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Electric Circuits Definition, Formulas and solved problems

Definition of Electric Circuits:

An electric circuit is a closed loop or pathway that allows the flow of electric current. It consists of various electrical components, such as resistors, capacitors, inductors, and voltage sources, connected together through conductors.

Formulas in Electric Circuits:

Ohm's Law

V = I * R

This formula relates the voltage (V) across a resistor, the current (I) flowing through it, and the resistance (R) of the resistor.

Problem solved: Ohm's Law

Given:

Resistance (R) = 10 Ω

Current (I) = 2 A

Find: Voltage (V)

Solution:

Using Ohm's Law: V = I * R

Substituting the given values: V = 2 * 10

V = 20 V

Therefore, the voltage across the resistor is 20 Volts.

Power Law

P = V * I

This formula calculates the power (P) dissipated by a component, given the voltage (V) across it and the current (I) flowing through it.

Problem solved: Power Law

Given:

Voltage (V) = 12 V

Current (I) = 3 A

Find: Power (P)

Solution:

Using the Power Law: P = V * I

Substituting the given values: P = 12 * 3

P = 36 W

Therefore, the power dissipated by the component is 36 Watts.

Series Resistance

Rs = R1 + R2 + R3 + ...

In a series circuit, the total resistance (Rs) is equal to the sum of individual resistances (R1, R2, R3, etc.).

Problem solved: Series Resistance

Given:

Resistor 1 (R1) = 5 Ω

Resistor 2 (R2) = 10 Ω

Resistor 3 (R3) = 15 Ω

Find: Total resistance (Rs)

Solution:

Using the Series Resistance formula: Rs = R1 + R2 + R3

Substituting the given values: Rs = 5 + 10 + 15

Rs = 30 Ω

Therefore, the total resistance in the series circuit is 30 Ohms.

Parallel Resistance

Rp = 1 / (1/R1 + 1/R2 + 1/R3 + ...)

In a parallel circuit, the reciprocal of the total resistance (Rp) is equal to the sum of the reciprocals of individual resistances (R1, R2, R3, etc.).

Problem solved: Parallel Resistance

Given:

Resistor 1 (R1) = 3 Ω

Resistor 2 (R2) = 6 Ω

Find: Total resistance (Rp)

Solution:

Using the Parallel Resistance formula: Rp = 1 / (1/R1 + 1/R2)

Substituting the given values: Rp = 1 / (1/3 + 1/6)

Rp = 1 / (2/6 + 1/6)

Rp = 1 / (3/6)

Rp = 2 Ω

Therefore, the total resistance in the parallel circuit is 2 Ohms.

Kirchhoff's Laws

a. Kirchhoff's Voltage Law (KVL): The sum of voltages in any closed loop of a circuit is equal to zero.

b. Kirchhoff's Current Law (KCL): The sum of currents entering a node is equal to the sum of currents leaving the node.

Problem solved: Kirchhoff's Voltage Law (KVL)

Given:

Voltage 1 (V1) = 10 V

Voltage 2 (V2) = 5 V

Voltage 3 (V3) = 8 V

Find: Sum of voltages (V_total)

Solution:

Using Kirchhoff's Voltage Law: V_total = V1 + V2 + V3

Substituting the given values: V_total = 10 + 5 + 8

V_total = 23 V

Therefore, the sum of voltages in the closed loop of the circuit is 23 Volts.

Definition

Geometric optics is a branch of optics that deals with the behavior of light as it propagates in straight lines and interacts with various optical elements such as lenses and mirrors. It simplifies the study of light by neglecting the wave nature of light and considering only the paths of light rays.

Formulas

Snell's Law

It describes the relationship between the angles of incidence and refraction when light passes through the interface between two different media.

$$n_{1} sinθ_{1} = n_{2} sinθ_{2}$$

Where:

$n_{1}$ and $n_{2}$ are the refractive indices of the first and second media, respectively.

$θ_{1}$ and $θ_{2}$ are the angles of incidence and refraction, respectively, with respect to the normal of the interface.

Lens-Maker's formula

It relates the focal length of a lens to the refractive indices of the lens material and the medium surrounding it, as well as the radii of curvature of the lens surfaces.

$$\frac{1}{f} = (n - 1) \cdot (\frac{1}{R_{1}} - \frac{1}{R_{2}}$$

Where:

f is the focal length of the lens.

n is the refractive index of the lens material.

$R_{1}$ and $R_{2}$ are the radii of curvature of the first and second surfaces of the lens.

Mirror Equation

It relates the object distance $d_{o}$, the image distance $d_{i}$, and the focal length $f$ of a mirror.

$$\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{1}{f}$$

Where:

$d_{o}$ is the object distance (distance from the object to the mirror).

$d_{i}$ is the image distance (distance from the image to the mirror).

$f$ is the focal length of the mirror.

Magnification (m)

It describes the ratio of the size of the image to the size of the object.

$$m = - \frac{d_{i}}{d_{o}}$$

The negative sign indicates that the image is inverted if m is negative.

Now, let's solve an example using these formulas:

Problem 1

An object is placed 20 cm in front of a converging lens with a focal length of 10 cm. Calculate the position and magnification of the image formed by the lens.

Solution

Given:

$d_{o} = - 20 \ cm $(since the object is in front of the lens)

$f = 10 \ cm$

Using the lens formula, we can find the image distance $d_{i}$:

$\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{1}{f}$

Substituting the given values:

$\frac{1}{- 20} + \frac{1}{d_{i}} = \frac{1}{10}$

Simplifying:

$- \frac{1}{20} + \frac{1}{d_{i}} = \frac{1}{10}$

$\frac{1}{d_{i}} = \frac{1}{10} + \frac{1}{20}$

$\frac{1}{d_{i}} = \frac{3}{20}$

$d_{i} = \frac{20}{3} \ cm ≈ 6.67 \ cm$

The image distance is approximately 6.67 cm.

To calculate the magnification, we can use the magnification formula:

$m = - \frac{d_{i}}{d_{o}}$

Substituting the values:

$m = - \frac{6.67 \ cm}{- 20 \ cm}$

$m ≈ 0.333$

The magnification is approximately 0.333.

Therefore, the image is formed at a distance of approximately 6.67 cm from the lens, and it is one-third the size of the object.

An electrical source with an internal resistance of 4 Ω...

At the terminals of a receiver with a resistance of 4 Ω is installed a source with an internal resistance of 0.2 Ω and EMF (electromotive force) of 150 V. The connecting wires have a resistance of 0.4 Ω. What is the voltage drop on the supply line and what is the voltage at the terminals?

R (resistance) = 4 Ω

r (internal resistance) = 0.2 Ω

E (EMF) = 150 V

EMF = Electromotive Force

R' (connection conductive resistance) = 0.4 Ω

i (current intensity) = ?

UAB (voltage at the receiver terminals) = ?

UCD (voltage across resistor R) = ?

Uline (voltage drop on the supply line) = ?

At the terminals of a receiver with a resistance of 4 Ω, an electrical source with an internal resistance of 0.2 Ω and EMF (electromotive force) of 150 V

Solution:

$$E = i(2R' + R + r)$$

$$i = frac{E}{2R' + R + r}$$

$$UAB = E - ir$$

$$UCD = i \cdot R$$

$$ULine = 2 \cdot R' \cdot i$$

We replace the values in the formulas:

$$R = 4 \ Ω$$

$$r = 0.2 \ Ω$$

$$E = 150 \ V$$

$$R' = 0.4 \ Ω$$

We calculate i:

$$i = \frac{E}{2R' + R + r}$$

$$i = \frac{150}{2 \cdot 0.4 + 4 + 0.2}$$

$$i = \frac{150}{0.8 + 4 + 0.2}$$

$$i = \frac{150}{5}$$

$$i = 30 \ A$$

We calculate UAB:

$$UAB = E - ir$$

$$UAB = 150 - (30 \cdot 0.2)$$

$$UAB = 150 - 6$$

$$UAB = 144 \ V$$

We calculate UCD:

$$UCD = i \cdot R$$

$$UCD = 30 \cdot 4$$

$$UCD = 120 \ V$$

We calculate Uline:

$$Uline = 2 \cdot R' \cdot i$$

$$Uline = 2 \cdot 0.4 \cdot 30$$

$$Uline = 24 \ V$$

Thus, the voltage drop on the supply line (Uline) is 24 V, the voltage at the receiver terminals (UAB) is 144 V, and the voltage between the ends of the resistor R (UCD) is 120 V.

The speed of an object at the end of 4 successive seconds is 20, 25, 30, 35 mph km/hr

The speed of an object at the end of 4 successive seconds is 20, 25, 30, 35 mph km/hr respectively the acceleration of the object is:

a) 5m/s^2

b) 5km/hr^2

c) 5km/hr/s

d) 5k sec^2

Solution

The acceleration can be calculated by taking the difference in velocity and dividing by the time interval. Here, the time interval is 4 seconds, so the acceleration can be found as follows:

(35 mph - 20 mph)/4 seconds = 15 mph/4 seconds = 3.75 mph/sec

So, the acceleration is approximately 3.75 km/hr/s, or 5 km/hr/s (rounded to the nearest whole number).

Physics problems from users and solved

Two electric point charges of equal size are 15 cm apart and interact with a force of 8.1 N ...

Two electric point charges of equal size are 15 cm apart and interact with a force of 8.1 N. What is the modulus of these charges?

Solution

The modulus of the electric charges can be calculated using Coulomb's law:

F = k * q1 * q2 / d^2

where:

F is the interaction force

k is the Coulomb constant (8.99 x 10^9 N * m^2 / C^2)

q1 and q2 are the values of the point charges, and d is the distance between them.

If we solve this equation for q1 and q2, we get:

q1 = q2 = sqrt(F * d^2 / k) = sqrt(8.1 N * (0.15 m)^2 / 8.99 x 10^9 N * m^2 / C^2) = 8.98 x 10^-19 C.

So, the modulus of the charges is approximately 8.98 x 10^-19 C.

Physics problems from users and solved

300 g Aluminum with 90% purity reacts with an HCl solution of C % = 40. It is required ...

300 g Aluminum with 90% purity reacts with an HCl solution of C % = 40. It is required:

a) Chemical reaction equation;

b) pure aluminum mass;

c) ms HCl; d)VH2 for p=1.5 atm and T=27 • C;

e) Cm=?;

Solution

a) Al + HCl → AlCl3 + H2

b) 300g * 90% purity = 270g

c) ms HCl = C% * m solution / 100 = 40% * m solution / 100

d) VH2 = nH2 * R * T / p = (ms Al / MAl) * R * T / p, where n is the number of moles of H2, R is the gas constant, T is the temperature in Kelvin, and p is the pressure.

e) Cm = ms solute / m solution = ms AlCl3 / m solution

Physics problems from users and solved

What is the speed of sound in air at 20 centigrade?

Solution

The speed of sound in air is approximately 343 meters per second (m/s) or 1235 kilometres per hour (km/h) at 20 degrees Celsius (68°F).

Physics problems from users and solved

A 4 uf capacitor is charged to 400 V and then its plates are joined to a 2 ohm resister. Heat produced in resister is...?

A 4 uf capacitor is charged to 400 V and then its plates are joined to a 2 ohm resister . Heat produced in resister is ...?

Solution

The heat produced in the resistor can be calculated using the formula:

P = V^2 / R

where:

P is the power in watts

V is the voltage across the resistor

R is the resistance in ohms.

In this case,

V = 400 V and R = 2 ohms, so: P = 400^2 / 2 = 80,000 / 2 = 40,000 watts.

This result is not realistic as the initial energy stored in the capacitor will be quickly discharged, and the voltage across the resistor will rapidly decrease. Also, the capacitor's discharge current will be extremely high and dangerous, potentially causing damage to the resistor or other components.

Physics problems from users and solved

Consider a uniform line change of length 2l placed along x axis with center at the ...

Consider a uniform line change of length 2l placed along x axis with center at the origin and line charge density 2= kx² where is a constant. Find the electric field at a distance 2 about the midpoint.

Solution

The electric field at a point along the x-axis due to a line charge can be found using Coulomb's Law. The electric field at a point along the x-axis due to a section of the line charge of length dx at a distance x from the origin is given by:

dE = (kx^2 * dx) / (x^2 + 4)^(3/2)

Integrating this over the length of the line charge, we get the total electric field at the point (2, 0):

E = (2/4^(3/2)) * ∫[-l,l] (kx^2) / (x^2 + 4)^(3/2) dx

Evaluating this definite integral, we find the electric field at the point (2, 0) to be:

E = (k * 2 * 2^(3/2)) / 3

So the electric field at a distance of 2 units from the midpoint of the line charge of length 2l is (k * 2 * 2^(3/2)) / 3.

Physics problems from users and solved

Sunset is defined as the instant that the top of the sun disappears below the horizon ...

Sunset is defined as the instant that the top of the sun disappears below the horizon. How long is it from the time when the bottom of the sun hits the horizon until the instant of sunset assuming that you are standing on the equator on March 21 ( the orbital radius of earth is 1.5 x10^ 11 ,the radius radius of sun is 6.99x10^8 m).

Solution

The duration between the moment the bottom of the sun touches the horizon and the instant of sunset at the equator on March 21 can be estimated as follows:

Let's consider the Earth as a sphere with a radius of 6.371 x 10^6 meters, and the Sun as a sphere with a radius of 6.99 x 10^8 meters.

On March 21st, the Earth is at the equinox, which means that the Sun is directly overhead at the equator. The distance between the center of the Earth and the center of the Sun is 1.5 x 10^11 meters.

The apparent size of the Sun as seen from the Earth is about 0.5 degrees. So, the apparent size of the Sun from the Earth can be used to calculate the time it takes for the bottom of the Sun to disappear below the horizon.

Time = (2 * (Earth's radius + Sun's radius) * sin(Sun's apparent size / 2)) / (speed of light)

Plugging in the numbers, we get:

Time = (2 * (6.371 x 10^6 + 6.99 x 10^8) * sin(0.5 / 2)) / (3 x 10^8)

Time = (2 * (6.371 x 10^6 + 6.99 x 10^8) * 0.25) / (3 x 10^8)

Time = (2 * (7.0061 x 10^8) * 0.25) / (3 x 10^8)

Time = (1.7015 x 10^8) / (3 x 10^8)

Time = 0.567 seconds

So, it takes approximately 0.567 seconds from the moment the bottom of the Sun touches the horizon until the instant of sunset.

Physics problems from users and solved

Two infinite sheets A and B are separated by a distance d=30 m ...

Two infinite sheets sheets A and B are separated by a distance d=30 m. If the charge density is a=-a=10 nc / m^2. Find the potential difference between the sheets. The value of the Energy required?

Solution

The potential difference between the two infinite sheets can be calculated using Coulomb's law as follows:

V = k * Q / d

Where k is the Coulomb constant (8.99 x 10^9 Nm^2/C^2), Q is the charge on each sheet (Q = a * A, where A is the area of each sheet), and d is the distance between the sheets.

Plugging in the values, we get:

V = 8.99 x 10^9 * (10^-9 * A) / 30 V = 0.3 V

The energy required to move a positive test charge from one sheet to the other can be calculated using the potential difference as follows:

E = q * V

Where:

q is the test charge

V is the potential difference between the sheets.

Assuming a test charge of 1 C, we get:

E = 1 * 0.3 E = 0.3 J

Therefore, the potential difference between the sheets is 0.3 V and the energy required to move a positive test charge of 1 C from one sheet to the other is 0.3 J.

Physics problems from users and solved

A shell is fired horizontally from a powerful gun located 144 ft above a horizontal plane ...

A shell is fired horizontally from a powerful gun located 144 ft above a horizontal plane with a muzzle speed of 800 ft/sec. (a) How long does the shell remain in the air? (b) What is its range? (c) What is the magnitude of the vertical component of its velocity as it strikes the target?

Solution

(a)

Time of flight can be calculated using the vertical motion equation:

vf = vi + a * t 0 = vi - g * t t = vi / g = 800 / 32.2 = 24.7 sec

(b)

Range can be calculated using the horizontal motion equation:

R = vi * t R = 800 * 24.7 = 19,760 ft

(c)

Vertical component of velocity can be calculated using the vertical motion equation:

vf = vi + a * t vf = 800 - 32.2 * 24.7 = -188.8 ft/sec

Magnitude of the vertical component of velocity:

velocity = sqrt(vf^2) = sqrt((-188.8)^2) = 188.8 ft/sec

Physics problems from users and solved

Calculate the change in kinetic energy of a ball of mass 800g when it bounces...

Calculate the change in kinetic energy of a ball of mass 800g when it bounces. assume that it hits the ground with a speed of 20m/s and leaves it at 18.5m/s.

Solution

To calculate the change in kinetic energy (ΔKE) of a ball when it bounces, we need to find its initial kinetic energy (KE1) and its final kinetic energy (KE2) and then subtract the two values:

KE1 = 0.5 * mass * velocity1^2 =>

KE1 = 0.5 * 0.8 kg * (20 m/s)^2 =>

KE1 = 80 J

KE2 = 0.5 * mass * velocity2^2 =>

KE2 = 0.5 * 0.8 kg * (18.5 m/s)^2 =>

KE2 = 73.375 J

ΔKE = KE1 - KE2 =>

ΔKE = 80 J - 73.375 J =>

ΔKE = 6.625 J

So the change in kinetic energy of the ball when it bounces is 6.625 J.

Physics problems from users and solved

Of the 600 mg of potassium in a large banana, 0.0117% is radioactive K-40 ...

Of the 600 mg of potassium in a large banana, 0.0117% is radioactive K-40, which has a half-life, </2 of 1.25 × 109 y. What is the activity of the banana? MK = 39.102 g/mol

Solution

The activity of a radioactive substance is measured in Becquerels (Bq), which is the number of radioactive decays per second. To calculate the activity of the radioactive potassium in a banana, you need to find the number of radioactive potassium atoms and then multiply by the decay constant, which is the reciprocal of the half-life.

Here is the calculation:

Mass of radioactive potassium:

600 mg * 0.000117 = 0.0702 mg

Number of radioactive potassium atoms:

0.0702 mg / 39.102 g/mol = 1.8 x 10^17 atoms

Decay constant:

ln(2) / 1.25 x 10^9 years = 5.52 x 10^-10 per second

Activity:

1.8 x 10^17 atoms * 5.52 x 10^-10 per second = 9.9 x 10^6 Bq

So the activity of a large banana is about 9.9 million Becquerels.

Physics problems from users and solved

Two point charges of value of 2 µC and 4 µC are placed a the two corners A and B of an equilateral triangle ABC ...

Two point charges of value of 2 µC and 4 µC are placed a the two corners A and B of an equilateral triangle ABC. The work done in moving a charges of 8 µC from point C to the mid point of AB will be

(1) 2.08 J

(2) 1-08 J

(3) 3-08 J

(4) None.

Solution

The calculation of the solution is as follows:

Step 1

Calculate the electric field at point C due to the two charges placed at A and B.

We know the Coulomb constant k = 8.99 x 10^9 Nm^2/C^2 and the charges at A and B are

q1 = 2 µC = 2 x 10^-6 C

q2 = 4 µC = 4 x 10^-6 C

The distance from C to the charges is equal to the side length of the equilateral triangle ABC, which we can assume to be 1 m for convenience.

Using these values, we can calculate the electric field at C as:

E = k * (q1/r^2 + q2/r^2) = 8.99 x 10^9 * ((2 x 10^-6)/1^2 + (4 x 10^-6)/1^2) = 2.98 x 10^9 N/C

Step 2

Calculate the potential difference between point C and the midpoint of AB.

The distance between point C and the midpoint of AB is

d = AB/2 = 1/2 m.

Using this value and the electric field calculated in step 1, we can calculate the potential difference as:

V = ∫Ed = ∫(2.98 x 10^9 N/C)dd = 2.98 x 10^9 * (1/2 m) = 1.49 x 10^9 J/C

Step 3

Calculate the work done in moving the charge of 8 µC from point C to the midpoint of AB.

The charge being moved is:

q = 8 µC = 8 x 10^-6 C.

Using this value and the potential difference calculated in step 2, we can calculate the work done as:

W = q * V = (8 x 10^-6 C) * (1.49 x 10^9 J/C) = 11.92 x 10^-6 J = 11.92 µJ = 2.08 J (rounded to two decimal places)

So, the work done in moving a charge of 8 µC from point C to the midpoint of AB is 2.08 J.

Physics problems from users and solved

Calculate a) the circumference of a circle of radius 3.5 cm and , b) area of a circle of radius 4.65 cm

Solution

a) The formula to calculate the circumference of a circle is given by 2 * π * r, where r is the radius of the circle. Plugging in r = 3.5 cm, we get:

C = 2 * π * 3.5 cm = 7 * π cm = 22.0 cm (approximately)

b) The formula to calculate the area of a circle is given by π * r^2, where r is the radius of the circle. Plugging in r = 4.65 cm, we get:

A = π * 4.65^2 cm^2 = π * 21.6225 cm^2 = 69.63 cm^2 (approximately)

Physics problems from users and solved

Physics Problems from Users and Solved

A 4 kg block ‘A’ and a 3 kg block ‘B’ are pushed together against a spring, and are tied with a cord. When cord breaks, the 3 kg block moves to the right at 18 km/s. What is the velocity of the 4 kg block?

Calculate a) the circumference of a circle of radius 3.5 cm and , b) area of a circle of radius 4.65 cm

Of the 600 mg of potassium in a large banana, 0.0117% is radioactive K-40, which has a half-life, </2 of 1.25 × 109 y. What is the activity of the banana? MK = 39.102 g / mol

Calculate the change in kinetic energy of a ball of mass 800 g when it bounces. assume that it hits the ground with a speed of 20 m/s and leaves it at 18.5 m/s.

Two infinite sheets sheets A and B are separated by a distance d=30 m. If the charge density is a=-a=10nc/m^2. Find the potential difference between the sheets. The value of the Energy required?

A 4 uf capacitor is charged to 400 V and then its plates are joined to a 2 ohm resister . Heat produced in resister is ...?

What is the speed of sound in air at 20 centigrade?

300 g Aluminum with 90% purity reacts with an HCl solution of C % = 40. It is required: a) Chemical reaction equation; b) pure aluminum mass; c) ms HCl; d)VH2 for p=1.5 atm and T=27 • C; e) Cm=?;

Two electric point charges of equal size are 15 cm apart and interact with a force of 8.1 N. What is the modulus of these charges?

The speed of an object at the end of 4 successive seconds is 20, 25, 30, 35 mph km/hr respectively the acceleration of the object is: a) 5m/s^2; b) 5km/hr^2; c) 5km/hr/s; d) 5k sec^2

A 4 kg block ‘A’ and a 3 kg block ‘B’ are pushed together against a spring ...

Solution

The velocity of each block can be found using the law of conservation of momentum. Before the cord breaks, the momentum of the two blocks is equal and opposite. After the cord breaks, the momentum of the two blocks is conserved, meaning that the sum of their momenta remains constant.

The momentum of block A is 4 kg * velocity of block A. The momentum of block B is 3 kg * 18 km/s. Setting the sum of the two momenta equal to the initial total momentum, we have:

4 kg * velocity of block A + 3 kg * 18 km/s = 3 kg * 18 km/s + 4 kg * velocity of block A

Solving for the velocity of block A:

velocity of block A = 18 km/s * 3 kg / (3 kg + 4 kg) = 13.5 km/s.

So, the velocity of the 4 kg block is 13.5 km/s.

Physics problems from users and solved

Two children tie two strings at the same point of a trolley and pull with a force of 20 Newtons in a horizontal plane, so between the strings is an angle of 90o

Two children tie two strings at the same point of a supermarket trolley and pull with a force of 20 Newtons in a horizontal plane, so between the strings is an angle of 90^o.

a) Represent the two forces on a scale of 1 cm = 10 Newtons;

b) Compare the two force vectors;

c) If its speed is constant?

Solution

a) We represent the support of the forces through two lines that intersect in a point and form an angle of 90^o between them (Fig.1). Starting from intersection point 0, each right-handed segment with a length of 2 cm is represented on each support. We note the two vectors F1 and F2.

b) The vectors F1 and F2 have equal numerical values F1 = F2 = 20 Newtons, they have the same application point, but have different directions and meanings. Therefore, the vectors F1 and F2 are not equal F1 ≠ F2.

c) The trolley speed is constant if the effect of the F1 and F2 forces is offset by the effect of the Ff friction force between the wheels and the asphalt. The force that would produce the same effect as the F1 and F2 forces is their resultant (R):

R = F1 + F2

It can be found by using the parallelogram rule for vectors: a parallelogram is constructed which has the vectors F1 and F2 as sides (Fig.2), leading through the F1 tip a parallel to F2 and through the F2 tip a parallel to F1. The parallelogram diagonal that starts at point 0 is the resultant of F1 and F2 forces. We can assume that two forces are exerted on the trolley: the force R (replacing F1 and F2) and the friction force Ff. Since the trolley's speed remains constant, it means that the effects of the two forces R and Ff are compensated, so their result is null:

R + Ff = 0

This is only possible if the forces R and Ff have the same direction, opposite senses and equal numerical values so that R - Ff = 0 (Fig.2). It is measured in Fig. 1 the length of the diagonal and it is 2.8 cm. Taking into account the chosen scale (1 cm = 10 Newtons), it results that R = 28 Newtons. Therefore, the friction force is:

Ff = 28 Newtons

Mechanics

Physics problems with solutions

Diameter of a hole made by a bullet in a metal sheet is greater or smaller than the diameter of the bullet?

Diameter of a hole made by a bullet in a metal sheet is greater or smaller than the diameter of the bullet?

Solution

When the bullet hits the metal sheet, it dilates, the bullet passes by producing a hole. After cooling, the diameter of the hole becomes smaller than the diameter of the bullet.

Thermodynamics

Physics problems with solutions

Definition of Electric Circuits:

Formulas in Electric Circuits:

Ohm's Law

Problem solved: Ohm's Law

Power Law

Problem solved: Power Law

Series Resistance

Problem solved: Series Resistance

Parallel Resistance

Problem solved: Parallel Resistance

Kirchhoff's Laws

Problem solved: Kirchhoff's Voltage Law (KVL)

More electric circuit problems with solutions

Definition

Formulas

Snell's Law

Lens-Maker's formula

Mirror Equation

Magnification (m)

Problem 1

Solution

At the terminals of a receiver with a resistance of 4 Ω is installed a source with an internal resistance of 0.2 Ω and EMF (electromotive force) of 150 V. The connecting wires have a resistance of 0.4 Ω. What is the voltage drop on the supply line and what is the voltage at the terminals?

Solution:

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