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Dovator [93]
3 years ago
13

A car moving at 95 km/h passes a 1.00-km-long train traveling in the same direction on a track that isparallel to the road. If t

he speed of the train is 75 km/h how long does it take the car to pass the train,and how far will the car have traveled in this time?
Physics
1 answer:
victus00 [196]3 years ago
7 0

Answer:

Time to pass the train=0.05 h

How far the car traveled in this time=4.75 Km

Explanation:

We have that the train and the car are moving in the same direction, the difference between the speed of the vehicles is:

\Delta V=V_{car}-V_{train}=95km/h-75km/h=20km/h

We will use this difference in the speed of the car an train to calculate how much time take the car to pass the train. For this we have that the train is 1km long and the car is moving with a speed of 20km/h (we use this value because is the speed that the car have in advantage of the train) then for a movement with a constant speed we have:

V=\dfrac{x}{t}

Where x is the distance, t is the time and v is the speed. using the data that we have:

V=\dfrac{x}{t}=\dfrac{1km}{20km/h}=0.05h

This is the time that the car take to pass the train. Now to calculate how far the car have traveled in this time we have to considered the speed of 95Km/h of the car, then:

V=\dfrac{x}{t}\\x=v\cdot t\\x=95km/h\cdot 0.05h\\x=4.75km

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The width of the central maxima, formed from light of wavelength 575 nm behind a single slit that has a width of 115 μm, is 1.15
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Answer:

 L  = 1.15 m

Explanation:

The diffraction phenomenon is described by the equation

        a sin θ = m λ

Where a is the width of the slit, λ  the wavelength and m is an integer, the order of diffraction is left.

The diffraction measurements are made on a screen that is far from the slit, and the angles in the experiment are very small, let's use trigonometry

          tan θ = y / L

          tan θ = sint θ / cos θ≈ sin θ

We substitute in the first equation

           a (y / L) = m λ

The first maximum occurs for m = 1

The distance is measured from the center point of maximum, which coincides with the center of the slit, in this case the distance is the total width of the central maximum, so the distance (y) measured from the center is

         y = 1.15 / 2 = 0.575 cm

         y = 0.575 10⁻² m

Let's clear the distance to the screen (L)

       L = a y / λ  

Let's calculate

     L = 115 10⁻⁶  0.575 10⁻² / 575 10⁻⁹

     L  = 1.15 m

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a gym consists of a rectangular region with a semi-circle on each end. if the perimeter of the room is to be a 200 m running tra
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The dimensions of the rectangle are:

l = 50 m

b = 100/\pi m

<h3>What is a perimeter in math?</h3>

The perimeter is the length of the outline of a shape. To find the perimeter of a rectangle or square you have to add the lengths of all the four sides.

<h3>How do we find a perimeter of a rectangle?</h3>

The perimeter of a rectangle,denoted by P is given by the formula, P=2l+2b, where l is the length and b is the breadth of the rectangle.

<h3>Given:</h3>

As per the question:

Perimeter of the room is given as P = 200 m

The region is rectangular having a semicircle at each end.

Now,

Let 'l' be the length of the rectangle, 'b' be its breadth and 'r' be the radius of the semi-circle at each end.

Then, Area of the given rectangle, A = lb

Perimeter of the room, P is =\pi r+l+\pi r+l=2\pi r+2l=\pi b+2l

Therefore,  \pi b+2l=200

b=(200-2l)/\pi

Now,

Area, A = l(200-2l)/\pi=(200l-2l^{2} )/\pi

Now, differentiate A w.r.t l:

Again differentiating w.r.t 'l', we get:

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(200-4l)/\pi=0

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r=b/2=50/\pi

To know more about area of a recatangle, visit the link

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