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Alinara [238K]
2 years ago
12

Assume you are in the car and the car is moving at a certain speed to

Physics
1 answer:
Setler79 [48]2 years ago
6 0
You are at rest with respect to the car.
You are in motion with respect to the School.

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It takes you 5 min to walk with an average velocity of .75 m/s to the north from the parking lot to the entrance of the amusemen
skad [1K]
A displacement is a vector quantity that takes into account the shortest distance from the starting point to the endpoint. 

The given above gave a time interval in minutes which needs to be converted to seconds. Given that each minute is 60 seconds, 5 minutes equal 300 seconds. To determine the distance, multiply time with speed. The product is 225 m. 

Thus, the displacement is 225 m. 
3 0
3 years ago
When you changed from low to high power, how did the change affect the working distance of the lens?
Basile [38]

The working distance gets shorter as the magnification gets bigger. In order to focus, the high-power objective lens must be significantly nearer to the specimen than the low-power lens. Magnification is negatively correlated with working distance.

Magnification change The magnification of a specimen is increased by switching from low power to high power. The magnification of an image is determined by multiplying the magnification of the objective lens by the magnification of the ocular lens, or eyepiece.

The geometry of the optical system connects the magnifying power, or how much the thing being observed seems expanded, and the field of view, or the size of the object that can be seen.

To know more about  working distance

brainly.com/question/13551539

#SPJ4

4 0
1 year ago
Which is not an example of a force?
NNADVOKAT [17]

Answer:

possibly A?

Explanation:

just feels right man

7 0
2 years ago
Two cars are moving towards each other and sound emitted by first car with real frequency of 3000 hertz is detected by a person
sertanlavr [38]

Answer:

 v ’= 21.44 m / s

Explanation:

This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s

           f ’= f (v + v₀) / (v-v_{s})

where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer

in this exercise both the source and the observer are moving, we will assume that both have the same speed,

                v₀ = v_{s} = v ’

we substitute

               f ’= f (v + v’) / (v - v ’)

               f ’/ f (v-v’) = v + v ’

               v (f ’/ f -1) = v’ (1 + f ’/ f)

               v ’= (f’ / f-1) / (1 + f ’/ f) v

               v ’= (f’-f) / (f + f’) v

let's calculate

                v ’= (3400 -3000) / (3000 +3400) 343

                v ’= 400/6400 343

                v ’= 21.44 m / s

3 0
2 years ago
Interference occurs with not only light waves but also all frequencies of electromagnetic waves and all other types of waves, su
telo118 [61]

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

4 0
3 years ago
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