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Colt1911 [192]
3 years ago
9

The diagram below shows the velocity vectors for two cars that are moving relative to each other.

Physics
1 answer:
scoundrel [369]3 years ago
7 0

Answer:

The answer is "5 \ \frac{m}{s} \ west"

Explanation:

\to \vec{V_1} = (25 \frac{m}{s}) (\hat{-i})\\\\\to  \vec{V_2} = (20 \frac{m}{s}) (\hat{-i})\\\\

velocity of car | respect to car :

\to \vec{V_{12}} = \vec{V_1} - \vec{V_2}\\\\

          =\vec{-25} \hat{i}+ \vec{20} \hat{i}\\\\= 5 \ \frac{m}{s} \ west

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dem82 [27]
Teacher -> teach the topic
Learner -> learn the topic
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5 0
3 years ago
A skier halfway between the top and bottom of a hill. Which statement best describes the skier?
Dvinal [7]

Answer: The skier has potential and kinetic energy.

Explanation: This is what I found from a different user on this website

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3 years ago
A high power line carries a current of 1.0 kA. What is the strength of the magnetic field this line produces at the ground, 10 m
solmaris [256]

Answer:

The strength of the magnetic field that the line produces is 2x10^{-5} Tesla.

Explanation:

From Biot-Savart law, the equation to determine the strength of the magnetic field for any straight wire can be deduced:

           

B = \frac{\mu_{0}I}{2\pi r} (1)      

                                     

Where \mu_{0} is the permiability constant, I is the current and r is the distance from the wire.    

             

Notice that it is necessary to express the current, I, from kiloampere to ampere.

I = 1.0kA \cdot \frac{1000A}{1kA} ⇒ 1000A

Finally, equation 1 can be used:

B = \frac{(4\pi x10^{-7}T.m/A)(1000A)}{2\pi (10m)}    

           

B = 2x10^{-5}T    

Hence, the strength of the magnetic field that the line produces is 2x10^{-5} Tesla.

         

8 0
3 years ago
A seagull flies at a velocity of 9.00 m/s straight into the wind. (a) if it takes the bird 20.0 min to travel 6.00 km relative t
enot [183]

Here we will the speed of seagull which is v = 9 m/s

this is the speed of seagull when there is no effect of wind on it

now in part a)

if effect of wind is in opposite direction then it travels 6 km in 20 min

so the average speed is given by the ratio of total distance and total time

v_{avg} = \frac{6000}{20*60}

v_{avg} = 5m/s

now since effect of wind is in opposite direction then we can say

V_{net} = v_{bird} - v_{wind}

5 = 9 - v_{wind}

v_{wind}= 4 m/s

Part b)

now if bird travels in the same direction of wind then we will have

v_{net}= v_{bird} + v_{wind}

v_{net} = 9 + 4 = 13 m/s

now we can find the time to go back

time = \frac{distance}{speed}

time = \frac{6000}{13}

time = 7.7 minutes

Part c)

Total time of round trip when wind is present

T = t_1 + t_2

T = 20 + 7.7 = 27.7 min

now when there is no wind total time is given by

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4 0
3 years ago
You are standing on a street corner with your friend. You then travel 14.0 m due west across the street and into your apartment
Margarita [4]

Answer:

Explanation:

We shall express each displacement vectorially , i for each unit displacement towards east , j for northward displacement and k for vertical displacement .

14 m due west = - 14 i

22.0 m upward in the elevator = 22 k

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Total displacement = - 14 i + 22 k + 12 j + 6 i

D = - 8 i + 12 j + 22 k

magnitude = √ ( 8² + 12² + 22² )

= √ ( 64 + 144 + 484 )

= √ 692

= 26.3 m

Net displacement from starting point = 26.3 m .

5 0
3 years ago
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