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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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Using the information from the previous problem (A 2kg ball rotates on the end of a 1.4m long string. The ball makes 5 revolutio
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The centripetal force on the ball is 140.9 N; option A

<h3>What is the centripetal force on the ball?</h3>

The centripetal force on the ball is given by the formula below:

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where;

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The linear velocity, v = wr

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2 years ago
In which medium does sound travel the fastest?
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Answer:

Solids

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Reflection off of a smooth surface like a mirror is an example of diffuse reflection.
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Answer:

true

Explanation:

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3 years ago
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A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out
anzhelika [568]

Answer:

The horizontal distance is 4.823 m

Solution:

As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

KE_{initial} + PE_{initial} = KE_{final} + PE_{final}

where

KE = Kinetic energy

PE = Potential energy

Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.

Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.

Therefore,

0 + mgH = \frac{1}{2}mv^{2} + 0

2gH = v^{2}

v = \sqrt{2\times 9.8\times 5} = 9.89\ m/s

where

v = velocity at the hill's bottom

Now,

Making use of the principle of conservation of momentum in order to calculate the velocity after the inclusion, v' of the backpack:

mv = (m + m')v'

65.0\times 9.89 = (65.0 + 20.0)v'

v' = 7.56\ m/s

Now, time taken for the fall:

h = \frac{1}{2}gt^{2}

t = \sqrt{\frac{2h}{g}}

t = \sqrt{\frac{2\times 2}{9.8} = 0.638\ s

Now, the horizontal distance is given by:

x = v't = 7.56\times 0.638 = 4.823\ m

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3 years ago
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