Ask question

Ask question

All categories

3 years ago

7

1. What is the relationship between the current through each resistor and the current through the battery?

1 answer:

Anastaziya [24]3 years ago

4 0

Answer:

1.Components in a series circuit share the same current.

2.The sum of individual voltage drop in a series circuit equals that supply voltage.

Explanation:

In a simple circuit, the amount of current is the same flowing through the resistor and the battery as long as it is a series circuit.Components in a series circuit share the same current.

In a series circuit, the total resistance equals the sum of individual resistance

The sum of individual voltage drop in a series circuit equals that supply voltage.

You might be interested in

What is energy efficiency

astra-53 [7]

Answer:

the easiest and most cost effective ways to combat climate change, clean the air we breathe, improve the competitiveness of our businesses and reduce energy costs for consumers.

Explanation:

6 0

3 years ago

If earth had no atmosphere , would a falling object ever reach terminal velocity?

Bas_tet [7]

The right answer is no.

6 0

3 years ago

Potential energy is related to the speed of an object.<br><br><br> True or False

adelina 88 [10]

Answer:

False

Explanation:

Potential energy is stored energy, so it is not in motion. Kinetic energy is the energy in motion.

4 0

3 years ago

A uniform, solid, 1800.0 kgkg sphere has a radius of 5.00 mm. Find the gravitational force this sphere exerts on a 2.30 kgkg poi

iVinArrow [24]

Answer

given,

Mass of the solid sphere = 1800 Kg

radius of the sphere,R = 5 m

mass of the small sphere, m = 2.30 Kg

when the Point is outside the sphere the Force between them is equal to

$F = \dfrac{GMm}{r^2}$ when r>R

When Point is inside the Sphere

$F = \dfrac{GMm}{R^2}\ \dfrac{r}{R}$ when r<R

where r is the distance where the point mass is placed form the center

Now Force calculation

a) r = 5.05 m

$F = \dfrac{GMm}{r^2}$ [/tex]

$F = \dfrac{6.67\times 10^{-11}\times 1800\times 2.3}{5.05^2}$

F = 1.082 x 10⁻⁸ N

b) r = 2.65 m

$F = \dfrac{GMm}{R^2}\ \dfrac{r}{R}$

$F = \dfrac{6.67\times 10^{-11}\times 1800\times 2.3}{5.05^2}\ \dfrac{2.65}{5.05}$

$F = 5.68\times 10^{-9}\ N$

3 0

3 years ago

In the Bohr’s model of the hydrogen atom, the electron moves in a circular orbit of radius 5.041 × 10−11 m around the proton. As

Ilia_Sergeevich [38]

Answer:

The orbital speed of the electron is 2.296 x 10⁶ m/s

Explanation:

Given;

radius of the circular orbit, r = 5.041 × 10⁻¹¹ m

In the Bohr’s model of the hydrogen atom, the velocity of the electron is given as;

$V = \frac{nh}{2\pi mr}$

where;

h is Planck's constant

m is mass of electron

r is the radius of the circular orbit

n is the energy level of hydrogen in ground state

Substitute in these values and solve for V

$V = \frac{nh}{2\pi mr} = \frac{1*6.626*10^{-34}}{2\pi *9.11 *10^{-31}*5.041*10^{-11}}\\\\V = 2.296*10^6 \ m/s$

Therefore, the orbital speed of the electron is 2.296 x 10⁶ m/s

4 0

4 years ago