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3 years ago

What kind of energy change occurs when a battery is operating a remote control toy?

Physics

1 answer:

solniwko [45]3 years ago

7 0

Answer:

Explanation:

The correct answer is option C.

When the battery is operating a remote control toy the energy is converted from potential energy to the kinetic energy.

A battery stores electrical potential from the chemical reaction.

When the circuit is connected to the potential energy of the battery helps in the movement of the toy.

The energy produced by the movement of the control toy is kinetic energy.

Hence, we can say that Potential energy is changed to kinetic energy

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HOLA, NECESITO AYUDA!

givi [52]

The electrostatic force between the two ions is $2.9\cdot 10^{-10} N$

Explanation:

The electrostatic force between two charged particle is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem, the ion of sodium has a charge of

$q_1 = +e = +1.6\cdot 10^{-19} C$

while the ion of chlorine has a charge of

$q_2 = -e = -1.6\cdot 10^{-19}C$

And the distance between the two ions is

$r=282 pm = 282\cdot 10^{-12} m$

Substituting, we find the electrostatic force between the two ions:

$F=(8.99\cdot 10^9) \frac{(1.6\cdot 10^{-19})(-1.6\cdot 10^{-19})}{(282\cdot 10^{-12})^2}=-2.9\cdot 10^{-10} N$

where the negative sign simply means that the force is attractive, since the two ions have opposite charge.

Learn more about electrostatic force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

6 0

3 years ago

A spring is 20cm long is stretched to 25cm by a load of 50N. What will be its length when stretched by 100N. assuming that the e

IgorLugansk [536]

Answer:

Final Length = 30 cm

Explanation:

The relationship between the force applied on a string and its stretching length, within the elastic limit, is given by Hooke's Law:

F = kΔx

where,

F = Force applied

k = spring constant

Δx = change in length of spring

First, we find the spring constant of the spring. For this purpose, we have the following data:

F = 50 N

Δx = change in length = 25 cm - 20 cm = 5 cm = 0.05 m

Therefore,

50 N = k(0.05 m)

k = 50 N/0.05 m

k = 1000 N/m

Now, we find the change in its length for F = 100 N:

100 N = (1000 N/m)Δx

Δx = (100 N)/(1000 N/m)

Δx = 0.1 m = 10 cm

but,

Δx = Final Length - Initial Length

10 cm = Final Length - 20 cm

Final Length = 10 cm + 20 cm

<u>Final Length = 30 cm</u>

6 0

3 years ago

The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in va

bonufazy [111]

Answer:

10000 V

0.00225988700565 m²

$8\times 10^{-12}\ F$

Explanation:

E = Electric field = $4\times 10^6\ V/m$

d = Gap = 2.5 mm

Q = Charge = 80 nC

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

Potential difference is given by

$V=Ed\\\Rightarrow V=4\times 10^6\times 2.5\times 10^{-3}\\\Rightarrow V=10000\ V$

The potential difference between the plates is 10000 V

Area is given by

$A=\dfrac{Q}{\epsilon_0E}\\\Rightarrow A=\dfrac{80\times 10^{-9}}{8.85\times 10^{-12}\times 4\times 10^6}\\\Rightarrow A=0.00225988700565\ m^2$

The area of the plate is 0.00225988700565 m²

Capacitance is given by

$C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.85\times 10^{-12}\times 0.00225988700565}{2.5\times 10^{-3}}\\\Rightarrow C=8\times 10^{-12}\ F$

The capacitance is $8\times 10^{-12}\ F$

4 0

3 years ago

A student pulls a cart across the floor with a force of 45 N. If the cart travels 10 m, how much work does the cart do on the st

bixtya [17]

I'm assuming we're applying the standard Integral form of the calculation of work. The solution is provided in the image.

6 0

3 years ago

Which of the following is most needed for cosmotologists to study the age of the universe

Hatshy [7]

If its not Distance traveled then its energy

5 0

3 years ago