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amm1812
3 years ago
15

5N of force is applied to move a large nail a distance of 10 cm from an electromagnet on a frictionless table. The nail is then

released. What is the best description of the change from potential to kinetic energy as the nail moves back toward the electromagnet?
A) All of the energy is kinetic as soon as the nail is released.
B) The rate of change from potential to kinetic energy is constant.
C) The rate of change from potential to kinetic energy is exponential.
D) The kinetic energy is 25 J when the nail is 5 cm from the electromagnet.
Physics
1 answer:
DedPeter [7]3 years ago
7 0

Answer:

C) The rate of change from potential to kinetic energy is exponential.

Explanation:

As we know that total mechanical energy must be conserved here

As we know that there is no friction force on this system of electromagnet and nail

So here we can say that that

Magnetic potential energy of the nail + electromagnet system will convert into kinetic energy of the nail as it is released.

So we will have

Initial potential energy = work done to move it away by 10 cm

U = 5(0.10)

U = 0.5 J

now as the nail is released then this potential energy will start to convert into kinetic energy

So here correct answer must be

C) The rate of change from potential to kinetic energy is exponential.

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What causes a solar eclipse? Select all that apply.
brilliants [131]

I think it is either D) or E)

But i am going to go with E)

4 0
3 years ago
Read 2 more answers
A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm
Sergio [31]

Answer:

a) k = 200 N/m

b) E = 4 J

c) Δx = 6.3 cm

Explanation:

a)

  • In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:

       F = - k * \Delta x (1)

  • where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
  • Solving for k:

       k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)

b)

  • Assuming no friction present, total mechanical energy mus keep constant.
  • When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:

        U = \frac{1}{2}* k* (\Delta x)^{2} (3)

  • Replacing k and Δx by their values, we get:

       U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)

c)

  • When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
  • We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
  • So, we can write the following expression:

        \frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2}  = \frac{1}{2}*k*\Delta x^{2}   (5)

  • Replacing the right side of (5) with (4), k, m, and v by the givens, and simplifying, we can solve for Δx₁, as follows:

        \frac{1}{2}* 200N/m* \Delta x_{1} ^{2} + \frac{1}{2} * 1.8kg* (-2.0m/s)^{2}  = 4J   (6)

⇒      \frac{1}{2}* 200N/m* \Delta x_{1} ^{2}   = 4J  - 3.6 J = 0.4 J (7)

⇒     \Delta x_{1}   = \sqrt{\frac{0.8J}{200N/m} } = 6.3 cm (8)

6 0
3 years ago
This problem is based on the whole idea of pressure but I’m having trouble on when the area circle formula is included.
Mice21 [21]

Answer:

6.23x10^6Pa

Explanation:

Data obtained from the question include:

F (force) = 490N

r (radius) = 0.005m

A (area of the circlular heel) =?

P (pressure) =.?

First, we'll begin by calculating the area of the circlular heel. This is illustrated below:

Area of circle = πr^2

Area = 22/7 x (0.00)^2

Area = 7.86x10^-5m^2

Pressure is simply force per unit area. It represented mathematically as

Pressure = Force /Area

Pressure = 490/7.86x10^-5

Pressure = 6.23x10^6N/m2

Recall: 1N/m2 = 1Pa

Therefore, 6.23x10^6N/m2 = 6.23x10^6Pa

Therefore, the woman exert a pressure of 6.23x10^6Pa on the floor

8 0
4 years ago
12 points , just need help with this question
tigry1 [53]

Answer:f 30

Explanation: I am not really sure but try this

3 0
3 years ago
How much heat is needed to change the temperature of 3 grams of gold (c = 0.129 ) from 21°C to 363°C? The answer is expressed to
Temka [501]

Q= mcΔT

Where Q is heat or energy

M is mass, c is heat capacitance and t is temperature

You have to convert Celsius into kelvin in order to use this formula I believe

Celsius + 273 = Kelvin

21 + 273 = 294K

363 + 273 = 636K

Now...

Q= (0.003)(0.129)(636-294)

Q= 0.132 J if you are using kilograms, in terms of grams which seems more appropriate the answer would be 132J of energy.  

3 0
3 years ago
Read 2 more answers
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