I really need help with this question someone plz help !

Answer:

0.173 N.

Explanation:

We will calculate the mass and then use the following calculations on the surface of planet X that is :

                           [tex]W=mg[/tex]

We would use the following equation to get the value of g for planet X that is :

                   [tex]y_f-y_i=v_{yi}t+\frac{1}{2}gt^2[/tex]

Then, put the values in the above equation.

                          [tex]16=0+\frac{1}{2}\times g\times(2.90)^2[/tex]

                           [tex]\bf\mathit{g=3.80\;m/s^2}[/tex]

Now, we will measure the ball weight on planet X's surface:

                          [tex]m=\frac{100}{1000} \;\;\;\;\;\;\;\;\;\;[1kg=1000g][/tex]

Then, we have to put the value in the above equation.

                        [tex]W=0.1\times 1.73=0.173\:N[/tex]

Answer:

Explanation:

There's a formula for this:

[tex]F = k*displacement[/tex]

F being force, k being the spring constant, and displacement being the change in x

We are given the force and the spring constant, so this is essentially isolating the Δx term. Do 60N/120N per meter. The newtons cancel out and you get a final answer of Δx = 0.5 meters

Answer:

[tex]\vec{L}=-30\frac{kgm^2}{s}\hat{k}[/tex]

Explanation:

In order to calculate the angular momentum of the particle you use the following formula:

[tex]\vec{L}=\vec{r}\ X\ \vec{p}[/tex]       (1)

r is the position vector respect to the point (0 , 5.0), that is:

r = 0m i + 5.0m j    (2)

p is the linear momentum vector and it is given by:

[tex]\vec{p}=m\vec{v}=(2.0kg)(3.0m/s)(\hat{i+\hat{j}})=6\frac{kgm}{s}(\hat{i}+\hat{j})[/tex]   (3)

the direction of p comes from the fat that the particle is moving along the i + j direction.

Then, you use the results of (2) and (3) in the equation (1) and solve for L:

[tex]\vec{L}=-30\frac{kgm^2}{s}\hat{k}[/tex]

The angular momentum is -30 kgm^2/s ^k

Answer:

none of the answers is correct, the time  is the same  t₁ = t₂ = 0.600 s

Explanation:

This is a kinematics exercise, analyze the situation a bit. The vertical speed in both cases is the same is zero, the horizontal speed in the second case is double (vₓ₂ = 2 vₓ₁)

let's find the time to hit the ground

     y = y₀ + I go t - ½ g t²

     0 = y₀ - ½ g t²

     t = √ 2y₀ / g

with the data from the first launch

     y₀i = ½ g t²

     y₀ = ½  9.8  0.6²

     y₀ = 1,764 m

with this is the same height the time to descend in the second case is the same

    t₂ = 0.600 s

this is because the horizontal velocity change changes the offset on the x axis, but does not affect the offset on the y axis

Therefore, none of the answers is correct, the time  is the same

t₁ = t₂ = 0.600 s

Answer:

19.872 degrees

Explanation:

Mathematically;

Using Snell’s law

n1 sin A = n2 sinB

Where ;

n1 = refractive index in air = 1

n2 is refractive index in water = 1.33

A = ?

B = 45

Substituting the values in the equation;

1 sin A = 1.33 sin45

Sin A = 1.33 sin 45

A = arc sin (1.33 sin 45)

A = 70.12

Thus, the actual direction of the Sun with respect to the horizon = 90-A = 19.872 degrees

Answer:

[tex]\vec{F_3}=(-34.0N)\hat{i}+(-12.0N)\hat{j}[/tex]

Explanation:

You have three forces F1, F2 an F3 that produce the following  acceleration:

a = −(8.00m/s^2)ˆi + (6.00m/s^2)ˆj

you know that force F1 and F2 are:

F1 = (30.0N)ˆi + (16.0N)ˆj

F2 = −(12.0N)ˆi + (8.00N)ˆj

and the force F3 is unknown:

F3 = F3x ˆi + F3y ˆj

The second Newton law is given by the following equation:

[tex]\vec{F}=m\vec{a}[/tex]

F: the total force = F1 +F2 + F3

m: mass of the object = 2 kg

By the properties of vectors you have:

[tex]\vec{F_1}+\vec{F_2}+\vec{F_3}=m\vec{a}\\\\(30.0-12.0+F_{3x})N\hat{i}+(16.0+8.00+F_{3y})N\hat{j}=(2.0kg)[(-8.00m/s^2)\hat{i}+(6.00m/s^2)\hat{j}]\\\\(18.0+F_{3x})N\hat{i}+(24.0+F_{3y})\hat{j}=(-16.00N)\hat{i}+(12.0N)\hat{j}[/tex]

Both x and y component must be equal in the previous equality, then you have:

[tex]18.0N+F_{3x}=-16.00N\\\\F_{3x}=-34.00N\\\\24.0N+F_{3y}=12.0N\\\\F_{3y}=-12.00N[/tex]

Hence, the vector F3 is:

[tex]\vec{F_3}=(-34.0N)\hat{i}+(-12.0N)\hat{j}[/tex]

Answer:

Explanation:

Assuming that the TV signal is sent in a straight line from the camera to the TV receiver, which is very far from the truth.

The reporter hears the sound is

4.1 / 343 = 0.01195 s later

The viewer hears the sound from the TV is

2.9 / 343 = 0.00845s

the difference is 0.00845 sec

the question is how far the TV signal can travel in that time.

the distance between politician and TV set is

= 0.00845 * 3*10^8 m

= 2536 km

d = 2536km

From Ohm's law . . . Resistance = (voltage) / (current)

Resistance = (120 volts) / (7.6 Amperes)

Resistance = 15.8 Ω

Answer:

5.1 rad/s

Explanation:

Mechanical energy of the system is conserved since no external work is done on the sphere.

[tex]mgh = mv^2/2 + I\omega^2/2[/tex]

Substituting v = ωr and I = 2 m r^2/5, we get,

=> [tex]mgh=m(\omega r)^2/2 + (2\omega r^2/5)\omega^2/2[/tex]

=> [tex]mgh = m\omega^2r^2/2 + m\omega^2r^2/5[/tex]

=> [tex]gh =\omega^2r^2/2+\omega^2r^2/5[/tex]

=>  [tex]gh = 7\omega^2 r^2/10[/tex]

=>  [tex]\omega r = (10gh/7)^{1/2}[/tex]

=> [tex]\omega = (1/r)(10gh/7)^{1/2} = (1 / 1.7)(10\times 9.8\times 5.3 / 7)^{1/2}[/tex] = 5.1 rad/s

Answer:

663N

Explanation:

We need to find the force that will overcome the frictional force.

The angle of the normal force is 15°.

The mass of the crate is 32 kg

The coefficient of static friction is 0.49

Frictional force is given in terms of Normal force as:

F = μNcosθ

where μ = coefficient of static friction

N = normal force

θ = angle of normal force

Frictional force is given as:

F = mg

=>mg = μNcosθ

=> N = mg/(μcosθ)

N = (32 * 9.8) / (0.49 * cos15)

N= 313.6 / 0.473

N = 663 N

The force needed to cause the box to move must be 663N or greater.

Answer:

it takes the shape and size of the container that it is in

Explanation:

Answer:

it takes the shape and size of a container

Answer:

The average upward force exerted by the water is 988.2 N

Explanation:

Given;

mass of the diver, m = 60 kg

height of the board above the water, h = 10 m

time when her feet touched the water, t = 2.10 s

The final velocity of the diver, when she is under the influence of acceleration of free  fall.

V² = U² + 2gh

where;

V is the final velocity

U is the initial velocity = 0

g is acceleration due gravity

h is the height of fall

V² = U² + 2gh

V² = 0 + 2 x 9.8 x 10

V² = 196

V = √196

V = 14 m/s

Acceleration of the diver during 2.10 s before her feet touched the water.

14 m/s is her initial velocity at this sage,

her final velocity at this stage is zero (0)

V = U + at

0 = 14 + 2.1(a)

2.1a = -14

a = -14 / 2.1

a = -6.67 m/s²

The average upward force exerted by the water;

[tex]F_{on\ diver} = mg - F_{ \ water}\\\\ma = mg - F_{ \ water}\\\\F_{ \ water} = mg - ma\\\\F_{ \ water} = m(g-a)\\\\F_{ \ water} = 60[9.8-(-6.67)]\\\\F_{ \ water} = 60 (9.8+6.67)\\\\F_{ \ water} = 60(16.47)\\\\F_{ \ water} = 988.2 \ N[/tex]

Therefore, the average upward force exerted by the water is 988.2 N

Answer:

A , D , E

Explanation:

Solution:-

- Consider the two identical objects with mass ( m ).

- The stiffness of the springs are ( k1 and k2 ).

- Both the spring store 55.0 J of potential energy.

- We will apply the principle of energy conservation on both the systems. In both cases the spring stores 55.0 Joules of energy. Once released, the objects gain kinetic energy with a consequent loss of potential energy in either spring.

- The maximum speed ( v ) is attained when all the potential energy is converted to kinetic energy.

- Apply Energy conservation for spring with stiffness ( k1 ).

                         ΔU = ΔEk

                         55.0 = 0.5*m*v^2

                         v = √ ( 110 / m )

- Apply Energy conservation for spring with stiffness ( k2 ).

                         ΔU = ΔEk

                         55.0 = 0.5*m*v^2

                         v = √ ( 110 / m )

Answer: Both objects will have the same maximum speed ( A )

- We are told that one spring is more stiff as compared to the other one. The measure of stiffness is proportionally quantified by the spring constant. To mathematically express we can write it as:

                         k1 > k2

Where,

                 k1: The stiff spring

                 k2: The flexible spring

Answer: The stiff spring has a larger spring constant than the flexible spring. ( D )

- We will assume that the spring with constant ( k1 ) undergoes a displacement ( x1 ) and the spring with constant ( k2 ) undergoes a displacement ( x2 ). The potential energy stored in both spring is the same. Hence,

                      U1 = U2

                      0.5*( k1 ) * ( x1 )^2 = 0.5*( k2 ) * ( x2 )^2

                      [ k1 / k2 ] = [ x2 / x1 ]^2

Since,

                     k1 > k2 , then [ k1 / k2 ] > 1    

Then,

                     [ x2 / x1 ]^2 > 1

                     [ x2 / x1 ] > 1

                     x2 > x1                  

Answer: The flexible spring ( x2 ) was compressed more than the stiff spring ( x1 ). ( E )

Answer:

Ff = 33.4N

Explanation:

To find the frictional force while the object is moving, you take into account that the friction force depends of the coefficient of kinetic friction.

The frictional force is given by:

[tex]F_f=\mu_kN[/tex]         (1)

Ff: frictional force = ?

µk: coefficient of kinetic friction = 0.167

N: normal force of the object = 200N

You replace the values of the parameters in the equation (1):

[tex]F_f=(0.167)(200N)=33.4N[/tex]

The frictional force, while the objects is moving, is 33.4N

Answer:

130.528779365 degrees

Explanation:

The angle of incidence is 30 degrees. From this, we can use Snell's Law to calculate the angle of refraction.

n1/n2 = sin(theta2)/sin(theta1)

let theta1 be 30 degrees, and n1 be the refractive index of air = 1

1/1.5 = sin(theta2)/sin(30deg)

solve:

sin(theta2) = 2/3 sin(30deg) = 1/3

theta2 = arcsin (1/3) = 19.4712206345 degrees

The angle of reflection will always be equal to the angle of incidence, in this case, 30 degrees.

Because these angles are measured relative to the normal, the angle formed between the two rays is the difference between the normal line (180 degrees) and the sum of the two angle measures.

Angle between = 180-30-19.4712206345 = 130.528779365 degrees

The angle between the reflected and transmitted rays 130.5287 degrees

The angle of incidence is 30 degrees. From this, we can use Snell's Law to calculate the angle of refraction.

[tex]\dfrac{n_1}{n_2} = \dfrac{sin(\theta_2)}{sin(\theta_1)}[/tex]

let [tex]\theta_1[/tex] be 30 degrees, and n1 be the refractive index of air = 1

[tex]\dfrac{1}{1.5} = \dfrac{sin(\theta_2)}{sin(30)}[/tex]

solve:

[tex]sin(\theta_2) = \dfrac{2}{3} sin(30) = \dfrac{1}{3}[/tex]

[tex]\theta_2 = sin ^{-1}\dfrac{1}{3} = 19.4712 \ degrees[/tex]

The angle of reflection will always be equal to the angle of incidence, in this case, 30 degrees.

Because these angles are measured relative to the normal, the angle formed between the two rays is the difference between the normal line (180 degrees) and the sum of the two angle measures.

Angle between = 180-30-19.4712206345 = 130.528779365 degrees

Hence the angle between the reflected and transmitted rays 130.5287 degrees

To know more about the Refraction of light follow

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Answer:

Explanation:

  a )

from lens makers formula

[tex]\frac{1}{f} =(\mu-1)(\frac{1}{r_1} -\frac{1}{r_2})[/tex]

f is focal length , r₁ is radius of curvature of one face and r₂ is radius of curvature of second face

putting the values

[tex]\frac{1}{1.8} =(1.38-1)(\frac{1}{.5} -\frac{1}{r_2})[/tex]

1.462 = 2 - 1 / r₂

1 / r₂ = .538

r₂ = 1.86 cm .

= 18.6 mm .

b )

object distance u = 25 cm

focal length of convex lens  f  = 1.8 cm

image distance  v   = ?

lens formula

[tex]\frac{1}{v} - \frac{1}{u} = \frac{1}{f}[/tex]

[tex]\frac{1}{v} - \frac{1}{-25} = \frac{1}{1.8}[/tex]

[tex]\frac{1}{v} = \frac{1}{1.8} -\frac{1}{25}[/tex]

.5555 - .04

= .515

v = 1.94 cm

c )

magnification = v / u

= 1.94 / 25

= .0776

size of image = .0776 x size of object

= .0776 x 10 mm

= .776 mm

It will be a real image and it will be inverted.

Answer:

The voltage is  [tex]V = 418.60 \ Volts[/tex]  

Explanation:

From the question we are told that

    The area of the both plate is  [tex]A = 7.00 *10^{-3} \ m^2[/tex]

    The distance between the plate is [tex]d = 4.80*10^{-4}\ m[/tex]

     The magnitude of the charge is  [tex]q = 5.40 *10^{-8} \ C[/tex]

The capacitance of the capacitor that consist of the two plates is mathematically represented as

        [tex]C = \frac{\epsilon _o A}{d}[/tex]

Where [tex]\epsilon_o[/tex] is the permitivity of free space with a value  [tex]e = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]

So

       [tex]C = \frac{8.85*10^{-12} * (7* 10^{-3})}{ 4.8*10^{-4}}[/tex]

        [tex]C = 1.29 *10^{-10} \ F[/tex]

The potential difference between the plate is mathematically represented as

      [tex]V = \frac{ Q}{C }[/tex]

     [tex]V = \frac{ 5.4*10^{-8}}{1.29 *10^{-10}}[/tex]

     [tex]V = 418.60 \ Volts[/tex]

Answer:

490 N

Explanation:

is the correct answer

If the up and down vectors are the same length. The right vector is longer than the left vector, then  the net force acting on Hector and the toboggan would be 490 Newtons.

Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum.

As given in the problem John pushes Hector on a plastic toboggan .The free-body diagram is shown. A free body diagram with 4 force vectors. The first vector is pointing downward, labeled F Subscript g Baseline = negative 490 N. The second vector is pointing right, labeled F Subscript t Baseline = 735 N. The third vector is pointing upward, labeled F Subscript N Baseline = 490 N. The fourth vector is pointing left, labeled F Subscript f Baseline = negative 245 N.

The net force acting on the vertical direction = 490-490

                                                                           =0

The net force acting on the horizontal direction = 735 -245

                                                                                =490 Newtons

Thus, the net force acting on Hector and the toboggan would be 490 Newtons.

Learn more about Newton's second law from here, refer to the link ;

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Answer:

The amount of water that will power a battery with that rating = 7.35 m³

Explanation:

The power rating for the battery is missing from the question.

Complete Question

Assuming 100% efficient energy conversion how much water stored behind a 50 centimeter high hydroelectric dam would be required to charged the battery with power rating, 12 V, 50 Ampere-minutes

Solution

Potential energy possessed by water at that height = mgH

m = mass of the water = ρV

ρ = density of water = 1000 kg/m³

V = volume of water = ?

g = acceleration due to gravity = 9.8 m/s²

H = height of water = 50 cm = 0.5 m

Potential energy = ρVgH = 1000 × V × 9.8 × 0.5 = (4900V) J

Energy of the battery = qV

q = 50 A.h = 50 × 60 = 3,000 C

V = 12 V

qV = 3,000 × 12 = 36,000 J

Energy = 36,000 J

At a 100% conversion rate, the energy of the water totally powers the battery

(4900V) = (36,000)

4900V = 36,000

V = (36,000/4900)

V = 7.35 m³

Hope this Helps!!!

Answer:

vBxf = 0.08625m/s

Explanation:

This is a problem about the momentum conservation law. The total momentum before equals the total momentum after.

[tex]p_f=p_i[/tex]

pf: final momentum

pi: initial momentum

The analysis of the momentum conservation is about a horizontal momentum (x axis). When the quarterback jumps straight up, his horizontal momentum is zero. Then, after the quarterback throw the ball the sum of the momentum of both quarterback and ball must be zero.

Then, you have:

[tex]m_Qv_{Qxi}+m_{Bxi}v_{Bxi}=m_Qv_{Qxf}+m_{Bxf}v_{Bxf}[/tex]    (1)

mQ: the mass of the quarterback = 80kg

mB: the mass of the football = 0.43kg

(vQx)i: the horizontal velocity of quarterback before throwing the ball = 0m/s

(vBx)i: the horizontal velocity of football before being thrown = 0m/s

(vQx)f: the horizontal velocity of quarterback after throwing the ball = ?

(vBx)f: the horizontal velocity of football after being thrown = 15 m/s

You replace the values of the variables in the equation (1), and you solve for (vBx)f:

[tex]0\ kgm/s=-(80kg)(v_{Bxf})+(0.46kg)(15m/s)\\\\v_{Bxf}=\frac{(0.46kg)(15m/s)}{80kg}=0.08625\frac{m}{s}[/tex]

Where you have taken the speed of the quarterback as negative because is in the negative direction of the x axis.

Hence, the speed of the quarterback after he throws the ball is 0.08625m/s

Answer:

The highest electric field is experienced by a 2 C charge acted on by a 6 N electric force. Its magnitude is 3 N.

Explanation:

The formula for electric field is given as:

E = F/q

where,

E = Electric field

F = Electric Force

q = Charge Experiencing Force

Now, we apply this formula to all the cases given in question.

A) A 2C charge acted on by a 4 N electric force

F = 4 N

q = 2 C

Therefore,

E = 4 N/2 C = 2 N/C

B) A 3 C charge acted on by a 5 N electric force

F = 5 N

q = 3 C

Therefore,

E = 5 N/3 C = 1.67 N/C

C) A 4 C charge acted on by a 6 N electric force

F = 6 N

q = 4 C

Therefore,

E = 6 N/4 C = 1.5 N/C

D) A 2 C charge acted on by a 6 N electric force

F = 6 N

q = 2 C

Therefore,

E = 6 N/2 C = 3 N/C

E) A 3 C charge acted on by a 3 N electric force

F = 3 N

q = 3 C

Therefore,

E = 3 N/3 C = 1 N/C

F) A 4 C charge acted on by a 2 N electric force

F = 2 N

q = 4 C

Therefore,

E = 2 N/4 C = 0.5 N/C

The highest field is 3 N, which is found in part D.

A 2 C charge acted on by a 6 N electric force

Answer:

6.4 m/s

Explanation:

From the question, we are given that

Speed of the river, v(r) = 5 m/s

velocity relative to the water, v(w) = 4 m/s

Width of the river, d = 780 m

The magnitude of his velocity relative to the earth is v(m)

v(m) can be gotten by using the relation

[v(m)]² = [v(w)]² + [v(r)]²

[v(m)]² = 4² + 5²

[v(m)]² = 16 + 25

[v(m)]² = 41

v(m) = √41

v(m) = 6.4 m/s

thus, the magnitude of the velocity relative to earth is 6.4 m/s

Answer:

Vf = 78.64 m/s

Explanation:

The rocket is travelling upward at a constant acceleration of 3.99 m/s² until it runs out of fuel. So, in order to calculate its velocity at the point, where it runs out of fuel, we can simply use 3rd equation of motion:

2as = Vf² - Vi²

where,

a = acceleration = 3.99 m/s²

s = distance or height covered by rocket till fuel runs out = 775 m

Vf = Final Velocity = ?

Vi = Initial velocity = 0 m/s   (Since, rocket starts from rest)

Therefore,

2(3.99 m/s²)(775 m) = Vf² - (0 m/s)²

Vf = √(6184.5 m²/s²)

Vf = 78.64 m/s

An object's resistance to any change in motion is the Inertia of the object.

Answer:

D

Explanation:

the other students are making fun of her most likely because they are jealous that she is successing in school. hope this helps :)

Answer:

D

Explanation:

Answer:

The correct answer is C 1, 4, and 5 tie, then 2 and 3 tie

Explanation:

Solution

The electric field due to sheets E₁ positive =б/2E₀

E₂ is negative = б/2E₀

Now,

At the point 1, 4, 5 the electric field due to the sheets are in the opposite direction

At the point 1, the net field = -E₁ + E₂ =0

At the point A, the net field = -E₁ - E₂ = 0

Now,

At nay point inside between them, the electric field is seen to be at the same direction.

At the 2, 3 points the field is seen at the right

Thus,

E net = E₁ + E₂

= б/2E₀ + σ/2E₀

=б/E₀

Note: Kindly find an attached copy of the complete question to the solution

The correct answer is option C

The rank of the points according to the magnitude of the electric field is 1, 4, and 5 tie, then 2 and 3 tie

Let sheet 1 has positive surface charge density and sheet 2 has a negative surface charge density

The electric field (without direction) due to sheets will be

E₁ =σ/2E₀

E₂= σ/2E₀

Now,

At the point 1, 4, 5 the electric field due to the sheets is given by:

E = E₁ - E₂

E = σ/2E₀ - σ/2E₀

since the positive charge plate will have electric field lines away from the sheet and the negative charge plate will have electric field lines towards the sheet

E = 0

Now,

At points 2, 3 which are between the plates,

The net electric field is:

E = E₁ + E₂

since the electric field due to both the plates will be from positive to negative ( towards the negatively charged plate)

E = σ/2E₀ + σ/2E₀

E = σ/E₀

Learn more about surface charge density:

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Answer:

a)   v = 75 ft / s , b)  v = 55 ft / s , c)   Δx = 1000 ft

Explanation:

We can solve this exercise with the expressions of kinematics

a) average speed is defined as the distance traveled in a given time interval

        v = (x₂-x₁) / (t₂-t₁)

         v = (550 - 400) / (10 -8)

         v = 75 ft / s

b) we repeat the calculations for this interval

   v = (550 - 0) / (10 -0)

   v = 55 ft / s

c)  we clear the distance from the average velocity equation

     Δx = v (t₂ -t₁)

     Δx = 100 (20-10)

     Δx = 1000 ft

Answer:

the answer is opposite.

plz mark brainliest

Explanation:

Answer:

The drag (air resistance) it experiences along its flight to the basket, due to the shape and surface area of the socks, the size of the sock (weight), and the speed with which the socks is tossed.

Explanation:

The socks, like every other particle or body travelling through air is met by a resistance that impedes its motion. This resistance is due to the air molecules around, that collide with the body as it travels through them. The resistance offered by this force is proportional to the surface area of the body that collides with the air molecule, so, rolling the socks into a ball reduces the effect of air resistance on the socks, compared to the one tossed without rolling. Air resistance is also largely dependent on the relative motion of the body and the air molecules, the density of the fluid (air), and the size of the body (weight).

Therefore, whether the socks lands in the basket or not is affected by the drag (air resistance) it experiences along its flight to the basket, due to the shape and surface area of the socks, size of the socks (weight), and the speed with which the socks is tossed.

Drag force opposes motion of objects through fluid with its magnitude depending on the velocity of the object in the fluid

The single parameter that effects whether or not the socks lands in the basket is the drag force, [tex]\mathbf{F_D}[/tex] acting on the socks

[tex]F_D = \mathbf{C_D \times A \times \dfrac{\rho \times v_r^2}{2}}[/tex]

The reason that drag force is the parameter that effects the landing point of the socks is as follows:

The parameters that effects whether or not the socks land in the basket or not are;

The parameters, R, u, and θ depends on the thrower, that parameter that effects the whether or not the socks lands in the basket that is independent of the thrower, is the drag, [tex]\mathbf{F_D}[/tex]

Drag is the force opposing (slows) the motion of an object in a fluid.

The drag force, [tex]\mathbf{F_D}[/tex], slowing down motion, is given by the following formula;

[tex]F_D = \mathbf{C_D \times A \times \dfrac{\rho \times v_r^2}{2}}[/tex]

Where;

[tex]v_r[/tex] = The velocity of flow of the fluid, relative to the object

ρ = The density of the fluid

[tex]C_D[/tex] = The drag coefficient

A = The cross sectional area of the fluid

Therefore, the independent parameter that effects whether or not the socks lands in the basket is the drag force on the socks

Learn more about drag force here:

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