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Flura [38]
2 years ago
13

To test the resilience of its bumper during low-speed collisions, a 4 060-kg automobile is driven into a brick wall. the car's b

umper behaves like a spring with a force constant 8.00 106 n/m and compresses 3.72 cm as the car is brought to rest. what was the speed of the car before impact, assuming no mechanical energy is transformed or transferred away during impact with the wall
Physics
1 answer:
Simora [160]2 years ago
4 0
Mass of car, m = 4060 kg
Spring constant, k = 8.00 x 10⁶ N/m
Spring compression, x = 3.72 cm = 3.72 x 10⁻² m

Let the car strike the wall with speed v m/s.

The kinetic energy of the car is released into the stored energy of the spring (if losses are ignored), so that
(1/2)mv² = (1/2)kx²
(4060 kg)*(v m/s)² = (8 x 10⁶ N/m)(3.72 x 10⁻² m)²
4060 v² = 1.1701 x 10⁴
v = 1.6513 m/s

Answer:  1.65 m/s (nearest hundredth)

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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
Lady bird [3.3K]

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

3 0
3 years ago
Describe the point at which the kinetic energy of the pendulum is the highest
Arlecino [84]
The highest point<span> of the </span>pendulums<span> swing is when the potential energy is at its </span>highest<span> and the </span>kinetic energy<span> is at its lowest.</span>
5 0
3 years ago
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
Nikolay [14]

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:

d^{2} A/dl^{2} =-4l/\pi< 0

Thus we get maximum are when dA/dl=0

Therefore,

(200-4l)/\pi=0

l = 50 m

Now, from

\pi b+2l=200

\pi b=200-2*50

b=100/\pi

r=b/2=50/\pi

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

brainly.com/question/20693059

#SPJ4

4 0
1 year ago
3. Si usted duplicara la amplitud de un M.A.S. ¿cómo cambiaría la frecuencia, velocidad máxima, la aceleración máxima y la energ
WITCHER [35]

Answer:

Explanation:

la frecuencia = ω/2π, nada cambio

v(max) = ωA → ω2Α = 2ωA  duplicara velocidad máxima

a(max) = ω²Α → ω²2Α = 2ω²Α duplicara la aceleración máxima

la energía total ½kA² → ½k(2Α)² = 4(½kA²) cuatro veces la energía

8 0
2 years ago
The melting point of a solid is 90.0C. What is the heat required to change 2.5 kg of this solid at 30.0C to a liquid? The specif
Neko [114]

Hey again!

Ok..

Now... The melting Point of this solid is 90°C.

Meaning That as soon as it gets to this temp... It STARTS Melting.

So at that temp... It still has some solid parts in it.

You can say its a Solid Liquid Mixture.

Additional Heat being applied at that point is not raising the temperature;rather its used in breaking the bonds in the solid. This is the Fusion stage.

After Fusion...It'd then Be a Pure Liquid with no solids in it.

So

Q'=MC∆0----- This is the heat needed to take the solid's temp from 30°c - 90°c

Q"=ml ----- This is the heat used in breaking the bonds holding the solids in the solid-liquid phase.

So

Q= Q' + Q"

Q= mc∆0 + ml

∆0 = 90°c - 30°c = 60°c

Q= 2.5(390)(60) + (2.5)(4000)

Q=6.9 x 10⁴Joules

7 0
3 years ago
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