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GREYUIT [131]
3 years ago
9

It takes 130 J of work to compress a certain spring 0.10m. (a) What is the force constant of this spring? (b) To compress the sp

ring an additional 0.10 m, does it take 130 J, more than 130 J or less than 130 J? Verify your answer with a calculation.
Physics
1 answer:
Ede4ka [16]3 years ago
6 0

Explanation:

Given that,

Work done to stretch the spring, W = 130 J

Distance, x = 0.1 m

(a) We know that work done in stretching the spring is as follows :

W=\dfrac{1}{2}kx^2\\\\k=\dfrac{2W}{x^2}\\\\k=\dfrac{2\times 130}{(0.1)^2}\\\\k=26000\ N/m

(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m

So,

W=\dfrac{1}{2}kx^2\\\\W=\dfrac{1}{2}\times 26000\times 0.2^2\\\\W=520\ J

So, the new work is more than 130 J.

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Consider this row in the periodic table of elements. Which element(s) are most likely to conduct electricity?
Sidana [21]

A) lithium and beryllium

Explanation:

From the given row on the periodic table, only lithium and beryllium will conduct electricity.

What makes a substance able to conduct electricity?

  • The presence of free mobile electrons and in some, ions allows them to carry electric currents.

Metals are generally known to be good conductors of heat and electricity. This is because, metals have a large pool of electrons i.e free mobile electrons. They are electropositive with a large size and readily release their electrons for conduction.

Lithium and Beryllium are in the metallic block on the periodic table.

Learn more:

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6 0
3 years ago
Read 2 more answers
Give two mathematical examples of Newton's third law and how you get the solution​
bagirrra123 [75]

Answer:

1) Any particle moving in a horizontal plane slowed by friction, deceleration = 32 μ

2) The particle moving by acceleration = P/m - 32μ OR The external force = ma + 32μm

Explanation:

* Lets revise Newton’s Third Law:

- For every action there is a reaction, equal in magnitude and opposite

 in direction.

- Examples:

# 1) A particle moving freely against friction in a horizontal plane

- When no external forces acts on the particle, then its equation of

  motion is;

∵ ∑ forces in direction of motion = mass × acceleration

∵ No external force

∵ The friction force (F) = μR, where μ is coefficient of the frictional force

   and R is the normal reaction of the weight of the particle on the

   surface

∵ The frictional force is in opposite direction of the motion

∴ ∑ forces in the direction of motion = 0 - F

∴ 0 - F = mass × acceleration

- Substitute F by μR

∴ - μR = mass × acceleration

∵ R = mg where m is the mass of the particle and g is the acceleration

  of gravity

∴ - μ(mg) = ma ⇒ a is the acceleration of motion

- By divide both sides by m

∴ - μ(g) = a

∵ The acceleration of gravity ≅ 32 feet/sec²

∴ a = - 32 μ

* Any particle moving in a horizontal plane slowed by friction,

 deceleration = 32 μ

# 2) A particle moving under the action of an external force P in a

  horizontal plane.

- When an external force P acts on the particle, then its equation

 of motion is;

∵ ∑ forces in direction of motion = mass × acceleration

∵ The external force = P

∵ The friction force (F) = μR, where μ is coefficient of the frictional force

   and R is the normal reaction of the weight of the particle on the

   surface

∵ The frictional force is in opposite direction of the motion

∴ ∑ forces in the direction of motion = P - F

∴ P - F = mass × acceleration

- Substitute F by μR

∴ P - μR = mass × acceleration

∵ R = mg where m is the mass of the particle and g is the acceleration

  of gravity

∴ P - μ(mg) = ma ⇒ a is the acceleration of motion

∵ The acceleration of gravity ≅ 32 feet/sec²

∴ P - 32μm = ma ⇒ (1)

- divide both side by m

∴ a = (P - 32μm)/m ⇒ divide the 2 terms in the bracket by m

∴ a = P/m - 32μ

* The particle moving by acceleration = P/m - 32μ

- If you want to fin the external force P use equation (1)

∵ P - 32μm = ma ⇒ add 32μm to both sides

∴ P = ma + 32μm

* The external force = ma + 32μm

7 0
3 years ago
A cylinder which is in a horizontal position contains an unknown noble gas at 4.63 × 104 Pa and is sealed with a massless piston
AleksandrR [38]

Answer:

The change in internal energy of the system is -17746.78 J

Explanation:

Given that,

Pressure P=4.63\times10^{4}\ Pa

Remove heat \Delta U= -1.95\times10^{4}\ J

Radius = 0.272 m

Distance d = 0.163 m

We need to calculate the internal energy

Using thermodynamics first equation

dU=Q-W...(I)

Where, dU = internal energy

Q = heat

W = work done

Put the value of W in equation (I)

dU=Q-PdV

Where, W = PdV

Put the value in the equation

dU=-1.95\times10^{4}-(4.63\times10^{4}\times3.14\times(0.272)^2\times(-0.163))

dU=-17746.78\ J

Hence, The change in internal energy of the system is -17746.78 J

3 0
4 years ago
A car achieves a velocity 50 meters per second after 5 seconds. What is the Car's acceleration?
Reptile [31]

Answer: 10 m/s

Explanation: Velocity/Time

50/5= 10

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7 0
3 years ago
In which of the basic regions of the galaxy is the sun located?
AlladinOne [14]

galactic disk


The galactic disk is a thinned, leveled out distribution of stars which includes the typical to the largest and brightest. The Sun is in the Milky Way and lies amongst the majority of the stars where it bulges.


8 0
4 years ago
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