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2 years ago

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

Physics

1 answer:

KengaRu [80]2 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 20 cm-radius ball is uniformly charged to 71 nC.

artcher [175]

Answer:

Part a)

$\rho = 2.12\mu C/m^3$

Part b)

$q_1 = 1.11 nC$

$q_2 = 8.88 nC$

$q_3 = 71 nC$

Part c)

$E_1 = 3996 N/C$

$E_2 = 7992 N/C$

$E_3 = 15975 N/C$

Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

So here the volume of the charge ball is given as

$V = \frac{4}{3}\pi R^3$

$V = \frac{4}{3}\pi(0.20)^3$

$V = 0.0335 m^3$

now the charge density of the ball is given as

$\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3$

Part b)

Now the charge enclosed by the surface is given as

$q = \rho V$

at radius of 5 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3$

$q = 1.11 nC$

at radius of 10 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3$

$q = 8.88 nC$

at radius of 20 cm

$q = 71 nC$

Part c)

As we know that electric field is given as

$E = \frac{kq}{r^2}$

so we have electric field at r = 5 cm

$E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}$

$E_1 = 3996 N/C$

electric field at r = 10 cm

$E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}$

$E_2 = 7992 N/C$

electric field at r = 20 cm

$E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}$

$E_3 = 15975 N/C$

3 0

3 years ago

The current that flows from and back to the power supply in a parallel circuit is called __________ current.

blagie [28]

Answer:

mainline current

Explanation:

Current that flows from and back to the power supply in a parallel circuit. Fuse. A type of circuit protection device.

6 0

2 years ago

What type of energy transformation is demonstrated in the above diagram?

Svetach [21]

Answer:

C :)

Explanation:

Trust me

6 0

3 years ago

Air "breaks down" when the electric field strength reaches 3 × 106 n/c, causing a spark. A parallel-plate capacitor is made from

Lera25 [3.4K]

Electric field between the plates of parallel plate capacitor is given as

$E = \frac{Q}{A\epsilon_0}$

here area of plates of capacitor is given as

$A = 0.055 * 0.055$

$A = 3.025 * 10^{-3}$

also the maximum field strength is given as

$E = 3 * 10^6 N/C$

now we will plug in all data to find the maximum possible charge on capacitor plates

$3 * 10^6 = \frac{Q}{3.025 * 10^{-3}*8.85 * 10^{-12}}$

$Q = 8.03 * 10^{-8} C$

so the maximum charge that plate will hold will be given by above

3 0

3 years ago

Read 2 more answers

The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .

ASHA 777 [7]

Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is $W=1.20 \cdot 10^{-3}J$

- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: $W_e=q\Delta V$

where

q is the charge

$\Delta V$ is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

$K_f - K_i = W + W_e = W+q\Delta V$

and since the charge starts from rest, $K_i = 0$ , so the formula becomes

$K_f = W+q\Delta V$

In this problem, we have

$W=1.20 \cdot 10^{-3}J$ is the work done by the external force

$q=-6.70 \mu C=-6.7\cdot 10^{-6}C$ is the charge

$K_f = 4.72\cdot 10^{-4}J$ is the final kinetic energy

Solving the formula for $\Delta V$ , we find

$\Delta V=\frac{K_f-W}{q}=\frac{4.72\cdot 10^{-4}J-1.2\cdot 10^{-3} J}{-6.7\cdot 10^{-6}C}=108.7 V$

4 0

3 years ago

Read 2 more answers