ratios = 1 : 4 : 2

fragment break fly off = 120°

α = β = γ = 120°

sin α = sin β = sin γ = 0.866

using lammi's theorem

$\dfrac{A}{sin\alpha}=\dfrac{B}{sin\beta}=\dfrac{C}{sin\gamma}$

A,B and C is momentum of the fragments

$\dfrac{m\times 6}{0.866}=\dfrac{4m\times v_2}{0.866}=\dfrac{2m\times 3}{0.866}$

4 x V₂ = 2 x 3

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3 0

1 year ago

A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the neg

WINSTONCH [101]

Answer:

$=2,012,319.36 \ m/s$

Explanation:

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$F = Eq$

E is the electric field and q is the magnitude of the charge.

#Since the electric field is the same in both cases, and the charge of the protons and electrons have the same magnitude, you can state that the magnitude of the electric forces acting in both proton and electron are the same.

$F_e = F_p\\\\F_e= Force \ on \ electron\\F_p = Force \ on \ proton$

-Applying Newton's 2nd Law:

$F=ma$

$F_e=M_ea_e$

$F_p=M_pa_p$

#equate the two forces:

$F_e = F_p\\\\M_ea_e=M_pa_p\\\\a_e=\frac{M_pa_p}{M_e}$

#The equations for velocity in uniform acceleration:

$V_f^2=V_o^2+2ad\\\\V_o^2=0\\\\\therefore V_f^2=2ad$

#For the proton:

$V_f^2=2a_pd\\\\a_p=\frac{V_f^2}{2d}\\\\a_p=\frac{47000m/s)^2}{2d}$

#For the electron:

$V_f^2=2{a_e}^2\times 2d\\\\A_e=M_p\times A_p/M_e\\\\V_f^2=M_p\times (47000m/s)^2/2d\times2d/M_e\\\\V_f^2=M_p\times (47000m/s)^2/M_e\\\\V_f=47000m/s\times\sqrt{\frac{M_p}{M_e}}$

The mass values of the proton and electron are:

$M_p=1.67\times 10^{-27} kg\\\\M_e=9.11\times10^{-31}kg$

The speed of the ion is therefore calculated as:

$V_f=47000m/s\times\sqrt{\frac{M_p}{M_e}}\\\\=47000m/s\times\sqrt{\frac{1.67\times10^{-27}}{9.11\times10^{-31}}\\\\=2,012,319.36 \ m/s$

Hence, the ion's speed at the negative plate is $=2,012,319.36 \ m/s$

7 0

1 year ago

A turntable rotates counterclockwise at 76 rpm . A speck of dust on the turntable is at 0.47 rad at t=0. What is the angle of th

icang [17]

To solve this exercise it is necessary to apply the kinematic equations of angular motion.

By definition we know that the displacement when there is constant angular velocity is

$\theta= \theta_0 +\omega t$

From our given data we know that,

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$\omega = 76\frac{rev}{min}(\frac{2\pi rad}{1rev})(\frac{1 min}{60s})$

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Moreover we know that

$\theta_0 = 0.47 rad$

Therefore for time t=8.1s we have,

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That number in revolution is:

$\theta = 64.9298rad(\frac{1rev}{2\pi})$

$\theta = 15.108 Revolutions$

Here, we see that there are 15 complete revolutions

And 0.108 revolutions i not complete, so the tunable rotation is

$\theta_{net} = 0.108*2\pi=0.216\pi$

Therefore the angle of the speck at a time 8.1s is $0.216\pi$

4 0

2 years ago