Ask question

Ask question

All categories

2 months ago

11

A hiker leaves her camp and walks 3.5 km in a direction of 55° south of west to the lake. After a short rest at the lake, she hi

kes 2.7 km in a direction of 16° east of south to the scenic overlook. What is the magnitude of the hiker’s resultant displacement? Round your answer to the nearest tenth. km What is the direction of the hiker’s resultant displacement? Round your answer to nearest whole degree. ° south of west

2 answers:

Softa [3K]2 months ago

6 0

The hiker’s overall displacement magnitude measures 5.6 km.
<span>The direction of this displacement is 77 degrees</span> south of west.

Maru [3.3K]2 months ago

5 0

1) Magnitude

Assigning south as the positive y-axis and east as the positive x-axis, we break down the two displacements into components:

$d_{1x} = -(3.5 km) cos 55^{\circ}=-2.0 km$

$d_{1y} = (3.5 km) sin 55^{\circ}=2.87 km$

$d_{2x} = (2.7 km) sin 16^{\circ}=0.74 km$

$d_{2y} = (2.7 km) cos 16^{\circ}=2.60 km$

The total displacement components resolve to:

$d_x = d_{1x}+d_{2x}=-2.0 km +0.74 km=-1.26 km$ east (equivalent to 1.26 km west)

$d_y=d_{1y}+d_{2y}=2.87 km + 2.60 km=5.47 km$ south

Calculating the resultant displacement's magnitude yields:

$d=\sqrt{d_x^2+d_y^2}=\sqrt{(1.26)^2+(5.47)^2}=5.61 km$

2) Direction

The displacement's direction is found to be:

$\theta= arctan(\frac{d_y}{d_x})=arctan(\frac{5.47}{1.26})=arctan(4.34)=77.0^{\circ}$ south of west.

You might be interested in

The mean free path of a helium atom in helium gas at standard temperature and pressure is 0.2 um.What is the radius of the heliu

Yuliya22 [3333]

Answer: $0.10233nm$

Explanation:

The mean free path of an atom can be calculated using the following equation:

$\lambda=\frac{RT}{\sqrt{2} \pi d^{2}N_{A}P}$ (1)

Where:

$\lambda=0.2\mu m=0.2(10)^{-6}m$

$R=8.3145J/mol.K$ is referred to as the Universal gas constant

$T=0\°C=273.115K$ represents the absolute standard temperature

$d$ denotes the diameter of helium atoms

$N_{A}=6.0221(10)^{23}/mol$ symbolizes Avogadro's number

$P=1atm=101.3kPa=101.3(10)^{3}Pa=101.3(10)^{3}J/m^{3}$ indicates absolute standard pressure

<pFrom this, we can solve for $d$ using (1), aiming to determine the radius $r$ of the helium atom:

$d=\sqrt{\frac{RT}{\sqrt{2}\pi\lambda N_{A}P}}$ (2)

$d=\sqrt{\frac{(8.3145J/mol.K)(273.115K)}{\sqrt{2}\pi(0.2(10)^{-6}m)(6.0221(10)^{23}/mol)(101.3(10)^{3}J/m^{3})}}$ (3)

$d=2.0467(10)^{-10}m$ (4)

If the radius equals half of that diameter:

$r=\frac{d}{2}$ (5)

Eventually:

$r=\frac{2.0467(10)^{-10}m}{2}$ (6)

$r=1.0233(10)^{-10}m$ (7)

Nonetheless, we were tasked with finding this radius in nanometers. Knowing $1nm=(10)^{-9}m$ :

$r=1.0233(10)^{-10}m.\frac{1nm}{(10)^{-9}m}=0.10233nm$ (8)

Ultimately:

$r=0.10233nm$ Represents the radius of the helium atom in nanometers.

5 0

2 months ago

If the position of an object is zero at one instant, what is true about the velocity of that object?

ValentinkaMS [3465]

If the position of an object is zero at a particular moment, this does not provide any indication about its velocity. It might simply be moving through that point, and you observed it exactly when it was at zero.

6 0

2 months ago

How many electrons must be removed from a neutral, isolated conducting sphere to give it a positive charge of 8.0 x 10 8 C? [Q=n

inna [3103]

The new charge of the ball will amount to 8x10^8C after removing 5x10^27 electrons.

Explanation:

Initially, if the sphere is electrically neutral, its charge stands at 0C.

When an electron with a charge of (-1.6*10^-19 C) is taken away, we effectively add a positive charge, leading to:

1.6*10^-19 C as the sphere's new charge.

For a total of N electrons removed, the sphere's overall charge now becomes:

N*1.6*10^-19 C.

To calculate N when:

N*1.6*10^-19 C = 8.0x 10^8 C.

We find that N is: (8.0/1.6)x10^(8 + 19) = 5x10^27 electrons.

7 0

2 months ago

Ice fishermen sit on top of frozen lakes in the winter and catch fish in the liquid water below through holes cut in the ice she

Yuliya22 [3333]

This is due to the fact that below 4°c, water behaves differently than other substances and decreases in density as its temperature drops further.

8 0

18 days ago

If you secure a refrigerator magnet about 2mmfrom the metallic surface of a refrigerator door and then move the magnet sideways,

Yuliya22 [3333]

Response:

(A) 4* 6 ^ ⁻6 T m² (B) 2 * 10 ^ ⁻6 v

Clarification:

Solution

Given that:

A refrigerator magnet with a depth of approximately 2 mm

The estimated magnetic field strength of the magnet is = 5 m T

The Area = 8 cm²

Now,

(A) The magnetic flux ΦB = BA

Therefore,

ΦB = (5 * 10^⁻ 3) ( 4 * 10 ^⁻2) * ( 2 * 10^ ⁻2) Tm²

Thus,

ΦB = 4* 6 ^ ⁻6 T m²

(B) By employing Faraday's Law, the subsequent equation applies:

Ε = Bℓυ

Where,

ℓ = 2 cm equals 2 * 10 ^⁻2 m

B = 5 m T = 5 * 10 ^ ⁻3 T

υ = 2 cm/s = 2 * 10 ^ ⁻2 m/s

Therefore,

Ε = (5 * 10 ^ ⁻3 T) * (2 * 10 ^ ⁻2) (2 * 10 ^ ⁻2) v

E =2 * 10 ^ ⁻6 v

7 0

1 month ago