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3 years ago

How did potential energy get stored in the spring/pom pom system?

Physics

1 answer:

KengaRu [80]3 years ago

4 0

The elastic potential energy of a spring is the energy stored in the spring due to its elastic nature.

The potential energy stored in a spring is known as elastic potential energy. The elastic potential energy of a spring is the energy stored in the spring due to its elastic nature.

For example, when a spring is compressed it acquires potential energy which will be converted to kinetic energy of the object impacted by the spring.

The elastic potential energy of spring is given as;

$U_{x} = \frac{1}{2}kx^{2}$

where;

k is the spring constant

x is the extension of the spring.

Thus, we can conclude that, the elastic potential energy of a spring is the energy stored in the spring due to its elastic nature.

Learn more here: brainly.com/question/17717335

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A rectangular loop of area A is placed in a region where the magnetic field is perpendicular to the plane of the loop. The magni

mina [271]

Answer:

Induced emf, $\epsilon=-A\dfrac{B_{max}e^{-t/\tau}}{\tau}$

Explanation:

The varying magnetic field with time t is given by according to equation as :

$B=B_{max}e^{-t/\tau}$

Where

$B_{max}\ and\ t$ are constant

Let $\epsilon$ is the emf induced in the loop as a function of time. We know that the rate of change of magnetic flux is equal to the induced emf as:

$\epsilon=-\dfrac{d\phi}{dt}$

$\epsilon=-\dfrac{d(BA)}{dt}$

$\epsilon=-A\dfrac{d(B)}{dt}$

$\epsilon=-A\dfrac{d(B_{max}e^{-t/\tau})}{dt}$

$\epsilon=A\dfrac{B_{max}e^{-t/\tau}}{\tau}$

So, the induced emf in the loop as a function of time is $A\dfrac{B_{max}e^{-t/\tau}}{\tau}$ . Hence, this is the required solution.

7 0

3 years ago

Car a runs a red light and broadsides car b, which is stopped and waiting to make a left turn. car a has a mass of 1,800kg. car

frez [133]

The law of conservation of momentum tells us that momentum is conserved, therefore total initial momentum should be equal to total final momentum. In this case, we can expressed this mathematically as:

mA vA + mB vB = m v

where, m is the mass in kg, v is the velocity in m/s

since m is the total mass, m = mA + mB, we can write the equation as:

mA vA + mB vB = (mA + mB) v

furthermore, car B was at a stop signal therefore vB = 0, hence

mA vA + 0 = (mA + mB) v

1800 (vA) = (1800 + 1500) (7.1 m/s)

vA = 13.02 m/s

7 0

3 years ago

The technology in which parabolic mirror arrays are placed in a desert with a water-filled pipe traversing the array is used to

andriy [413]

Answer:

The sun

Explanation:

In this system the energy of the sun heats the water in the pipe, producing a high pressured steam, which is used for moving a turbine and producing electricity, is a transformation of energy from solar to thermal, then to mechanical to electrical.

4 0

4 years ago

Which of the following is a force acting between objects that do not touch?

kirill [66]

The answer is B friction force

3 0

3 years ago

Read 2 more answers

Topic Gravitational force amd firld strength.. help me please

I am Lyosha [343]

The gravitational force between m₁ and m₂ has magnitude

$F_{1,2} = \dfrac{Gm_1m_2}{x^2}$

while the gravitational force between m₁ and m₃ has magnitude

$F_{1,3} = \dfrac{Gm_1m_3}{(15-x)^2}$

where x is measured in m.

The mass m₁ is attracted to m₂ in one direction, and attracted to m₃ in the opposite direction such that m₁ in equilibrium. So by Newton's second law, we have

$F_{1,2} - F_{1,3} = 0$

Solve for x :

$\dfrac{Gm_1m_2}{x^2} = \dfrac{Gm_1m_3}{(15-x)^2} \\\\ \dfrac{m_2}{x^2} = \dfrac{m_3}{(15-x)^2} \\\\ \dfrac{(15-x)^2}{x^2} = \dfrac{m_3}{m_2} = \dfrac{60\,\rm kg}{40\,\rm kg} = \dfrac32 \\\\ \left(\dfrac{15-x}x\right)^2 = \dfrac32 \\\\ \left(\dfrac{15}x-1\right)^2 = \dfrac32 \\\\ \dfrac{15}x - 1 = \pm \sqrt{\dfrac32} \\\\ \dfrac{15}x = 1 \pm \sqrt{\dfrac32} \\\\ x = \dfrac{15}{1\pm\sqrt{\dfrac32}}$

The solution with the negative square root is negative, so we throw it out. The other is the one we want,

$x \approx 6.74\,\mathrm m$

5 0

3 years ago