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NISA [10]
3 years ago
15

A boat travels in a straight path that is 25° west of north. Which describes the values of the west and north components of the

boat’s displacement?
Both components are positive numbers. Both components are negative numbers. The west component is a negative number, and the north component is a positive number. The west component is a positive number, and the north component is a negative number
Physics
2 answers:
goldfiish [28.3K]3 years ago
5 0
The answer to the question would be the second choice: <span>The west component is a negative number, and the north component is a positive number. It can be based upon in a Cartesian coordinate system wherein the coordinate is located in the second quadrant. </span>
skelet666 [1.2K]3 years ago
5 0

Answer:

The west component is a negative number, and the north component is a positive number.

Explanation:

For this problem, we take the standard Cartesian axis, so that:

- along the vertical axis, the north direction is positive, and the south direction is negative

- along the horizontal axis, the east direction is positive, and the west direction is negative

The boat travels in a direction which is 25° west of north. This means that its vertical component is in the north direction, while the horizontal component is in the west direction. Since north = positive and west = negative, the correct option is

The west component is a negative number, and the north component is a positive number.

You might be interested in
A 15.0 mW laser puts out a narrow beam 2.00 mm indiameter.
Murljashka [212]

Answer:

1341.03 V/m

Explanation:

The power output per unit area is the intensity and also the is the magnitude of the Poynting vector.

                                 S = \frac{P}{A} = cε₀E^{2} _{rms}

                             ⇒ \frac{P}{A} = cε₀E^{2}_{rms}

Where;

P is the power output

A is the area of the beam

c is speed of light

ε₀ is permittivity of free space 8.85 × 10⁻¹² F/m

E_{rms} is the average (rms) value of electric field

Making electricfield E_{rms} the subject of the equation

                                 E^{2}_{rms} = P / Acε₀

                                 E_{rms} = √(P / Acε₀)

But area A = πr²

                                 E_{rms} = √(P / πr²cε₀)                    

Given:

Output power, P = 15 mW = 0. 015 W

Diameter, d = 2 mm = 0.002 m

⇒ Radius, r = \frac{d}{2} = \frac{0.002}{2} = 0.001 m

Solving for average (rms) value of electric field;     

E_{rms} = \sqrt{\frac{0.015 W}{\pi * (0.001 m)^2 * (3 * 10^8 m/s) * (8.85 * 10^-12) C^2/Nm^2} }

                                E_{rms} = 1341.03 V/m

                             

                         

                                 

                                 

                                 

6 0
3 years ago
Pls answer……………………..
UkoKoshka [18]

2nd egg experienced more impulse

Hope this helps! :)

7 0
3 years ago
g Two cars, car 1 and car 2 are traveling in opposite directions, car 1 with a magnitude of velocity v1=13.0 m/s and car 2 v2= 7
bogdanovich [222]

Answer:

When they are approaching each other

    f_a = 2228.7 \  Hz

When they are passing  each other

    f_a = 2100Hz

 When they are retreating  from each other

     f_a =  1980.7 Hz

Explanation:

From the question we are told that

     The velocity of car one is  v_1 = 13.0 m/s

      The velocity of car two is  v_2 = 7.22 m/s

     The frequency of sound from car one is  f_e = 2.10 kHz

Generally the speed of sound at normal temperature is  v = 343 m/s

  Now as the cars move relative to each other doppler effect is created and this  can be represented  mathematically  as

              f_a = f_o [\frac{v \pm v_o}{v \pm v_s} ]

Where v_s is the velocity of the source of sound

            v_o is the velocity of the observer of the sound

            f_o is the actual frequence

             f_a  is the apparent frequency

Considering the case when they are approaching each other

        f_a = f_o [\frac{v +  v_o}{v -  v_s} ]

          v_o = v_2  

         v_s = v_1

         f_o = f_e

Substituting value

            f_a = 2100  [\frac{343 +  7.22}{ 343  -  13} ]

              f_a = 2228.7 \  Hz

Considering the case when they are passing  each other    

At that instant

                  v_o = v_s = 0m/s

                   f_o = f_e

               f_a = f_o [\frac{v }{v } ]

              f_a = f_o

Substituting value

             f_a = 2100Hz

Considering the case when they are retreating  from each other    

                f_a = f_o [\frac{v -  v_o}{v +   v_s} ]

          v_o = v_2  

         v_s = v_1

         f_o = f_e      

Substituting value

         f_a = 2100  [\frac{343 -  7.22}{343 +   13} ]    

          f_a =  1980.7 Hz    

7 0
3 years ago
A football player kicks a football in a field goal attempt. When the football reaches its maximum height, what is the relationsh
Stels [109]

At position of maximum height we know that the vertical component of its velocity will become zero

so the object will have only horizontal component of velocity

so at that instant the motion of object is along x direction

while if we check the acceleration of object then it is due to gravity

so the acceleration of object is vertically downwards

so it is along y axis

so here these two physical quantities are perpendicular to each other

so correct answer would be

<em>C)At the maximum height, the velocity and acceleration vectors are perpendicular to each other. </em>

3 0
3 years ago
Read 2 more answers
In the Olympic shot-put event, an athlete throws the shot with an initial speed of 12.0m/s at a 40.0? angle from the horizontal.
HACTEHA [7]

A) Horizontal range: 16.34 m

B) Horizontal range: 16.38 m

C) Horizontal range: 16.34 m

D) Horizontal range: 16.07 m

E) The angle that gives the maximum range is 41.9^{\circ}

Explanation:

A)

The motion of the shot is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion is a uniformly accelerated motion, so we can use the following suvat equation to find the time of flight:

s=u_y t + \frac{1}{2}at^2 (1)

where

s = -1.80 m is the vertical displacement of the shot to reach the ground (negative = downward)

u_y = u sin \theta is the initial vertical velocity, where

u = 12.0 m/s is the initial speed

\theta=40.0^{\circ} is the angle of projection

So

u_y=(12.0)(sin 40.0^{\circ})=7.7 m/s

a=g=-9.8 m/s^2 is the acceleration due to gravity (downward)

Substituting the numbers, we get

-1.80 = 7.7t -4.9t^2\\4.9t^2-7.7t-1.80=0

which has two solutions:

t = -0.21 s (negative, we ignore it)

t = 1.778 s (this is the time of flight)

The horizontal motion is instead uniform, so the horizontal range is given by

d=u_x t

where

u_x = u cos \theta=(12.0)(cos 40^{\circ})=9.19 m/s is the horizontal velocity

t = 1.778 s is the time of flight

Solving, we find

d=(9.19)(1.778)=16.34 m

B)

In this second case,

\theta=42.5^{\circ}

So the vertical velocity is

u_y = u sin \theta = (12.0)(sin 42.5^{\circ})=8.1 m/s

So the equation for the vertical motion becomes

4.9t^2-8.1t-1.80=0

Solving for t, we find that the time of flight is

t = 1.851 s

The horizontal velocity is

u_x = u cos \theta = (12.0)(cos 42.5^{\circ})=8.85 m/s

So, the range of the shot is

d=u_x t = (8.85)(1.851)=16.38 m

C)

In this third case,

\theta=45^{\circ}

So the vertical velocity is

u_y = u sin \theta = (12.0)(sin 45^{\circ})=8.5 m/s

So the equation for the vertical motion becomes

4.9t^2-8.5t-1.80=0

Solving for t, we find that the time of flight is

t = 1.925 s

The horizontal velocity is

u_x = u cos \theta = (12.0)(cos 45^{\circ})=8.49 m/s

So, the range of the shot is

d=u_x t = (8.49)(1.925)=16.34 m

D)

In this 4th case,

\theta=47.5^{\circ}

So the vertical velocity is

u_y = u sin \theta = (12.0)(sin 47.5^{\circ})=8.8 m/s

So the equation for the vertical motion becomes

4.9t^2-8.8t-1.80=0

Solving for t, we find that the time of flight is

t = 1.981 s

The horizontal velocity is

u_x = u cos \theta = (12.0)(cos 47.5^{\circ})=8.11 m/s

So, the range of the shot is

d=u_x t = (8.11)(1.981)=16.07 m

E)

From the previous parts, we see that the maximum range is obtained when the angle of releases is \theta=42.5^{\circ}.

The actual angle of release which corresponds to the maximum range can be obtained as follows:

The equation for the vertical motion can be rewritten as

s-u sin \theta t + \frac{1}{2}gt^2=0

The solutions of this quadratic equation are

t=\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g}

This is the time of flight: so, the horizontal range is

d=u_x t = u cos \theta (\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g})=\\=\frac{u^2}{-2g}(1+\sqrt{1+\frac{2gs}{u^2 sin^2 \theta}})sin 2\theta

It can be found that the maximum of this function is obtained when the angle is

\theta=cos^{-1}(\sqrt{\frac{2gs+u^2}{2gs+2u^2}})

Therefore in this problem, the angle which leads to the maximum range is

\theta=cos^{-1}(\sqrt{\frac{2(-9.8)(-1.80)+(12.0)^2}{2(-9.8)(-1.80)+2(12.0)^2}})=41.9^{\circ}

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

8 0
3 years ago
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