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N76 [4]
3 years ago
5

A certain spring is found not to obey Hooke’s law; it exerts a restoring force Fx(x)=−αx−βx2 if it is stretched or compressed, w

here α=60.0N/m and β=18.0N/m2. The mass of the spring is negligible. (a) Calculate the potential-energy function U(x) for this spring. Let U = 0 when x = 0. (b) An object with mass 0.900 kg on a frictionless, horizontal surface is attached to this spring, pulled a distance 1.00 m to the right (the +x-direction) to stretch the spring, and released. What is the speed of the object when it is 0.50 m to the right of the x = 0 equilibrium position?
Physics
1 answer:
Masteriza [31]3 years ago
4 0

Answer:U(x) = 30x^2 +6x^3

V^2=8.28m/s

Explanation:The law of conservation of energy is given by K1+U1= K2+U2 ...eq 1

Kinetic energy K.E= 1/2 mv^2

Restoring force function F(x)= -60x - 18x^2

But F(x)= -dU/dx

dU(x)=-F(x)dx

Integrating U(x)= -integral F(x)dx + U(0)

Substituting, we get

U(x) = - integral(-60x-18x^2)dx+U(0)

U(x)= 30x^2+6x^3+U(0)

U=0 at x=0

Therefore U(x)= 30x^2+6x^3

b) Given : x1=1.00m,x2= 0.50m ,V1=0, V2=?

Substituting into eq (a)

U1= 30(1.00)^2+6(1.00)^3=36J

Using x2=0.5 into eq(a)

U2=30(0.50)^2+6(0.50)^3=8.25J

Object at rest K1=0

0+36=K2+8.25

K2=27.75J

Given; m =0.900kg, V2=?

27.75=1/2×0.900×V2^2

V2= SQRT(2×27.75)/0.81

V2= 8.28m/s

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During a plane showcase, a pilot makes circular "looping" with a speed equal to the sound speed (340 m/s). However, the pilot ca
Dmitriy789 [7]

Answer:

1472.98 m

Explanation:

Data provided:

Speed of circular looping, v = 340 m/s

Acceleration, a = 8g

here,

g is the acceleration due to the gravity = 9.81 m/s²

Now,

the centripetal acceleration is given as,

a=\frac{v^2}{r}

r is the radius of the loop

on substituting the respective values, we get

8\times9.81=\frac{340^2}{r}

or

r = 1472.98 m

5 0
3 years ago
After you enlarge a map, which one of the following scale remains correct?
9966 [12]

Answer:

None

Explanation:

An scale is the factor by which actual features on ground are enlarged or reduced for representing on a plane. There are different kinds of scales:

  • Verbal scale use of words to represent scale information on the map.  The distance or linear units are used for depicting this scale on the map.  For example: 1 inch = 1 Kilo meter.
  • Fractional scale uses the numbers or values for showing the scale instead of words. As the name says, it is represented using a fraction or ratio.  Example: 1: 10,000 or 1/10,000
  • In large scale more details are shown in a map, however, less area coverage will be shown in a single map as the scale is large and more details are given.  Example: 1:500
  • Small scale is exactly opposite to the large scale, less details are shown as magnification is not enough, however a large amount of area can be shown in a single map.  Example: 1:25,000
  • A graphic scale is a bar that has been calibrated to show map distances. On maps that have been reduced or enlarged the original ratio and written scales are incorrect, since the relationship between map distance and real world distance has been altered, graphic scale is enlarged or reduced to the same extent as the map, this makes it the right option.

I hope you find this information useful and interesting! Good luck!

6 0
3 years ago
According to Dalton’s atomic theory
NARA [144]
Dalton's atomic theory was the first complete attempt to describe all matter in terms of atoms and their properties. ... The first part of his theory states that all matter is made of atoms, which are indivisible. The second part of the theory says all atoms of a given element are identical in mass and properties.
6 0
2 years ago
Please help me !! i’ll mark brainliest if you’re right!
navik [9.2K]
Option 2nd is the answer
3 0
2 years ago
A. How far does a 100-newton force have to move to do 1,000 joules
Aloiza [94]

Work done by a force is given as the product of force and the distance moved by the force.

<h3>What is work done?</h3>

Work done is the product of force and the distance moved by the the force.

  • Work done = Force × distance

Thus, distance required by the 100 N force is given as:

  • Distance = work done/force

Distance = 1000/100 = 10 m

Distance to be moved is 10 m.

Force applied = work done/ distance

Force applied by the hoist = 500/2

Force applied by the hoist = 250 N

Distance moved in one push-up = 25 cm = 0.25 m

Work done by the athlete after one push-up = 250 × 0.25 m

Work done by the athlete = 62.5 J

Distance moved by the force = 0 m

Work done = 500 × 0 = 0 N

Therefore, for work to be done, force has to move a distance.

Learn more about work done at: brainly.com/question/25573309

5 0
2 years ago
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