All categories

2 years ago

If a person walks first 70 m in the direction 37° north of east, and then walks 82 m in the

Physics

1 answer:

Mrac [35]2 years ago

5 0

a. Aster is 56.3 m at 3.16° north-east from her initial position

b. She has to head to 183.16° or 86.84° south of west to return to her initial position

<h3>a. How to calculate how far Aster's final position from her initial position?</h3>

Let Aster's initial position be represented by the vector r = 0i + 0j

Since she then walks walks first 70 m in the direction 37° north of east, let this displacement be represented by the vector u = (70sin37°)i + (70cos37°)j

= (70 0.6018)i + (70 0.7986)j

= 42.13i + 55.9j

Also, she then walks 82 m in the direction 20° south of east. Let this displacement be represented by the vector v = (82sin20°)i - (82cos20°)j

= -(82 0.3420)i + (82 0.9397)j

= 28.05i - 77.05j

Finally, she walks 28 m in the direction 30° west of north. Let this displacement be represented by the vector, w = -(28sin30°)i + (28cos30°)j = -(28 0.5)i + (28 0.8660)j

= -14i + 24.25j m

So, the total displacement is R = r + u + v + w

= 42.13i + 55.9j + 28.05i + (-77.05)j + (-14)i + 24.25j m

= 56.18i + 3.1j

So, how far she walks is the magnitude of R. The magnitude of a vector Z = xi + yj is Z = √(x² + y²)

So, the magnitude of R = √((56.18)² + (3.1)²)

= √(3156.19 + 9.61)

= √3165.8

= 56.3 m

<h3>Her direction from final position to initial position</h3>

The direction of a vector Z = xi + yj is given by Ф = tan⁻¹ (y/x)

So, the direction of R is Ф' = tan⁻¹ (3.1/56.18)

= tan⁻¹ (0.0552)

= 3.16°

So, Aster is 56.3 m at 3.16° north-east from her initial position

<h3>b. What direction would she has to head to return to her initial position?</h3>

To return to her original position, the displacement vector is V = r - R

= 0i + 0j - (56.18i + 3.1j)

= -56.18i - 3.1j

So, the direction of V is Ф" = tan⁻¹ (-3.1/-56.18)

= tan⁻¹ (0.0552)

= 3.16°

Since this is in the third quadrant, we have that the direction she must go to return to her original position is α = 180° + 3.16°

= 183.16°

or 90° - 3.16°

= 86.84° south of west

So, she has to head to 183.16° or 86.84° south of west to return to her initial position

Learn more about direction of a vector here:

brainly.com/question/27854247

#SPJ1

You might be interested in

An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi

Veronika [31]

Answer:

$s=5.79\ km$

$\theta=47^{\circ}$ east of south

Explanation:

Given:

distance of the person form the initial position, $d'=8.4\ km$

direction of the person from the initial position, $47^{\circ}$ north of east

distance supposed to travel form the initial position, $d=5.3\ km$

direction supposed to travel from the initial position, due North

<u>Now refer the schematic for visualization of situation:</u>

$y=d'.\sin47^{\circ}-d$

$y=8.4\times \sin47-5.3$ ...............(1)

$x=d'.\cos47^{\circ}$

$x=8.4\times \cos47^{\circ}$ .................(2)

<u>Now the direction of the desired position with respect to south:</u>

$\tan\theta=\frac{y}{x}$

$\tan\theta=\frac{8.4\times \sin47}{8.4\times \cos47}$

$\theta=47^{\circ}$ east of south

<u>Now the distance from the current position to the desired position:</u>

$s=\sqrt{x^2+y^2}$

$s=\sqrt{(8.4\times \cos47)^2+(8.4\times \sin47-5.3)^2}$

$s=5.79\ km$

4 0

3 years ago

All of the planets and their satellites orbit the Sun in the same direction, and all their orbits lie near the same plane. True

MissTica

Answer:

True

Explanation:

The Sun rotates in the counterclockwise (CCW) direction when seen from its north pole. Since, the planets revolve around the Sun because of its gravity, the revolution of all the planets and their moons as seen from the north of the Sun is in CCW direction.

In fact most of the solar system bodies rotate in the same direction that is CCW. Some major exceptions to this are Venus and Uranus.

Almost all the planets and moons were made from the planetary disk around the Sun. Thus, they lie nearly in the same plane.

4 0

2 years ago

A car of mass m goes around a banked curve of radius r with speed v. If the road is frictionless due to ice, the car can still n

Sholpan [36]

Answer:

horizontal component of normal force is equal to the centripetal force on the car

Explanation:

As the car is moving with uniform speed in circle then the force required to move in the circle is towards the center of the circle

This force is due to friction force when car is moving in circle with uniform speed

Now it is given that car is moving on the ice surface such that the friction force is zero now

so here we can say that centripetal force is due to component of the normal force which is due to banked road

Now we have

$N sin\theta = \frac{mv^2}{R}$

$N cos\theta = mg$

so we have

$v = \sqrt{Rg tan\theta}$

so this is horizontal component of normal force is equal to the centripetal force on the car

5 0

3 years ago

An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p

igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Explanation:

Given:

Distance from the wire to the field point $r = 2.83 \times 10^{-2}$ m

Speed of electron $v = 35.5 \%c$

Current $I = 17.7$ A

For finding the acceleration,

First find the magnetic field due to wire,

$B = \frac{\mu _{o}I }{2\pi r }$

Where $\mu_{o} = 4\pi \times 10^{-7}$

$B = \frac{4\pi \times 10^{-7} \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }$

$B = 12.50 \times 10^{-5}$ T

The magnetic force exerted on the electron passing through straight wire,

$F = qvB$

$F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}$

$F = 21.3 \times 10^{-16}$ N

From the newton's second law

$F = ma$

Where $m =$ mass of electron $= 9.1 \times 10^{-31}$ kg

So acceleration is given by,

$a = \frac{F}{m}$

$a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }$

$a = 2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Therefore, the magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

7 0

2 years ago

A person is riding on a Ferris wheel. When the wheel makes one complete turn, the net work done on the person by the gravitation

lara31 [8.8K]

Answer:

Explanation:

m = Mass of person

g = Acceleration due to gravity = 9.81 m/s²

d = Vertical height from the ground

F = Force = Weight = mg

Net work done would be

$W_n=W_{up}+W_{down}\\\Rightarrow W_n=Fdcos180+Fdcos0\\\Rightarrow W_n=-mgd+mgd\\\Rightarrow W_n=0$

Hence, the work done on the person by the gravitational force is 0

7 0

3 years ago