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11111nata11111 [884]

3 years ago

12

A 2.0 kg hanging mass stretches a coiled spring by 0.15 m. The spring constant, k, is: (A) 0.075 N/m, (B) 2.9 N/m (C) 131 N/m, (

D) 1,742 N/m, (E) none of the above.

1 answer:

Alla [95]3 years ago

3 0

Answer:

C

Explanation:

Givens

m = 2 kg

F = 2 * 9.81

F = 19.62 N

x = 0.15 m

Formula

F = k*x

Solution

19.62 = k*0.15

k = 19.62/0.15

k = 130.8 which rounded to the nearest given answer is C

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Suppose a 1300 kg car is traveling around a circular curve in a road at a constant

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Answer:

F = 4212 N

Explanation:

Given that,

Mass of a car, m = 1300 kg

Speed of car on the road is 9 m/s

Radius of curve, r = 25 m

We need to find the magnitude of the unbalanced force that steers the car out of its natural straight- line path. The force is called centripetal force. It can be given by :

$F=\dfrac{mv^2}{r}\\\\F=\dfrac{1300\times 9^2}{25}\\\\F=4212\ N$

So, the force has a magnitude of 4212 N

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Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w

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Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

$U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}$

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

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$u(x) = \frac{-2}{x} + \frac{4}{x^2}$

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$u(1) = \frac{-2}{1} + \frac{4}{1^2}$

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$= \frac{1}{2} * 4(3)^2$

= 18J

Total energy will be =

4J + 18J = 22J

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$u(5) = \frac{-2}{5} + \frac{4}{5^2}$

$= \frac{4-10}{25} = \frac{-6}{25} J$

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$\frac{1}{2} * 4Vf^2$

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

$Vf = \sqrt{frac{22.4}{2}$

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b) magnitude of force when x = 5.0m

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$\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}$

$= \frac{2}{x^2} - \frac{8}{x^3}$

At x = 5.0 m

$\frac{2}{5^2} - \frac{8}{5^3}$

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= 0.016N

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The metric unit of force is _____. kg kg/cm3 kg/cm2 N

ioda

The metric unit of force is Newton or N. The Newton unit is also equal to kilogram per meter per second squared. The Newton name came from the late physicist Isaac Newton. It is also based on the second law of motion.

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3 years ago

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