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3 years ago

Two students suggest different functions for the battery plays a flashlight.

Physics

2 answers:

KiRa [710]3 years ago

7 0

Answer:

Student B is correct

Explanation:

The battery acts like a pump to drive the electric charge. The electric charge comes from the conducting material but the battery provides a potential difference or a gradient of flow for this charge which is necessary for the movement of the charge along the conductor.

irga5000 [103]3 years ago

3 0

Answer:

Student b

Explanation:

Because, Battery pumps the charges around the circuit. Here, charges are the free electrons. And this charges are already present in the conductor. When we connect this conductor with the battery - It set up an electric field and The diffrence between the potential between the ends of the wire pushes this free charges to move in perticular direction. (This is called electric energy)

This electric energy is converted by chemical reactions in the battery. When the battery is dead - It means that the chemical energy producing chemicals has been consumed. thereafter there will be no more chemical reactions takes place to produce electric energy. Hence we say that the battery is dead.

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What force, in newtons, must you exert on the balloon with your hands to create a gauge pressure of 62.5 cm H2O, if you squeeze

yulyashka [42]

Answer:54.70 N

Explanation:

Given

Gauge Pressure of $62.5 cm of H_2O$

i.e. $h=62.5 cm =0.625 m$

Effective area $A=51 cm^2$

initial Pressure $= 1 atm=101.325 kPa$

Gauge Pressure $P=\rho gh$

$\rho =density\ of\ water =1000 kg/m^3$

$P_{gauge}=1000\times 9.8\times 0.625=5.937 kPa$

Force creates a pressure of $P_1$ which will be equal to Gauge Pressure

$P_1=\frac{F}{A}$

$P_1=P_{gauge}$

$\frac{F}{A}=5.937 kPa$

$F=5.937\times 51\times 10^{-4}\times 10^3$

$F=30.27 N$

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3 years ago

A force that is doing work on a ball when that ball is falling through the air.

Ronch [10]

Answer:

i think the answer is gravity

3 0

3 years ago

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch

Leokris [45]

a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: $-3.62\cdot 10^7 N/C$

c) Electric field 8.00 cm outside the paint layer: $-7.27\cdot 10^7 N/C$

Explanation:

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

$\int EdS = \frac{q}{\epsilon_0}$

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

$\epsilon_0 = 8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius $r$ as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

$q=0$

So we have

$\int E dS=0$

which means that the electric field inside the paint layer is zero.

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

$r=R=0.065 m$

The area of the surface is

$A=4\pi R^2$

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

$E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}$

The charge included within the sphere in this case is the charge on the paint layer, therefore

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

So, the electric field is:

$E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C$

where the negative sign means the direction of the field is inward, since the charge is negative.

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

$r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m$

Gauss theorem this time becomes

$E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}$

And the charge included within the sphere is again the charge on the paint layer,

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

Therefore, the electric field is

$E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C$

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3 years ago

Consider the third period of the periodic table. Which element has the smallest atomic radius?

Anettt [7]

According to periodic trends in the periodic table, the atomic radius decreases from left to right.

In period three, the element with the smallest atomic radius would be the element in the rightmost area. Protons increase as it goes to the right, which would mean they pull in electrons closer which decreases the size.

So in period 3, the element with the smallest atomic radius is Argon (Ar).

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3 years ago

Question

nataly862011 [7]

The density of a material is constant and it is given by the ratio of the mass to the volume of the material

The mass of the liquid and the full bottle ae

The mass of the liquid is 450 g

The mass of the filled bottle is 530 grams

The reason the above values are correct are as follows:

The given parameters are;

Volume of the bottle, V = Half litre

Mass of the bottle, $m_b$ = 80 g

The volume of liquid in the bottle when filled, V = 500 cm³

The density of the olive oil with which the bottle is filled, ρ = 0.9 g/cm³

a. Required:

To calculate the mass of oil in the bottle

Solution:

The volume of oil in the bottle when the bottle is filled, V = 500 cm³

$Density, \ \rho = \dfrac{Mass}{Volume} = \dfrac{m}{V}$

The mass of the liquid, m = ρ × V

∴ m = 0.9 g/cm³ × 500 cm³ = 450 g

The mass of the liquid, m = 450 g

b. The mass of the oil in the bottle, m = grams

The mass of the full bottle, $m_{filled}$ = m + $m_b$

∴ $m_{filled}$ = 450 g + 80 g = 530 g

The mass of the full bottle, $m_{filled}$ = 530 grams

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3 years ago