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Goryan [66]
3 years ago
13

car rides on four wheels that are connected to the body of the car by spring assemblies that let the wheels move up and down ove

r bumps and dips in the road. When an 80 kg person sits on the left front fender of a small car, this corner of the car dips by about 1 cm. Treating the spring assembly as a single spring, what is the approximate spring constant
Physics
2 answers:
Kryger [21]3 years ago
4 0

Answer:

approximate spring constant = 78.5 KN/m

Explanation:

From the question,

Mass of person = 80kg

Dip of car = 1cm

From Hooke's law, we know that;

F = kx = mg

Where ;

k is combined spring constant

x = 1cm = 0.01m

g = 9.81 m/s²

Thus,

Kx = mg

K = mg/x

k = (80x9.81)/0.01

k = 78.48 KN/m

This is approximately 78.5 KN/m:

Tomtit [17]3 years ago
3 0

Answer:

78.4 KN/m

Explanation:

Given

mass of person 'm' =80 kg

car dips about i.e spring stretched 'x'=   1 cm  => 0.01m

acceleration due to gravity 'g'= 9.8 m/s^2

as we know that,in order to find approximate spring constant we use Hooke's Law i.e  F=kx

where,

F = the force needed

x= distance the spring is stretched or compressed beyond its natural length

k= constant of proportionality called the spring constant.

F=kx ---> (since f=mg)

mg=kx

k=(mg)/x

k=(80 x 9.8)/ 0.01

k=78.4x10^3

k=78.4 KN/m

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A spherical helium filled balloon (B) with a hanging passenger cage being held by a single vertical cable (C) attached to Earth
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Answer:

The tension is  T  = 4326.7 \  N

Explanation:

From the question we are told that

   The  total mass is  m  =  200 \  kg

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     The  density of air is  \rho_a  =  1.225 \ kg/m^3

Generally the upward  force acting on the balloon is mathematically represented as

        F_N  =   T  + mg

=>     (\rho_a  *  V  *  g ) =   T  + mg

=>   T  =  (\rho_a  * V  *  g   )  - mg

Here V is the volume  of the spherical helium filled balloon which is mathematically represented as

      V  =  \frac{4}{3}  * \pi r^3

=>   V  =  \frac{4}{3}  * 3.142 *(5)^3

=>   V  = 523.67\  m^3

So

    T  = (1.225 *  523.67*  9.8 ) -  200 *  9.8

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5 0
3 years ago
The spring is unstretched at the position x = 0. under the action of a force p, the cart moves from the initial position x1 = -8
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Missing figure and missing details can be found here:
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Solution:
(a) The work done by the spring is given by
W= \frac{1}{2} k (\Delta x)^2 &#10;
where k is the elastic constant of the spring and \Delta x is the stretch between the initial and final position. Since x1=-8 in=-0.203 m and x2=5 in=0.127 m, we have
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(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:
W_W = -F_{//} (x_2 -x_1)
where  the negative sign is given by the fact that F_{//} points in the opposite direction of the displacement of the cart, and where
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8 0
3 years ago
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vitfil [10]

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From the equation of motion, it is possible to find an expression for the range (the total horizontal distance covered) of a projectile, which is given by:

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