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Artyom0805 [142]
2 years ago
14

Car enthusiasts often lower their cars closer to the ground as a matter of style. James wants to lower his car by replacing all

four of his original coil springs with new ones offered as an option by the car manufacturer. Note that changing the coil springs on a car can unsafely affect its handling. Consult your national or state laws before altering the suspension on any car. The new springs will be identical to the original springs, except the force constant will be 5355.00 N/m smaller. When James removes the original springs, he discovers that the length of each spring expands from 8.55 cm (its length when installed) to 12.00 cm (its length with no load placed on it). If the mass of the car body is 1455.00 kg, by how much will the body be lowered with the new springs installed, compared to its original height?
Physics
1 answer:
Alexandra [31]2 years ago
5 0

Answer:

\Delta h=0.0364\ m=3.64\ cm

Explanation:

Given:

  • change in stiffness constant of the spring on replacing the original springs, \delta k=5355\ N.m^{-1}
  • mass of the car, m=1455\ kg
  • initial length of the original car-spring before compression, l_i=12\ cm=0.12\ m
  • final length of the original car-spring after compression, l_f=8.55\ cm=0.0855\ m

So, weight of the car:

w=m.g

w=1455\times 9.81

w=14273.55\ N

<u>Now the spring constant of original spring:</u>

w=4k_o.(l_i-l_f) (since 4 springs are in parallel)

14273.55=4k_o\times (0.12-0.0855)

k_o=103431.522\ N.m^{-1}

<u>So the stiffness constant of the new springs:</u>

k_n=k_o-\delta k

k_n=103431.522-5355

k_n=98076.522\ N.m^{-1}

<u>Now the height lowered:</u>

w=k_n.4\Delta h (since 4 springs are in parallel)

14273.55=4\times98076.522\times \Delta h

\Delta h=0.0364\ m=3.64\ cm

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2 years ago
1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n
sveta [45]

Answer:

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(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

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(2).\: y= 15m-\dfrac{1}{2}gt^2

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(a).

When the projectile hits the 50m mark, y=0; therefore,

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solving for t we get:

t= 1.75s.

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\boxed{v_0 = 28.58m/s.}

(b).

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v_x = 28.58m/s.

the vertical component of the velocity is

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which gives a speed v of

v = \sqrt{v_x^2+v_y^2}

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1 year ago
Microwave ovens emit microwave energy with a wavelength of 12.6 cm. what is the energy of exactly one photon of this microwave r
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We need the frequency of the photon, it is v = c/ λ

Where c is 3 x 10^8 ms^-1 and λ is the wave length

We also need the expression of connecting frequency to energy of photon 

which is E = hv where h is Planck’s constant

Combining the two equations will give us:

E = h x c/λ

Inserting the values, we will have:

E = 6.626 x 10^-34 x 3 x 10^8 / 0.126

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2 years ago
Read 2 more answers
Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1
7nadin3 [17]

Answer:

A). σ = 3.823 x 10^{-5} C^{2}/N-m^{2}

B). \sigma ^{'}=2.76\times 10^{-5} C/m^{2}

C). U=10.322 J

Explanation:

A). We know magnitude of charge per unit area for a conducting plate is given by

\sigma =k.\varepsilon _{0}.E

where, E is resultant electric field = 1.2 x 10^{6} V/m

           \varepsilon _{0} is permittivity of free space = 8.85 x 10^{-12} C^{2}/N-m^{2}

           k is dielectric constant = 3.6

∴\sigma =k.\varepsilon _{0}.E

                     = 3.6 x 8.85 x10^{-12} x 1.2 x 10^{6}

                    = 3.823 x 10^{-5} C^{2}/N-m^{2}

B).Now we know that the magnitude of charge per unit area on the surface of the dielectric plate is given by

\sigma ^{'}=\sigma\left ( 1-\frac{1}{k} \right )

\sigma ^{'}=3.823\times 10^{-5}\left ( 1-\frac{1}{3.6} \right )

\sigma ^{'}=2.76\times 10^{-5} C/m^{2}

C).

Area of the plate, A = 2.5 cm^{2}

                                 = 2.5 x 10^{-4}m^{2}

diameter of the plate, d = 1.8 mm

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U=\frac{1}{2}\times 3.6\times8.85 \times10^{-12}\times\left ( 1.2\times 10^{6} \right ) ^{2}\times 2.5\times 10^{-4}\times 1800

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2 years ago
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