This problem, a squid at rest suddenly sees a predator coming toward it and needs to escape. Assume the following:______.
(i) Including the water in its internal cavity, the squid has a total mass of 6.50 kg.
(ii) The mass of the water in its cavity is 1.75 kg.
(iii) In order to escape its predators, the squid needs to achieve an escape speed of 2.5 m/s.

The charge required for the fixed point charge is approximately \(4.8032 * 10^{-10} C\).

The electrostatic force between the fixed point charge and the electron is equal in magnitude to the gravitational force between the electron and the Earth, so we can set these two forces equal to each other and solve for the charge of the fixed point charge.

Electrostatic force between two point charges is given by Coulomb's law:

\(F = k * q1 * q2 / r^2\)

The gravitational force between two masses is given by Newton's law of gravitation:

\(F = G * m1 * m2 / r^2\)

Setting these two forces equal to each other and solving for q2, we get:

\(k * q1 * q2 / r^2 = G * m_e * m\_earth / r^2\)

Solving for q2, we get:

\(q2 = G * m_e * m\_earth / k\)

Substituting the given values, we get:

\(q2 = (6.67430 * 10^{-11} N * m^2 / kg^2) *\\ (9.10938356 *10^{-31} kg) *\\ (5.9722 *10^{24} kg) / (8.98755 * 10^9 N * m^2 / C^2)\)

\(q2 = 4.8032 * 10^{-10} C\)

Therefore, the charge required for the fixed point charge is approximately \(4.8032 * 10^{-10}\)C.

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Answer:

Determine how to separate gasoline from other substances in petroleum

The MA, VR and efficiency of the screw nail will be 20, 176 and 11.36%, respectively.

Mechanical advantage (MR) is the ratio between the load and effort.

Velocity ratio (VR) of a simple machine is the ratio of distance travelled by effort to the distance travelled by the load in the machine.

The efficiency of a simple machine can be defined as the ratio of useful work done by the machine (output work) to the total work put into the machine (input work).

We need to make 14 cm of diameter of the screw. Then, we can solve the problem.

Suppose the effort applied on screw is (E) = x

Then, The force is (F) = 20x

MA = F/E

MA = 20x/x (x and x are cancelled)

MA = 20

Then,

VR = 2πR/ p

VR = 2×22×0.14/7×0.05

VR = 176

and

efficiency = MA/VR × 100

efficiency = 20/176 × 100 = 11.36

Therefore, the efficiency is 11.36.

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The force exerted by the bottom block on the upper block is of a magnitude equivalent to that exerted by the Earth on the top block.

Every action has an equivalent and opposite response states Newton's Third Law of Motion. According to Newton's Third Law, the top block is inflicting an equivalent and opposing force on the bottom block in this situation where the bottom block is exerting a force on the top block. This indicates that the force exerted by the bottom block on the top block is equivalent to the force exerted by the top block on the bottom block.

Additionally, due to the interaction between the two items, the force exerted by the Earth on the top block is equivalent to and in opposition to the force exerted by the top block on the Earth. As a result, the force exerted by the lower block on the top block is of the same size as the force exerted by the Earth on the top block.

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Answer:

Atoms are the thing that make up molecules and compounds. Molecule. Two or more atoms joined together with covalent bonds. Molecules contain two or more atoms and are held together by covalent bonds, whereas compounds are held together by ionic bonds. Compound. Two or more elements bonded together through ionic attraction.

The length of side YZ (c) is approximately 119.13 miles.

To find the length of side c, we can use the Law of Cosines, which states:

c² = a² + b² - 2ab * cos(C)

Plugging in the given values:

a = 222 miles

b = 150 miles

C = 30 degrees

We need to convert the angle from degrees to radians to use it in the cosine function. The conversion is as follows:

θ (radians) = θ (degrees) * π / 180

C (radians) = 30 degrees * π / 180 = π / 6 radians

c² = 222² + 150² - 2 * 222 * 150 * cos(π / 6)

c² = 49284 + 22500 - 66600 * cos(π / 6)

c² = 49284 + 22500 - 66600 * (√3 / 2)

c² = 71784 - 66600 * (√3 / 2)

c² = 71784 - 66600 * 0.866

c² = 71784 - 57600

c² = 14184

c = √14184

c ≈ 119.13 miles (rounded to two decimal places)

Therefore, the length of side YZ (c) is approximately 119.13 miles.

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The velocity of the car will be   \(V=18.18\ \frac{m}{s}\)

As velocity is the movement of the car with respect to time.

It is given that

Mass m  =  1100kg

Momentum= 22000 \(\frac{kgm}{s}\)

The momentum is given as

\(P=mv\)

\(v=\dfrac{p}{m}\)

\(v=\dfrac{22000}{1100} =18.18 \ \dfrac{m}{s}\)

Thus  the velocity of the car will be   \(V=18.18\ \frac{m}{s}\)

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Explanation:

where I is the current, L is the length of the wire, and B is the magnetic field.

If we point our right-hand thumb in the direction of the current (to the right), and our fingers in the direction of the magnetic field (away from us), then our palm will point in the direction of the force.

Answer:

88 ft / sec = 60 mph

Thus 97/60 * 88 = 142 ft/sec      maximum speed of cannonball

R = V^2 sin 2 θ / g = 142^2 / 32 = 630 ft

Using 3.28 ft / m

630 ft / 3.28 f/m = 192 m is the maximum range of the cannonball

Vy = 142 ft / sec * sin 45 = 100 ft/sec    vertical speed at 45 deg

Tup = 100 ft/sec / 32 ft/sec^2 = 3.12 sec    time to reach height

T = 2 * 3.12 = 6.24      total time in air when fired at 45 deg

Answer:

There is less gravitational force in space.

Explanation:

Gravity doesn't exist as it does on Earth. Earth's gravity is as 6 times stronger as it is on the moon.

The scientific study of physics focuses on the underlying concepts that underpin the laws of nature. It investigates how matter, energy, space, and time behave and interact. Physics' fundamental goal is to comprehend the underlying laws and forces that create our universe.

Physics has developed theories and rules to explain a wide variety of events, from the motion of celestial bodies to the behaviour of subatomic particles, via meticulous observation, investigation, and mathematical analysis.

These theories offer a framework for comprehending and forecasting the behaviour of physical systems, such as Newton's laws of motion and Einstein's theory of relativity.

From the tiniest particles to the biggest cosmic structures, physics has enhanced our understanding of the world and sparked a host of technological advances.

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Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

Answer:

Explanation:

We shall apply law of conservation of momentum .

Momentum before collision = momentum after collision .

Momentum before collision = 400 kg m/s

Momentum after collision = 5  x v + 11 x 15

where v is velocity of A after the collision .

5  x v + 11 x 15 = 400

5 v = 400 - 165

5v = 235

v = 47 m /s .

The distance (in meters) that Savanna run during this time will be 13.56 meters.

From the information, we are told that during a run near the end of Cross-Country practice, Savanna maintained an average speed of 6.78 m/s for 2.00 minutes.

Average speed is calculated by dividing total distance that something has traveled by the total amount of time that it took it to travel that distance. The average speed is how fast something is going at a particular moment. Average speed simply measures the average rate of speed over the extent of a trip.

It should be noted that the distance will be calculated as:

Distance = Speed × Time

Distance = 6.78 × 2

Distance = 13.56

Therefore, based on the information given it can be seen that the distance (in meters) that Savanna run during this time will be 13.56 meters.

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The minimum mass required to be placed on the 0.00 cm mark of the meter stick to maintain equilibrium is 0.120 kg.

To maintain equilibrium, the torques acting on the meter stick must balance each other. The torque is given by the formula:

τ = r * F * sin(θ)

where τ is the torque, r is the distance from the pivot point to the point where the force is applied, F is the force applied, and θ is the angle between the force vector and the lever arm.

In this case, the meter stick is in equilibrium when the torques on both sides of the pivot point cancel each other out. The torque due to the weight of the meter stick itself is acting at the center of mass of the meter stick, which is at the 50.0 cm mark.

Let's denote the mass to be placed on the 0.00 cm mark as M. The torque due to the weight of M can be calculated as:

τ_M = r_M * F_M * sin(θ)

where r_M is the distance from the pivot point to the 0.00 cm mark (which is 30.0 cm), F_M is the weight of M, and θ is the angle between the weight vector and the lever arm.

Since the system is in equilibrium, the torques on both sides of the pivot point must be equal:

τ_M = τ_stick

r_M * F_M * sin(θ) = r_stick * F_stick * sin(θ)

Substituting the given values:

30.0 cm * F_M = 20.0 cm * (0.0400 kg * 9.8 m/s^2)

Solving for F_M:

F_M = (20.0 cm / 30.0 cm) * (0.0400 kg * 9.8 m/s^2)

F_M = 0.0264 kg * 9.8 m/s^2

F_M = 0.25872 N

Finally, we can convert the force into mass using the formula:

F = m * g

0.25872 N = M * 9.8 m/s^2

M = 0.0264 kg

Therefore, the minimum mass required to be placed on the 0.00 cm mark of the meter stick to maintain equilibrium is 0.120 kg.

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Given data:

* The mass of the cannon is 52.07 kg.

* The recoil velocity of the cannon is 7.75 m/s.

* The mass of the cannon ball is 9.06 kg.

Solution:

By the law of conservation of the momentum,

The momentum of the cannon ball before the collision is the same as the momentum of the cannon ball from the cannon.

The momentum of the cannon ball from the cannon is,

where m is the mass of the cannon and v is the recoil velocity of the cannon,

Substituting the known values,

Thus, the momentum of the cannon ball before the collision is 403.54 kgm/s.

Answer:

The power dissipated if the three resistors were connected in parallel across the same potential difference is 405 W

Explanation:

Given;

three identical resistors connected in series

let the first resistor = R₁

let the second resistor = R₂

let the third resistor = R₃

Rt = R₁ + R₂ + R₃

Since the resistors are identical, thus, R₁ = R₂ = R₃ = R

Rt = 3R

Power is given as;

P = IV = V² / R

\(P = \frac{V^2}{R_t} = \frac{V^2}{3R} \\\\3P = \frac{V^2}{R} ------equation(i)\)

If the 3 identical resistor connection were changed to parallel, then the equivalent resistance in the circuit will be;

\(\frac{1}{R_t} = \frac{1}{R_1} +\frac{1}{R_2} + \frac{1}{R_3} \\\\But, R_1 = R_2 = R_3\\\\\\frac{1}{R_t} = \frac{1}{R} +\frac{1}{R} + \frac{1}{R} \\\\\frac{1}{R_t} =\frac{3}{R} \\\\R_t = \frac{R}{3} \\\\P = \frac{V^2}{R_t} = \frac{3V^2}{R} \\\\P_{parallel} = \frac{3V^2}{R} ---------equation (ii)\\\\From \ equation \ (i), 3P_{series} = \frac{V^2}{R}, Substitute \ this \ into \ equation \ (ii)\\\\P = 3(\frac{V^2}{R} )\\\\P = 3(3P)\\\\P_{parallel} = 9P_{series}\\\\P_{parallel} = 9(45)\\\\\)

\(P_{parallel} = 405 \ W\)

Therefore, the power dissipated if the three resistors were connected in parallel across the same potential difference is 405 W

The point through which this ray will pass, after refraction by the lens is point D.

The refraction of light refers to the bending or change in direction that occurs when light passes from one medium to another. It is a phenomenon that happens due to the difference in the speed of light in different substances.

From the ray diagram given, after the light incident from point O, it will pass the converging at point D which is the focal length of the lens after refraction.

Thus, based on the converging lens given in the ray diagram, we can conclude that, the point through which this ray will pass, after refraction by the lens is point D.

So point D is the correct answer.

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Answer:

c

Explanation:

Answer:

D.matter has changed size,shape, or form

Explanation:

hope it helps<333

Answer:

Chemists make a distinction between two different types of changes that they study—physical changes and chemical changes. Physical changes are changes that do not alter the identity of a substance. Chemical changes are changes that occur when one substance is turned into another substance

Explanation:

please mark as brainliiest please

try to appreciate my work

Answer:

dino :)

Explanation:

Answer:

v = 28.98 ft / s

Explanation:

For this problem we must solve it in parts, let's start by looking for the speed of the two cars after the collision

In the exercise they indicate the weight of each car

          Wₐ = 1500 lb

          W_b = 1125 lb

Car B's velocity from v_b = 42.0 mph westward, car A travels east

let's find the mass of the vehicles

             W = mg

             m = W / g

             mₐ = Wₐ / g

             m_b = W_b / g

             mₐ = 1500/32 = 46.875 slug

             m_b = 125/32 = 35,156 slug

Let's reduce to the english system

             v_b = 42.0 mph (5280 foot / 1 mile) (1h / 3600s) = 61.6 ft / s

We define a system formed by the two vehicles, so that the forces during the crash have been internal and the moment is preserved

we assume the direction to the east (right) positive

initial instant. Before the crash

           p₀ = mₐ v₀ₐ - m_b v_{ob}

final instant. Right after the crash

           p_f = (mₐ + m_b) v

the moment is preserved

           p₀ = p_f

           mₐ v₀ₐ - m_b v_{ob} = (mₐ + m_b) v

           v = \(\frac{ m_a \ v_{oa} - m_b \ v_{ob} }{ m_a +m_b}\)

we substitute the values

           v = \(\frac{ 46.875}{82.03} \ v_{oa} - \frac{35.156}{82.03} \ 61.6\)

           v = 0.559 v₀ₐ - 26.40                  (1)

Now as the two vehicles united we can use the relationship between work and kinetic energy

the total mass is

              M = mₐ + m_b

              M = 46,875 + 35,156 = 82,031 slug

starting point. Jsto after the crash

              K₀ = ½ M v²

final point. When they stop

             K_f = 0

The work is

             W = - fr x

the negative sign is because the friction forces are always opposite to the displacement

Let's write Newton's second law

Axis y

           N-W = 0

           N = W

the friction force has the expression

            fr = μ N

we substitute

            -μ W x = Kf - Ko

            -μ W x = 0 - ½ (W / g) v²

            v² = 2 μ g x  

            v = \(\sqrt{ 2 \ 0.750 \ 32 \ 17.5}\)Ra (2 0.750 32 17.5  

            v = 28.98 ft / s

D. The frequency of the wave is determined 1.2 Hz.

The frequency of a wave describes the number of waves that pass a fixed place in a given amount of time.

General wave equation;

y = A cos(ωt)

where;

From the given wave equation;

x(t) = 0.65 cos(7.4t)

ω = 2πf

7.4 = 2πf

f = 7.4/2π

f = 1.18 Hz

f = ≈ 1.2 Hz

Thus, the frequency of the wave is determined 1.2 Hz.

The correct option is D.

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The potential energy of the object of mass 5.2 kg is 295.568 J and the kinetic energy as it strikes the ground is 295.568 J.

potential energy is the energy that is stored in an object due to its position relative to some zero position.

To calculate the potential energy of the object, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

As the object is just about to hit the ground, all the potential energy is converted to kinetic energy which is 295.568 J.

Hence, the potential energy of the object is 295.568 J and the kinetic energy as it strikes the ground is 295.568 J.

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The back emf at 2500 rpm if the magnetic field is unchanged is 100 V for the back emf in a motor is 72 V when operating at 1800 rpm.

The back emf in a motor is proportional to the speed of the motor. Therefore, we can use the following formula to determine the back emf at 2500 rpm:

E2 = E1 × (N2 / N1)

where E1 is the back emf at 1800 rpm, N1 is the speed at which the back emf was measured, E2 is the back emf at 2500 rpm, and N2 is a new speed.

Plugging in the values we get:

E2 = 72 V × (2500 rpm / 1800 rpm)

E2 = 100 V

Therefore, the back emf at 2500 rpm of the motor would be 100 V if the magnetic field is unchanged.

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If the object starts at position 12 on a horizontal line with a reference point of 0, then its initial position relative to the reference point is:

Initial position = Reference point + Object's position = 0 + 12 = 12

If the object then moves 14 units to the left, its new position relative to the reference point is:

New position = Initial position - Distance moved to the left = 12 - 14 = -2

Therefore, the position of the object is -2 units from the reference point, which means that it has moved 2 units to the left of the reference point.

The block accelerates down the ramp due to the component of its weight that acts parallel to the ramp; this force has magnitude

(100 kg) g sin(26.6°) ≈ 438 N

(Note that the positive sign here means we take "down the ramp" to be the positive direction.)

Since the ramp is frictionless, this is the only force acting on the block in this direction. By Newton's second law, the block's acceleration is a such that

438 N ≈ (100 kg) a   ⇒   a ≈ 4.38 m/s²

The block accelerates uniformly, so that it attains a speed v as it moves 50 m such that

v² = 2a (50 m)

Solve for v :

v = √(2a (50 m)) ≈ 20.9 m/s

Answer:

A prism is a clear crystal that refracts light

Explanation:

(1) the pressure of the skier is 416.67 N/m².

(2) the pressure exerted by the wheel is 76.67 N/m².

(3) the surface area of the knife is 0.03 m².

The pressure of the skier can be calculated using the formula:

pressure = force / area

Substituting the given values, we get:

pressure = 500 N / 1.2 m²

pressure = 416.67 N/m²

The pressure exerted by the wheel can be calculated using the same formula:

pressure = force / area

Substituting the given values, we get:

pressure = 23 N / 0.3 m²

pressure = 76.67 N/m²

The pressure exerted by the knife is given as 200 N/m² when a force of 6 N is applied. We can use the same formula to calculate the surface area of the knife:

pressure = force / area

Rearranging the formula, we get:

area = force / pressure

Substituting the given values, we get:

area = 6 N / 200 N/m²

area = 0.03 m²

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