Net force is the sum of all the forces acting on an object. If a spring balance pulls on a body with a force of 10 N, and friction acts on the body in the opposite direction with a force of 1 N, the net force would be 9 N in the direction of the spring balance (10 N – 1 N = 9 N).What is the net force acting on the object when the spring balance pulls the rope with a force of 25 N and friction acts on the body with a force of 20N?


Answer 1




(25 N - 20 N = 5 N)

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A 220 kg crate hangs from the end of a rope of length L = 14.0 m. You push horizontally on the crate with a varying force F to move it 4.00 m to the side.(a) What is the magnitude of F when the crate is in this final position? During the crate's displacement, what are (b) the total work done on it, (c) the work done by the gravitational force on the crate, and (d) the work done by the pull on the crate from the rope? (e) Knowing that the crate is motionless before and after its displacement, use the answers to (b), (c), and (d) to find the work your force F does on the crate. (f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)?

A small metal bead, labeled A, has a charge of 28 nC .It is touched to metal bead B, initially neutral, so that the two beads share the 28 nC charge, but not necessarily equally. When the two beads are then placed 5.0 cmapart, the force between them is 4.8×10^−4 N . Assume that A has a greater charge. What is the charge qA and qB on the beads?




Let the charge on bead A be q nC  and the charge on bead B be 28nC - qnC

Force F between them

4.8\times10^{-4} = (9*10^9* q*(28-q)*10^(-18))/((5*10^(-2))^2)

=120 x 10⁻⁸ = 9 x q(28 - q ) x 10⁻⁹

133.33 = 28q - q²

q²- 28q +133.33 = 0

It is a quadratic equation , which has two solution

q_A = 21.91 x 10⁻⁹C or q_B = 6.09 x 10⁻⁹ C

In a region where there is a uniform electric field that is upward and has magnitude 3.80x104 N/C a small object is projected upward with an initial speed of 2.32 m/s The object travels upward a distance of 5.98 cm in 0 200 s. What is the object's charge-to-mass ratio q/m (magnitude and sign)?Assume g 9.80 m/s and ignore air resistance E3? C/kg q/m



6.03 x 10^-3 C/Kg


E = 3.8 x 10^4 N/C, u = 2.32 m/s, s = 5.98 cm = 0.0598 m, t = 0.2 s, g = 9.8 m/s^2

Acceleration on object is a .

Use second equation of  motion.

S = u t + 1/2 a t^2

0.0598 = 2.32 x 0.2 + 0.5 x a x 0.2 x 0.2

0.0598 = 4.64 + 0.02 x a

a = - 229 m/s^2

Now, F = ma = qE

q / m = a / E = 229 / (3.8 x 10000)

q / m = 6.03 x 10^-3 C/Kg

In order to work well, a square antenna must intercept a flux of at least 0.040 N⋅m2/C when it is perpendicular to a uniform electric field of magnitude 5.0 N/C.



L > 0.08944 m or L > 8.9 cm



- Flux intercepted by antenna Ф = 0.04 N.m^2 / C

- The uniform electric field E = 5.0 N/C


- What is the minimum side length of the antenna L ?


- We can apply Gauss Law on the antenna surface as follows:

                             Ф = \int\limits^S {E} \, dA

- Since electric field is constant we can pull it out of integral. The surface at hand is a square. Hence,

                             Ф = E.(L)^2

                             L = sqrt (Ф / E)

                             L > sqrt (0.04 / 5.0)

                             L > 0.08944 m

Final answer:

The area of a square antenna needed to intercept a flux of 0.040 N⋅m2/C in a uniform electric field of magnitude 5.0 N/C is 0.008 m². Consequently, each side of the antenna must be about 0.089 meters (or 8.9 cm) long.


The question pertains to the relationship between electric field and flux. The electric flux through an area is defined as the electric field multiplied by the area through which it passes, oriented perpendicularly to the field.

We are given that the electric field (E) is 5.0 N/C and the flux Φ must be 0.040 N⋅m2/C.

Hence, to intercept this amount of flux, the antenna must have an area (A) such that A = Φ / E.

That is, A = 0.040 N⋅m2/C / 5.0 N/C = 0.008 m².

Since the antenna is square, each side will have a length of √(0.008) ≈ 0.089 meters (or 8.9 cm).

Learn more about Electric Flux here:


A train accelerates at -1.5 m/s2 for 10 seconds. If the train had an initialspeed of 32 m/s, what is its new speed?
A. 17 m/s
B. 15 m/s
C. 47 m/s
D. 32 m/s



17 m/s


Using formula a = (v-u) /t

acceleration a =  -1.5 m/s2

final velocity v = unknown

initial velocity u = 32 m/s

time t = 10s

-1.5 = (v-32)/10

-15 = v - 32

-15 + 32 = v

v = 17 m/s

If the diameter of the black marble is 3.0cm, and bye using the formula for volume, what is a good approximation if it’s volume? Record to the ones place


Complete question is;

If the diameter of the black marble is 3.0 cm, and by using the formula for volume, what is a good approximation of its volume?


14 cm³


We will assume that this black marble has the shape of a sphere from online sources.

Now, volume of a sphere is given by;

V = (4/3)πr³

We are given diameter = 3 cm

We know that radius = diameter/2

Thus; radius = 3/2 = 1.5 cm

So, volume = (4/3)π(1.5)³

Volume ≈ 14.14 cm³

A good approximation of its volume = 14 cm³

An electric generator contains a coil of 140 turns of wire, each forming a rectangular loop 71.2 cm by 22.6 cm. The coil is placed entirely in a uniform magnetic field with magnitude B = 4.32 T and initially perpendicular to the coil's plane. What is in volts the maximum value of the emf produced when the loop is spun at 1120 rev/min about an axis perpendicular to the magnetic field?





To solve this problem it is necessary to use the concept related to induced voltage or electromotive force measured in volts. Through this force it is possible to maintain a potential difference between two points in an open circuit or to produce an electric current in a closed circuit.

The equation that allows the calculation of this voltage is given by,

\epsilon = BAN \omega


B = Magnetic field

A= Area

N = Number of loops

\omega= Angular velocity

Our values previously given are:

N = 140

A = 71.2*10^(-2)m*22.6*10^(-2)m=0.1609m^2

B = 4.32 T

\omega = 1120 rev / min

We need convert the angular velocity to international system, then

\omega = 1120 rev/min

\omega = 1120rev/min*(2\pi)/(1rev)*(1min)/(60sec)

\omega = 117.2rad/s

Applying the equation for emf, we replace the values and we will obtain the value.

\epsilon = BAN \omega

\epsilon = (4.32)(0.1609)(140)*117.2

\epsilon = 11405Volt

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