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AVprozaik [17]
2 years ago
6

Aisha used a compass to walk 59° south of west for 3.50 miles, then stopped for lunch. Assume that the point from which she star

ted her hike corresponds to the origin of the coordinate system.
(a) Find the polar coordinates corresponding to the place where Aisha stopped for lunch. (Enter your answer for r in miles and counterclockwise from the +x-axis. Round your answers to at least th ree significant figures. as the smallest positive angle in degrees r =26
The radial polar coordinate does not coincide with x. mi =
(b) Find the Cartesian coordinates (in miles) for the same point.
x = 2.6
Use the relations between polar and Cartesian coordinates to find the coordinate x. mi
y = 4.24
Use the relations between polar and Cartesian coordinates to find the coordinate y. mi
Physics
1 answer:
Tatiana [17]2 years ago
8 0

Answer:

Explanation:

a )

59° south of west  means (180 + 59 )° counterclockwise from the +x-axis.

so angle θ = 239 ° .

r = 3.5 miles .

polar coordinates ( r ,  θ )

= ( 3.50 , 239° )

As the smallest positive angle in degree is 26 then 26 x 9 = 234

so rounding off 239 to 234 , the angle will be 234

polar coordinates ( 3.50 ,  234 )

b )

Cartesian coordinates (in miles) for the same point

x coordinate = r cosθ = 3.5 cos 234 =   -  2.06 mi

y coordinate  = r sinθ = 3.5 sin 234 = - 2.83  mi

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Mars has two moons, Phobos and Deimos. Phobos orbits Mars at a distance of 9380 km from Mars's center, while Deimos orbits at 23
Sloan [31]

Answer:

The ratio is   \frac{T_1}{T_2}  = 3.965

Explanation:

From the question we are told that

   The  radius of Phobos orbit is  R_2 =  9380 km

    The radius  of Deimos orbit is  R_1  =  23500 \  km

Generally from Kepler's third law

    T^2 =  \frac{ 4 *  \pi^2 *  R^3}{G * M  }

Here M is the mass of Mars which is constant

        G is the gravitational  constant

So we see that \frac{ 4 *  \pi^2  }{G * M  } =  constant

   

    T^2 = R^3   *  constant      

=>  [\frac{T_1}{T_2} ]^2 =  [\frac{R_1}{R_2} ]^3

Here T_1 is the period of Deimos

and  T_1 is the period of  Phobos

So

      [\frac{T_1}{T_2} ] =  [\frac{R_1}{R_2} ]^{\frac{3}{2}}

=>    \frac{T_1}{T_2}  =  [\frac{23500 }{9380} ]^{\frac{3}{2}}]

=>    \frac{T_1}{T_2}  = 3.965

   

8 0
2 years ago
When a perfume bottle is opened, some liquid changes to gas and the fragrance spreads around the room. Which sentence explains t
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Chromatic aberration comes from the fact that different wavelengths of light travel at different speeds through the material of
gtnhenbr [62]

Answer:

 y_red / y_blue = 1.11

Explanation:

Let's use the constructor equation to find the image for each wavelength

         1 /f = 1 /o + 1 /i

Where f is the focal length, or the distance to the object and i the distance to the image

Red light

           1 / i = 1 / f - 1 / o

           1 / i_red = 1 / f_red - 1 / o

           1 / i_red = 1 / 19.57 - 1/30

           1 / i_red = 1,776 10-2

           i_red = 56.29 cm

Blue light

            1 / i_blue = 1 / f_blue - 1 / o

            1 / i_blue = 1 / 18.87 - 1/30

            1 / i_blue = 1,966 10-2

            i_blue = 50.863 cm

Now let's use the magnification ratio

             m = y ’/ h = - i / o

             y ’= - h i / o

Red Light

            y_red ’= - 5 56.29 / 30

            y_red ’= - 9.3816 cm

Light blue

            y_blue ’= 5 50,863 / 30

            y_blue ’= - 8.47716 cm

The ratio of the height of the two images is

            y_red ’/ y_blue’ = 9.3816 / 8.47716

            y_red / y_blue = 1,107

            y_red / y_blue = 1.11

5 0
2 years ago
Your friend says, “chemical changes are caused by an input in energy. In physical changes, there is no transfer of energy” is yo
nalin [4]

Answer:

Ok, let's suppose the simplest of the physical changes:

We have an object that is not moving (so it is not accelerated)

and there is change, now the object moves.

Because there was a change, means that there was an acceleration, and by the second Newton's law.

Force equals mass times acceleration:

F = m*a

There must be a force.

So suppose that you pushed the object, then some energy that you had, you transferred it to the object, that now is moving and now has kinetic energy.

Now, is kinda true that in a closed system the total energy is always constant, but it depends on what is our system.

So if we think in our system as you and the object, then in the whole system the energy does not change because the energy that you lost is now on the object, but again, there was a transfer of energy.

So no, your friend is not correct.

3 0
2 years ago
In the late 19th century, great interest was directed toward the study of electrical discharges in gases and the nature of so-ca
gizmo_the_mogwai [7]

Answer:

A) He finds the same value of q / m for different materials , B)      y = ½ (q / m) E L² / v₀ₓ² , C) v = E / B , D)   B = 2.13 10⁻⁶ T, E) For the first part I have two off-center points., For the second part I can center one point but the other is off center

Therefore the third statement is correct

Explanation:

Part A

Thomson's experiments are the first proof that the atoms that until now were considered indivisible were constituted by different elements, in these experiments Thomson himself the ratio q / m of several cathodes and always found the same value, which allowed to establish that In atoms there are two types of particles, some of which are mobile and others are still.

When examining his statements the correct one is: He finds the same value of q / m for different materials

Part B

For this part let's use Newton's second law

        F = ma

        q E = m a

        a = q / m E

We use the kinematic relationship

          y = voy t - ½ to t²

          x = v₀ₓ t

The initial vertical velocity of electrons is zero

           y = ½ a (x / v₀ₓ)²

We replace

           y = ½ (q / m) E L² / v₀ₓ²

Part C

If there is no deflection, the electric and magnetic forces are the same and in the opposite direction

         Fm = Fe

         q v B = q E

          v = E / B

Part D

       

        We replace

        y = ½ (q / m) E L² / (E / B)²

         y = ½ (q / m) L² B² / E

If we do not want any deflection the magnetic field has to return the electrons the amount that they lower y = -4.12 cm

      -4.12 10⁻² = ½ q / m 0.12² B² / 1.1 10³

       -16.97 10⁻⁴ = 6.54 10⁻³ B² q / m

      B² = -2.59 10⁻¹ q / m

      q / m = -1.758 10¹¹ C/ kg

      B = √ 0.259 1.758 10¹¹ = √ 4.55 10⁻¹²

      B = 2.13 10⁻⁶ T

Part E

As the charge that the two particles is different

For the first part I have two off-center points.

For the second part I can center one point but the other is off center

Therefore the third statement is correct

8 0
2 years ago
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