EXPLANATION ⛔

A 20 gram mass is suspended from meter rod at 20cm. The meter rod is balanced on 40cm mark. Weight of meter rod is

A. 0.4N
B. 0.6N
C. 6N
D. 60N​

Answer:

a) Fs = (μs*mp*g)/2  |  b) τ = Fs*lp  |  c) Wτ,constant = τΘ

Explanation:

a) Fs = (μs*mp*g)/2

b) τ = Fs*lp

c) Wτ,constant = τΘ

Answer:

887.2 N

Explanation:

Given that

Mass of Lara, m = 62 kg

Length of the vine, l = 8.8 m

Speed of the swing, v = 6.3 m/s Then,

We start by calculating her weight.

w = mass * acceleration

w = mg

w = 62 * 9.8

w = 607.6 N

F(c) = mv²/l, on substituting

F(c) = (62 * 6.3²) / 8.8

F(c) = 2460.78 / 8.8

F(c) = 279.6 N

T - 607.6 N = 279.6 N

T = 607.6 N + 279.6 N

T = 887.2 N

Thus, the minimum breaking strength of the vine is 887.2 N

The minimum breaking strength of the vine is about 900N.

Tension is a force developed by a rope, string, or cable when stretched under an applied force.

Given:

Mass (m) = 62 kg, length (l) = 8.8 m, velocity (v) = 6.3 m/s, g = 10 m/s².

Weight(W) = m * g = 62 * 10 = 620N

centripetal force = F(c) = mv²/l = 62 * 6.3² / 8.8 = 279.6N

T - W = F(c)

T - 620 = 279.6

T = 900N

The minimum breaking strength of the vine is about 900N.

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Answer: 36, 200 feet deep according to information on google

Explanation:

A small submarine, the bathyscape Trieste, made it to 10,916 meters (35,813 feet) below sea level in the deepest point in the ocean, the Challenger Deep in the Marianas Trench, a few hundred miles east of the Philippines. This part of the ocean is 11,034 m (36,200 ft) deep, so it seems that a submarine can make it as deep as it's theoretically possible to go

Complete Question

A lawn mower has a flat, rod-shaped steel blade that rotates about its center. The mass of the blade is 0.65 kg and its length is 0.55 m. You may want to review (Pages 314 - 318) .

Part A What is the rotational energy of the blade at its operating angular speed of 3510 rpm

Part B

If all of the rotational kinetic energy of the blade could be converted to gravitational potential energy, to what height would the blade rise?

Answer:

Part A  

    [tex]R = 1081 \ J[/tex]

Part B  

     [tex]h = 169.7 \ m[/tex]

Explanation:

From the question we are told that

  The mass of the blade is  [tex]m_b = 0.65 \ kg[/tex]

   The length is  [tex]l = 0.55 \ m[/tex]

   The angular speed is  [tex]w = 3510 rpm = 3510 * \frac{2 \pi }{60} = 367.6 \ rad/sec[/tex]

Generally the moment of inertia of the of this mower is mathematically evaluated as

         [tex]I = \frac{m_b * l^2 }{12}[/tex]

substituting values

         [tex]I = \frac{0.65 * 0.55^2 }{12}[/tex]

         [tex]I = 0.016 \ kg m^2[/tex]

Generally the rotational kinetic energy of the bland is  

        [tex]R = \frac{1}{2} * I * w^2[/tex]

substituting values

       [tex]R = \frac{1}{2} * 0.016 * 367.6^2[/tex]

     [tex]R = 1081 \ J[/tex]

At point where the gravitational potential energy is equal to the rotational kinetic energy  we have that

       [tex]P = R = m_b * h * g[/tex]

Where P is the  gravitational potential energy

substituting values

          [tex]1081 = 0.65 * 9.8 * h[/tex]

=>       [tex]h = 169.7 \ m[/tex]

Answer:

Stretch in the spring = 0.1643 (Approx)

Explanation:

Given:

Mass of the sled (m) = 9 kg

Acceleration of the sled (a) = 2.10 m/s ²

Spring constant (k) = 115 N/m

Computation:

Tension force in the spring  (T) = ma

Tension force in the spring  (T) = 9 × 2.10

Tension force in the spring  (T) = 18.9 N

Tension force in the spring = Spring constant (k) × Stretch in the spring

18.9 N = 115 N  × Stretch in the spring

Stretch in the spring = 18.9 / 115

Stretch in the spring = 0.1643 (Approx)

Answer:

A. 2.82 eV

B. 439nm

C. 59.5 angstroms

Explanation:

A. To calculate the energy of the photon emitted you use the following formula:

[tex]E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})[/tex]     (1)

n1: final state = 5

n2: initial state = 2

Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

[tex]E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV[/tex]

B. The energy of the emitted photon is given by the following formula:

[tex]E=h\frac{c}{\lambda}[/tex]   (2)

h: Planck's constant = 6.62*10^{-34} kgm^2/s

c: speed of light = 3*10^8 m/s

λ: wavelength of the photon

You first convert the energy from eV to J:

[tex]2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J[/tex]

Next, you use the equation (2) and solve for λ:

[tex]\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm[/tex]

C. The radius of the orbit is given by:

[tex]r_n=n^2a_o[/tex]   (3)

where ao is the Bohr's radius = 2.380 Angstroms

You use the equation (3) with n=5:

[tex]r_5=5^2(2.380)=59.5[/tex]

hence, the radius of the atom in its 5-th state is 59.5 anstrongs

A) The energy E of the emitted photon in electron volts is; E = 2.856 eV

B) The wavelength in nanometers of the emitted photon is; λ = 434.4nm

C) The radius of the hydrogen atom in nanometers in its initial 5th energy level is; rₙ = 1.323 nm

A) Formula for the energy E of the emitted photons is;

E = -13.6([tex]\frac{1}{n_{2}^2} - \frac{1}{n_{1}^2}[/tex])

We are given;

n₂ = 5

n₁ = 2

Thus;

E = -13.6([tex]\frac{1}{5^2} - \frac{1}{2^2}[/tex])

E = 2.856 eV

B) The formula for the wavelength is;

λ = hc/E

where;

h is Planck's constant = 6.626 × 10⁻³⁴ m².kg/s

c is speed of light = 3 × 10⁸ m/s

E is energy of photon

λ is wavelength of the photon

Earlier we saw that E = 2.856 eV. Converting to Joules gives;

E = 4.5758 × 10⁻¹⁹ J

Thus;

λ = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(4.5758 × 10⁻¹⁹)

λ = 4.344 × 10⁻⁷ m

Converting to nm gives;

λ = 434.4nm

C) Formula for the radius of the hydrogen atom is;

rₙ = n²a₀

where;

a₀ is bohr's radius = 5.292 × 10⁻¹¹ m

n = 5

Thus;

rₙ = 5² × 5.292 × 10⁻¹¹

rₙ = 1.323 × 10⁻⁹

rₙ = 1.323 nm

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Answer:

1.48 g

Explanation:

A = A₀ (½)^(t / T)

where A is the final amount,

A₀ is the initial amount,

t is time,

and T is the half life.

A = (5.00 g) (½)^(5.00 a / 2.85 a)

A = 1.48 g

Answer:

B

Explanation:

Because the force has 2 components (horizontal and vertical), the horizontal component must be smaller than the total force. The Pythagorean theorem only adds positive values (because they're squared) so it makes sense. Using trigonometry, 100*cos(-20) yields a horizontal force of around 939.7N, which is less than 1000N.

Answer:[tex]8.062\ m/s[/tex]

Explanation:

Given

masss of football player [tex]M=110\ kg[/tex]

Velocity of football player [tex]u_1=8\ m/s[/tex]

mass of football [tex]m=0.41\ kg[/tex]

velocity of football [tex]u_2=25\ m/s[/tex]

Final velocity will be given by applying conservation of linear momentum

After catching the ball Player and ball moves with same velocity

[tex]\Rightarrow Mu_1+mu_2=(M+m)v[/tex]

[tex]\Rightarrow 110\times 8+0.41\times 25=(110+0.41)v[/tex]

[tex]\Rightarrow 880+10.25=110.41\times v[/tex]

[tex]\Rightarrow v=\frac{890.25}{110.41}=8.063\ m/s[/tex]

So, final velocity will be [tex]8.062\ m/s[/tex]

Answer:

A13 mm is about 1/4 A5 mm

Explanation:

Find the attachment

Answer:

25m

Explanation:

Let's assume the Jeep attains a velocity of 36km/h ; a constant speed same with that of the car.

While the Jeep is accelerating to that speed, the car with that speed passes it.

Now we can calculate the time taken for the Jeep to attain the velocity of 36km/h on her constant acceleration.

This time is t = v/a; from Newton's Law of Motion:

a = V-U / t ; a-acceleration

V is final velocity = 36km/h

U is initial velocity 0 since the body starts from rest.

Hence t = 36000/3600 ÷ 4 = 2.5s

Note conversting from km/h to m/s we multiply by 1000/3600.

But the distance covered by the car while the Jeep just accelerates is

S = U × t = 10× 2.5 = 25m.

Note From Newton's law of Motion, distance for constant speed is defined as: U × t

Hence the Car would be 25m off the starting point just as the Jeep accelerates. It would overtake the Jeep when it just covers 25m from the Jeep starting point.

Answer:

B. revolve around the sun

   rotate on an axis

   generally have moons

Explanation:

edge 2021

Answer:

The actual height is  [tex]A =3.158 \ m[/tex]

Explanation:

From the question we are told that

   The depth of the person is  [tex]d = 1.1 \ m[/tex]

    The apparent height  is  [tex]D = 4.2 \ m[/tex]

Generally

     The refractive index of water is  [tex]n_w = 1.33[/tex]

      The refractive index of the air is  [tex]n_a = 1[/tex]

The apparent depth is mathematically represented as

      [tex]D = A [\frac{n_w}{n_a} ][/tex]  

substituting values

     [tex]4.2 = A [\frac{1.33}{1} ][/tex]  

=>   [tex]A = \frac{4.2 }{1.33}[/tex]

      [tex]A =3.158 \ m[/tex]

The ball was dropped at the height of "3.158 m". To understand the calculation, check below.

According to the question,

Water's refractive index, [tex]n_w[/tex] = 1.33

Air's refractive index, [tex]n_a[/tex] = 1

Apparent height, D = 4.2 m

Person's depth, d = 1.1 m

We know the relation,

→ D = A[[tex]\frac{n_w}{n_a}[/tex]]

By substituting the values, we get

4.2 = A[[tex]\frac{1.33}{1}[/tex]]

By applying cross-multiplication,

  A = [tex]\frac{4.2}{1.33}[/tex]

      = 3.158 m

Thus the approach above is correct.

Find out more information about refractive index here:

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Answer:

239 N/C

Explanation:

Electric field strength at distance R from a charge Q is given by the expression

E = k Q / R² where Q is charge , R is distance of charge from the point . k is a constant .

R = 40 cm , Q = 4.25 x 10⁻⁹

Putting the given values

E = 9 x 10⁹ x 4.25 x 10⁻⁹ / ( 40 x 10⁻²)²

= 239 N/C .

Answer:

Inner radius = 2 mm

Explanation:

In a coaxial cable, series inductance per unit length is given by the formula;

L' = (µ/(2π))•ln(R/r)

Where R is outer radius and r is inner radius.

We are given;

L' = 50 nH/m = 50 × 10^(-9) H/m

R = 2.6mm = 2.6 × 10^(-3) m

Meanwhile µ is magnetic constant and has a value of µ = µ_o = 4π × 10^(−7) H/m

Plugging in the relevant values, we have;

50 × 10^(-9) = (4π × 10^(−7))/(2π)) × ln(2.6 × 10^(-3)/r)

Rearranging, we have;

(50 × 10^(-9))/(2 × 10^(−7)) = ln((2.6 × 10^(-3))/r)

0.25 = ln((2.6 × 10^(-3))/r)

So,

e^(0.25) = (2.6 × 10^(-3))/r)

1.284 = (2.6 × 10^(-3))/r)

Cross multiply to give;

r = (2.6 × 10^(-3))/1.284)

r = 0.002 m or 2 mm

Answer:

the kinetic energy of the 2 kg mass after the 100 m is equal to 1962 J

Explanation:

mass of block A = 2 kg

mass of block B = 4 kg

distance the blocks were pushed = 100 m

NB: Blocks were pushed the same distance at the same equal time period. And the ground is without friction.

Work done in moving the 2 kg mass along the 100 m distance is,

work = force x distance moved

force exerted by the 2 kg mass = 2 x 9.81 m/s^2(acceleration due to gravity)

force = 19.62 N

therefore,

work done = 19.62 x 100 = 1962 Joules of work.

According to energy conservation principles, the kinetic energy impacted of the 2 kg mass through this distance will be equal to the work done in moving the 2 kg mass through this distance.

Therefore, the kinetic energy of the 2 kg mass after the 100 m is equal to 1962 J

Answer:

17.85°

Explanation:

To find the angle to the normal in which the light travels in the aqueous fluid you use the Snell's law:

[tex]n_1sin\theta_1=n_2sin\theta_2[/tex]

n1: index of refraction of Sophia's cornea = 1.387

n2: index of refraction of aqueous fluid = 1.36

θ1: angle to normal in the first medium = 17.5°

θ2: angle to normal in the second medium

You solve the equation (1) for θ2, next, you replace the values of the rest of the variables:

[tex]\theta_2=sin^{-1}(\frac{n_1sin\theta_1}{n_2})\\\\\theta_2=sin^{-1}(\frac{(1.387)(sin17.5\°)}{1.36})=17.85\°[/tex]

hence, the angle to normal in the aqueous medium is 17.85°

Speed = (distance) / (time)

Speed = (

Velocity = speed, and its direction

The velocity of the plane is 10.2 miles per second East.

(about 48 times the speed of sound)

Answer:

  (possibly) Box D

Explanation:

The one that has balanced forces will not accelerate. The forces are unbalanced in figures A, B, C. We cannot tell about figure D, because the downward force is not shown. If that force is 20 N, box D will not accelerate.

Answer:

(a) f = 185 Hz

(b) v = 266.4 m/s

Explanation:

(a) The lowest frequency can be calculated by using the following formula for the calculation of the modes (resonant frequencies) in a string:

[tex]f_n=\frac{nv}{2L}[/tex]

[tex]f_n=nf[/tex]

n: order of the mode

v: velocity of the waves in the string

L: length of the string = 72.0cm = 0.72m

fn: frequency of the n-th mode

With the information about two consecutive modes you can find the lowest resonant frequency. First you find the resonant mode n:

[tex]f_n=nf\\\\f_{n-1}=(n-1)f\\\\\frac{f_n}{f_{n-1}}=\frac{n}{n-1}[/tex]

you solve the previous equation for n:

[tex](n-1)f_n=nf_{n-1}\\\\555n-555=370n\\\\n=3[/tex]

With this information you can calculate the lowest resonant frequency:

[tex]f_n=nf\\\\f=\frac{f_n}{n}=\frac{555}{3}=185Hz[/tex]

b) You have information about two consecutive modes fn, fn-1. Then, you can calculate the velocity of the waves:

[tex]f_{n}-f_{n-1}=n\frac{v}{2L}-(n-1)\frac{v}{2L}\\\\f_n-f_{n-1}=\frac{v}{2L}\\\\v=2L(f_n-f_{n-1})[/tex]

fn = 555 Hz

fn-1: 370 Hz

[tex]v=2(0.72m)(555-370)Hz=266.4\frac{m}{s}[/tex]´

hence, the velocityof the waves in the string is 266.4 m/s

Answer:

The average velocity of Jackson is 18.056 m/s South

The average velocity of Hunter is 10.65 m/s East

Explanation:

initial velocity of Jackson, u = 25 km/h east = 6.944 m/s east

time for this motion, [tex]t_i[/tex] = 20 minutes = 1200 seconds

⇒initial displacement of Jackson, [tex]x_i[/tex] = (6.944 m/s) x (1200 s) = 8332.8 m

Final velocity of Jackson, v =  45 km/h South = 12.5 m/s South

time at Jackson's final position, [tex]t_f[/tex] = 20 minutes + [tex]t_i[/tex] = 20 minutes + 20 minutes

time at Jackson's final position, [tex]t_f[/tex] = 40 minutes = 2400 s

⇒Final displacement of Jackson,[tex]x_f[/tex] = (12.5 m/s) x (2400 s) = 30,000m

Average velocity of Jackson;

[tex]= \frac{x_f-x_i}{t_f-t_i} \\\\= \frac{30,000-8332.8}{2400-1200} \\\\= 18.056 \ m/s \ South[/tex]

initial velocity of Hunter, u = 45 km/h South = 12.5 m/s South

time for this motion, [tex]t_i[/tex] = 10 minutes = 600 seconds

⇒initial displacement of Hunter, [tex]x_i[/tex] = (12.5 m/s) x (600 s) = 7500 m

Final velocity of Hunter, v =  40 km/h east = 11.11 m/s east

time at Hunter's final position, [tex]t_f[/tex] = 30 minutes + [tex]t_i[/tex] = 30 minutes + 10 minutes

time at Hunter's final position, [tex]t_f[/tex] = 40 minutes = 2400 s

⇒Final displacement of Hunter,[tex]x_f[/tex] = (11.11 m/s) x (2400 s) = 26,664m

Average velocity of Hunter;

[tex]= \frac{x_f-x_i}{t_f-t_o} \\\\= \frac{26,664-7500}{2400-600} \\\\= 10.65 \ m/s \ east[/tex]

Answer:

(a) 498.6 Hz

(b) 534.6 Hz

Explanation: Please see the attachments below

Answer:

  C. her moment of inertia increases and her angular speed decreases

  D. her moment of inertia increases and her angular speed decreases

Explanation:

The moment of inertia of a body is the sum of the products of an increment of mass and the square of its distance from the center of rotation. When a spinning person extends her arms, part of her mass increases its distance from the center of rotation, so increases the moment of inertia.

The kinetic energy of a spinning body is jointly proportional to the moment of inertia and the square of the angular speed. Hence an increase in moment of inertia will result in a decrease in angular speed unless there is a change in the rotational kinetic energy.

This effect is used by figure skaters to increase their spin rate by drawing their arms and legs closer to the axis of rotation. Similarly, they can slow the spin by extending arms and legs.

When the person extends her arms, her moment of inertia increases and her angular speed decreases.

_____

Note to those looking for a letter answer

Both choices C and D have identical (correct) wording the way the problem is presented here. You will need to check carefully the wording in any problem you may think is similar.

Answer:

11.8 Joules

Explanation:

Given:-

- The height of the target ball, si = 0.860 m

- The mass of target and steel ball, m = 0.012 kg

- The target ball travels a distance ( x ) after being struck = 1.40 m

Find:-

What is the kinetic energy (in joules) of the target ball just after it is struck?

Solution:-

- We are given the initial distance of the target ball as 0.86 m above the floor which travels a distance ( x ) after being struck.

- We will employ the one dimensional kinematic equation of motion to determine the initial velocity ( vi ) of the target ball as follows:

                        [tex]vf^2 = vi^2 - 2*g*x[/tex]

Where,

                  vf: The final velocity of target ball at maximum height = 0

                  g: The gravitational acceleration constant = 9.8 m/s^2

- Plug in the required parameters and evaluate the ( vi ) as follows:

                      [tex]0^2 = vi^2 - 2*( 9.80 )*( 1.40 )\\\\vi^2 = 27.44\\\\vi = \sqrt{27.44} = 5.24 m/s[/tex]

- The kinetic energy ( Ek ) of an object with mass ( m ) and initial velocity ( vi ) is expressed as:

                       [tex]E_k = 0.5*m*(vi)^2\\\\E_k = 0.5*0.86*27.44\\\\E_k = 11.8 J[/tex]

Answer: The kinetic energy of the target ball just after it is struck is 11.8 Joules.

Answer:

Coefficient of kinetic friction = 0.146

Explanation:

Given:

Mass of sled (m) = 18 kg

Horizontal force (F) = 30 N

FInal speed (v) = 2 m/s

Distance (s) = 8.5 m

Find:

Coefficient of kinetic friction.

Computation:

Initial speed (u) = 0 m/s

v² - u² = 2as

2(8.5)a = 2² - 0²

a = 0.2352 m/s²

Nweton's law of :

F (net) = ma

30N - μf = 18 (0.2352)

30 - 4.2336 = μ(mg)

25.7664 =  μ(18)(9.8)

μ = 0.146

Coefficient of kinetic friction = 0.146

Before she jumped from the cliff, her gravitational potential energy was

GPE = (her weight) x (height of the cliff) .

That's exactly the GPE that she lost on the way down to the water.  So we can write

24,500 J = (her weight) x (44.0 m)

Divide each side by 44.0 m:

Her weight = 24,500 J / 44 m

Her weight = 556.8 Newtons

(about 125 pounds)

Answer:

65 m/h

Explanation:

Let the distance of the car moving south be y.

Let the distance of the car moving west be x.

Let the distance between the two cars be a.

These three distances can be represented as a right angled triangle. So we can say:

[tex]a^2 = x^2 + y ^2[/tex]

Let us differentiate with respect to time, since the distances are changing with respect to time:

[tex]2a\frac{da}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt} \\\\=>a\frac{da}{dt} = x\frac{dx}{dt} + y\frac{dy}{dt}[/tex]__________(1)

da/dt = rate of change of distance between two cars

The speed of the car moving south (dy/dt) is 60 m/h and the speed of the car moving west (dx/dt) is 25 m/h.

Therefore:

dy/dt = 60 m/h and dx/dt  = 25 m/h

After two hours, the distance of the two cars will be:

y = 2 * 60 = 120 miles

x = 2 * 25 = 50 miles

Therefore:

[tex]a^2 = 50^2 + 120^2\\\\a^2 = 2500 + 14400 = 16900\\\\a = \sqrt{16900}\\ \\a = 130 miles[/tex]

From (1):

130(da/dt) = 50(25) + 120(60)

130(da/dt) = 1250 + 7200 = 8450

da/dt = 8450/130 = 65 m/h

Therefore, after two hours, the distance between the two cars is changing at a rate of 65 m/h.

The ducks' flight path as observed by someone standing on the ground is the sum of the wind velocity and the ducks' velocity relative to the wind:

ducks (relative to wind) + wind (relative to Earth) = ducks (relative to Earth)

or equivalently,

[tex]\vec v_{D/W}+\vec v_{W/E}=\vec v_{D/E}[/tex]

(see the attached graphic)

We have

[tex]\vec v_{D/W}=\left(7.0\dfrac{\rm m}{\rm s}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)[/tex]

[tex]\vec v_{W/E}=\left(5.0\dfrac{\rm m}{\rm s}\right)(\cos0^\circ\,\vec\imath+\sin0^\circ\,\vec\jmath)[/tex]

[tex]\vec v_{D/E}=v(\cos270^\circ\,\vec\imath+\sin270^\circ\,\vec\jmath)[/tex]

Then by setting components equal, we have

[tex]\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta+5.0\dfrac{\rm m}{\rm s}=0[/tex]

[tex]\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v[/tex]

We only care about the direction for this question, which we get from the first equation:

[tex]\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta=-5.0\dfrac{\rm m}{\rm s}[/tex]

[tex]\cos\theta=-\dfrac57[/tex]

[tex]\theta=\cos^{-1}\left(-\dfrac57\right)\text{ OR }\theta=360^\circ-\cos^{-1}\left(-\dfrac57\right)[/tex]

or approximately 136º or 224º.

Only one of these directions must be correct. Choosing between them is a matter of picking the one that satisfies both equations. We want

[tex]\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v[/tex]

which means θ must be between 180º and 360º (since angles in this range have negative sine).

So the ducks must fly (relative to the air) in a direction 224º relative to the positive horizontal direction, or about 44º South of West.

Answer:

for an object to be in equilibrium, it must be experiencing no acceleration. Both the net force and the net torque must be zero.

Hope I helped

Answer:

An object in static equilibrium has zero net force acting upon it.

The First Condition of Equilibrium is that the vector sum of all the forces acting on a body vanishes. This can be written as

 F = F1+ F2+ F3+ F4+. . . = 0

Answer:

I can help! What level of physics is it and what are your main topics?