an unsuccesstul life-historv strategv would be associated with bopulations that: decrease in size over time increase in size over time either decrease or do not change in size over time either increase or do not change in size over time.T/F

The force acting on the system of two blocks connected by a rope passing over a pulley is -13.26 N.

The system of two blocks connected by a rope passing over a pulley are M1 and M2, where M1 rests on the triangle edge to the left of the pulley, which makes an angle of theta subscript 1 with the base of the triangle. The coefficient of friction between box M1 and the surface is mu subscript 1. M2 rests on the triangle edge to the right of the pulley, which makes an angle of theta subscript 2 with the base of the triangle.

The coefficient of friction between box M2 and the surface is mu subscript 2. The system sits atop a scalene triangle whose long edge forms the base. The pulley is attached to the apex of the triangle.M1 has a mass of 2.25 kg and is on an incline of angle 1=42.5∘ that has a coefficient of kinetic friction 1=0.205. M2 has a mass of 5.55 kg and is on an incline of angle 2=33.5∘ that has a coefficient of kinetic friction 2=0.105.The free-body diagram of M1 shows that the weight of M1 acts straight downwards (vertically) and the normal force acts perpendicular to the slope.

The force of friction opposes the motion and acts opposite to the direction of motion.M1 = 2.25 kgTheta subscript 1 = 42.5 degreesMu subscript 1 = 0.205g = 9.81 m/s²In the free-body diagram of M2, the normal force acts perpendicular to the incline of the slope, the weight of the object acts vertically downwards and parallel to the incline, and the force of friction opposes the motion and acts opposite to the direction of motion.M2 = 5.55 kgTheta subscript 2 = 33.5 degreesMu subscript 2 = 0.105g = 9.81 m/s²The tension in the string is the same throughout the rope. Since the masses are being pulled by the same rope, the acceleration of the objects is the same as the acceleration of the rope.

The tension in the string is directly proportional to the acceleration of the objects and the rope.A system of two blocks connected by a rope passing over a pulley has a total mass of M. The acceleration of the system is given by the formula below:a = [(m1-m2)gsin(θ1) - μ1(m1+m2)gcos(θ1)] / (m1 + m2)Where, μ1 = 0.205 is the coefficient of friction of block M1θ1 = 42.5 degrees is the angle of the incline of block M1M1 = 2.25 kg is the mass of block M1M2 = 5.55 kg is the mass of block M2g = 9.81 m/s² is the acceleration due to gravitysinθ1 = sin 42.5 = 0.67cosθ1 = cos 42.5 = 0.75The acceleration of the system is:a = [(2.25-5.55)(9.81)(0.67) - (0.205)(2.25+5.55)(9.81)(0.75)] / (2.25 + 5.55)a = -1.7 m/s² (the negative sign indicates that the system is accelerating in the opposite direction).

The force acting on the system is given by:F = MaWhere M is the total mass of the system and a is the acceleration of the system. The total mass of the system is:M = m1 + m2M = 2.25 + 5.55M = 7.8 kgThe force acting on the system is:F = 7.8(-1.7)F = -13.26 N (the negative sign indicates that the force is acting in the opposite direction).

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The magnitude of the vector is approximately 35.85 m.

The angle that this vector makes with the positive x-axis is approximately 32 degrees.

To find the magnitude of the vector with components Ax=29 m and Ay=18 m, we use the Pythagorean theorem:

|A| = √(Ax^2 + Ay^2)

|A| = √(29^2 + 18^2)

|A| = √(841 + 324)

|A| = √1165

|A| =  34.13 m

To find the angle that this vector makes with the positive x-axis, we can use the inverse tangent function:

θ = tan^-1(Ay/Ax)

θ = tan^-1(18/29)

θ = 31.82 degrees

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Answer:

  0.017 N

Explanation:

The relevant relation is ...

  F = GMm/r²

where G is the universal gravitational constant, 6.67408 × 10^-11 m^3·kg^-1·s^-2, M and m are the masses of the objects, and r is the distance between them.

__

Filling in the given numbers, we find the force to be ...

  F = (6.67408 × 10^-11 m^3·kg^-1·s^-2)(8.7 × 10^20 kg)(77 kg)/(1.6 × 10^7 m)^2

where m in this expression is the unit "meters".

  F = 6.67408 · 8.7 · 77/2.56 × 10^(-11 +20 -2·7) N ≈ 0.017 N

The asteroid exerts a force of about 0.017 N on Sally.

__

Additional comment

That's about 0.000023 times the force of Earth's gravity.

The ecological methods allows to find the correct answer to the question of how to prevent river pollution is:

   b)  Maintain zones of grass at least wide on the sides of streams in the area.

Stream pollution from human activities can occur in a number of ways:

Let's analyze the different claims.

a) False. It is too expensive and the slopes cannot be changed due to possible flooding problems.

b) True. Maintaining a protective zone with grass on the sides of the rivers, prevents the waste of the frarmer from reaching the river, and the problems of the debris reaching the river bank. This un  ecological methods.

c) False. The dam controls flooding but does not eliminate the problems of river pollution.

d) False. Many lands are not acts for agriculture, but for livestock.

In conclusion we can find the correct answer to the question of how to prevent river pollution is:

   b) Maintain zones of grass at least wide on the sides of streams in the area.

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To calculate the torque required to create an angular acceleration, we can use the formula:

Torque = Moment of Inertia × Angular Acceleration

The moment of inertia of a disk can be calculated using the formula:

Moment of Inertia = (1/2) × Mass × Radius^2

Given:

Mass = 15,000 kg

Radius = 6.14 m

Angular Acceleration = 0.0500 rad/s^2

First, calculate the moment of inertia:

Moment of Inertia = (1/2) × Mass × Radius^2

Moment of Inertia = (1/2) × 15,000 kg × (6.14 m)^2

Next, calculate the torque:

Torque = Moment of Inertia × Angular Acceleration

Torque = Moment of Inertia × 0.0500 rad/s^2

Now, let's plug in the values and calculate:

Moment of Inertia = (1/2) × 15,000 kg × (6.14 m)^2

Moment of Inertia ≈ 283,594.13 kg·m^2

Torque = 283,594.13 kg·m^2 × 0.0500 rad/s^2

Torque ≈ 14,179.71 N·m

Rounding to three significant figures, the torque required to create an angular acceleration of 0.0500 rad/s^2 is approximately 14,180 N·m.

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A. P=IV

Explanation

Electric power is the rate, per unit time, at which electrical energy is transferred by an electric circuit.

the electric powe can be written as follows:

therefore, according to the given choices, the answer is

A. P=IV

I hope this helps you

The work required to move the crate vertically at a constant speed of 5.0 m is approximately 7000 Joules (J).

To determine the work required to move the crate vertically, we need to calculate the gravitational potential energy change. The work done is equal to the change in potential energy.

The formula for gravitational potential energy is given by:

Potential energy = mass * acceleration due to gravity * height

In this case, the mass of the crate is not provided, but we can use the given weight of the crate to find the mass. Weight is equal to mass multiplied by the acceleration due to gravity (W = mg).

Given:

Weight of crate (W) = 1400 N

Acceleration due to gravity (g) = 9.8 m/s^2

Vertical distance (height) = 5.0 m

First, calculate the mass of the crate:

1400 N = m * 9.8 m/s^2

m = 1400 N / 9.8 m/s^2 ≈ 143 kg

Now we can calculate the work:

Work = Potential energy = mass * g * height

Work = 143 kg * 9.8 m/s^2 * 5.0 m ≈ 7000 J

Therefore, the work required to move the crate vertically at a constant speed of 5.0 m is approximately 7000 Joules (J).

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A cell of inter resistance of 0.5 ohm is connected to coil of resistance 4 ohm and 8 ohm joined in parallel.If there is current of 2A in 8 ohm, the electromotive force (emf) of the cell is approximately 14.5 volts.

To find the emf of the cell, we can apply Ohm's Law and Kirchhoff's laws to analyze the circuit.

Given:

Resistance of the coil, R1 = 4 ohm

Resistance of the other resistor, R2 = 8 ohm

Current passing through the 8-ohm resistor, I = 2A

First, let's analyze the parallel combination of the 4-ohm and 8-ohm resistors.

The total resistance of two resistors in parallel can be calculated using the formula:

1/Rp = 1/R1 + 1/R2

Substituting the given values, we have:

1/Rp = 1/4 + 1/8

1/Rp = 2/8 + 1/8

1/Rp = 3/8

Rp = 8/3 ohm

Now, let's consider the total resistance in the circuit, which includes the internal resistance of the cell (0.5 ohm) and the parallel combination of the resistors (8/3 ohm).

R_total = R_internal + Rp

R_total = 0.5 + 8/3

R_total = 1.833 ohm

Now, we can find the emf of the cell using Ohm's Law:

emf = I * R_total

emf = 2 * 1.833

emf ≈ 3.667 volts

Therefore, the emf of the cell is approximately 3.667 volts.

However, it is worth noting that the given current of 2A passing through the 8-ohm resistor does not affect the emf calculation since the emf of the cell is independent of the current in the circuit.

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1,000 J of energy are needed to melt 10 g of a solid substance that is already at its melting point ,  the heat of fusion of the substance is 548 joules .

Heat of fusion, also known as enthalpy of fusion or latent heat of fusion, is the amount of energy required to melt or freeze a substance under constant pressure conditions. When it comes to chemistry, "fusion" is basically synonymous with "melting." In the classroom, heat of fusion is typically used when a substance is at its melting or freezing point. In such instances, most people consider heat of fusion to be a constant.

Water, for example, has a heat of fusion of 334 J/g at its melting point of 0°C. At 0°C, one grams  of liquid water requires 334 Joules of energy to completely freeze into ice. In addition, one grams of ice requires 334 Joules of energy to melt entirely.

q = m×∆Hf

q: Total change in heat energy (in Joules)

∆Hf: Heat of fusion of substance (in Joules per gram)

m: Mass of substance (in grams)

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Given data

*The given power is P = 2000 MW = 2000 × 10^6 W

*The given distribution voltage is V = 22 kV = 22 × 10^3 V

The formula for the current needed to carry the 2000 MW generated at a large power station is given as

Substitute the known values in the above expression as

Hence, the current needed to carry the 2000 MW generated at a large power station is I = 90.90 × 10^3 A

When studying the properties of light, for example, reaction time would not matter much. Light moves at such a fast rate that reaction time has no effect on the results of experiments measuring the speed of light, the reflection of light, or any other property related to light.

What is reflection?

Reflection is when a wave, such as a light wave or sound wave, bounces off a surface and returns to the source. This happens when the wave encounters an obstacle such as a wall or other barrier, and the wave is redirected back towards the source. Reflection is one of the fundamental principles of physics, and is used to explain many phenomena such as the fact that light is reflected off of a mirror. Reflection is also used to help explain phenomena such as sound reverberation and the formation of shadows.

What is sound reverberation ?

Sound reverberation is the effect created when a sound is reflected off of surfaces in a room or area. This effect is created when the sound bounces off of walls, ceilings, and other objects, and then is heard again by the listener. This creates an echo-like effect, and can be used to create a more acoustically ‘live’ sound.

Therefore, One example of where reaction time wouldn't matter very much for the results is when studying the properties of light.

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The net force on particle q₂, located between particles q₁ and q₃, is approximately 189000 N. The force exerted by particle q₁ on q₂ is positive and equals 252000 N, while the force exerted by particle q₃ on q₂ is negative and equals -63000 N.

To find the net force on particle q₂, we need to calculate the individual forces exerted on q₂ by particles q₁ and q₃ and then determine their sum.

The force between two charged particles can be calculated using Coulomb's law:

F = k * |q₁ * q₂| / r²

Where F is the force between the particles, k is the electrostatic constant (k ≈ 9.0 x \(10^9\) Nm²/C²), q₁ and q₂ are the charges of the particles, and r is the distance between them.

First, let's calculate the force exerted on q₂ by q₁:

F₁₂ = k * |q₁ * q₂| / r₁₂²

F₁₂ = (9.0 x \(10^9\) Nm²/C²) * |(8.0 μC) * (3.5 μC)| / (0.10 m)²

F₁₂ ≈ 252000 N

The force is positive because q₁ and q₂ have opposite charges.

Next, let's calculate the force exerted on q₂ by q₃:

F₂₃ = k * |q₂ * q₃| / r₂₃²

F₂₃ = (9.0 x \(10^9\)Nm²/C²) * |(3.5 μC) * (-2.5 μC)| / (0.15 m)²

F₂₃ ≈ -63000 N

The force is negative because q₂ and q₃ have the same charge.

Finally, we can find the net force on q₂ by summing the individual forces:

Net force = F₁₂ + F₂₃

Net force = 252000 N + (-63000 N)

Net force ≈ 189000 N

The net force on particle q₂ is approximately 189000 N.

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Answer:

Using the pythagoras theorem

S²=9²+4²

S²=81+16

S²=97

S=9.85km.

In finding the direction

tan□=opposite/Adjacent

=4/9

□=23.96

¤=90-23.96

=66.03 degrees

9.85, N 66.03 E

Answer:

14,432 g

Explanation:

PE=mh

360,800 J = (m)(25)

\(\frac{360,800 J}{25 m}\) = m

14,432 g = m

In series, both bulbs have the same current flowing through them. The bulb with the higher resistance will have a greater voltage drop across it and therefore have a higher power dissipation and brightness. In parallel, both bulbs have the same voltage across them.

The impulse exerted by Earth on the apple during the first 0.50 m of its fall is 0.74 Ns, and during the next 0.50 m, it is 0.37 Ns.

Using the equation for impulse, which is impulse = force x time, we can calculate the impulse that Earth exerts on the apple during the first 0.50 m and the next 0.50 m of its fall.

First, we need to calculate the force of gravity acting on the apple, which is given by the equation F = mg, where m is the mass of the apple and g is the acceleration due to gravity (approximately 9.81 m/s^2).

The mass of the apple is 150 g, which is 0.15 kg. Therefore, the force of gravity acting on the apple is:

F = mg = (0.15 kg)(9.81 m/s^2) = 1.47 N

Now, we can calculate the impulse exerted by Earth on the apple during the first 0.50 m of its fall. Since the force of gravity is constant, we can use the equation impulse = force x distance, where distance is the distance over which the force is applied.

Impulse during first 0.50 m = force x distance = (1.47 N)(0.50 m) = 0.74 Ns

For the next 0.50 m of the apple's fall, we need to consider that the velocity of the apple is increasing, so the force of gravity is no longer constant. However, we can approximate the average force over this distance as half the force at the start of the fall, or 0.5(1.47 N) = 0.74 N.

Using the same equation impulse = force x distance, we can calculate the impulse exerted by Earth on the apple during the next 0.50 m of its fall:

Impulse during next 0.50 m = force x distance = (0.74 N)(0.50 m) = 0.37 Ns.

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The object of mass of 10.3 kg subjected to a non-zero net force moves with an acceleration of 1.3 m/s², so the net force acting on the object is 13.39 N.

According to physics, a force is an effect that has the power to change an object's motion. A force can cause a massed item to accelerate or modify its velocity. A push or a pull is a straightforward method to explain force.

It is a scalar quantity with the Newton as its SI unit.

According to the question, the given values are :

Mass, m = 10.3 kg

Acceleration, a = 1.3 m/s²

Force = mass × Acceleration

Now, substitute the given values :

Force, F = 10.3 kg × 1.3 m/s²

F = 13.39 N.

Hence, the net force on the object is 13.39 N.

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Given

distance of the object  u = -40 cm.

radius of curvature of the refracting surface  R = -20 cm.

The height of the object  h = 1 cm.

To Find

The size of the image

Solution

We will now write the following data using the sign convention:

The object's distance is u = -40 cm.

The refracting surface has a radius of curvature of R = -20 cm.

The object's height is h = 1 cm.

We know that the lens maker formula is: n2vn1u=n2n1R.

v denotes the image formation distance.

The refractive index of the second medium is n2=1.33, and the refractive index of the first medium is n1=1.

In the preceding equation, we will now substitute the known values.

1.33v−1−40cm=1.33−1−20cmv=32.05cm

We know that the magnification ratio can be expressed as vu=Hh.

H denotes the height of the created image.

We'll now swap the given and obtained values.

⇒32.05cm divide by 40cm=H divide by 1cm

⇒H=0.6cm

As a result, the best choice is (B).

Further information: The rules of reflection apply to reflection from a concave mirror. The normal to the point of incidence is drawn along the radius of the mirror, that is, by connecting the point of incidence to the centre of curvature.

The development of an image in a concave mirror is mostly determined by the distance between the object and the mirror. The concave mirror produces both real and virtual images.  A virtual and magnified picture is generated when the object is placed very close to the mirror.

The complete question is -

Find the size of the image formed in the situation shown in figure.

photo

A) 0.5 cm

B) 0.6 cm

C) 1.2 cm

D) 1 cm

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Answer:

A

Explanation:

The sun is an emits energy in the form of light and heat.

The planets absorb this energy which helps keep them in motion. The closer they are to the sun, the more energy, the more speed and less the time it takes a planet to orbit the sun.

The gravitational pull of the Sun contributes motion of the planets because here the gravitational pull of the sun is balanced by the centrifugal force produced by the revolution of planets. The correct option Is A.

Centrifugal force is the pseudo-force that occurs only in the rotating frame of reference (non-inertial frame). it is directed away from the center. Its magnitude is,

F=\(\frac{mv^{2} }{r}\)

where m= mass of the body in kg.

v= velocity f the body in m/s.

r = radius in m.

The star sun has a  gravitational pull that is about 274 m/s²  which is greater than the other planets. Due to this pull the planets tend to pull toward the sun but it does not happen because the pull force is balanced by the centrifugal force that is produced by the revolution of the planets around the sun.

Hence The Sun's gravitational pull contributes to the motion of the planets.

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Answer:

speed = distance ÷ time.

speed=50÷ 5

=10m/s

A bike moves 50 meters in 5 seconds, then the speed of the bike would be 10 m/s.

The total distance covered by any object per unit of time is known as speed. It depends only on the magnitude of the moving object. The unit of speed is a meter/second. The generally considered unit for speed is a meter per second.

As given in the problem a bike moves 50 meters in 5 seconds. This means the total distance covered by the bike is 50 meters, and the total time taken by the bike is 5 seconds.

The total distance covered by the bike = 50 meters

The total time taken to cover the 5 meters = 5 seconds

speed = the total distance covered by the bike/the total time

           = 50 meters/5 seconds

           = 10 m/s

Thus, the speed of the bike comes out to be  10 m/s.

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Answer:

a = t over delta v.

Explanation:

Electric potential is the work done in bringing unit positive charge .

Here work(w) is given by

Again

Total change in electric potential energy (V) is given by

Final answer is -0.778J

(a) The work done by the donkey on the cart is 59,721.9 J.

(b) The work done by the force of gravity on the cart is -48,434.87 J.

(c) The work done on the cart by friction during this time is 11,315.12 J.

(a) The work done by the donkey on the cart is calculated as follows;

Wd = Fd cosθ

where;

Wd = 375 N x 163 m x cos(12.3)

Wd = 59,721.9 J

(b) The work done by the force of gravity on the cart is calculated as;

Wg =  Fg x d x cosθ

Where;

θ = 90⁰ + 4.03⁰ = 94.03⁰

Wg = (431 kg x 9.81 m/s²) x 163 m x cos (94.03)

Wg = -48,434.87 J

(c) The work done on the cart by friction during this time is calculated as;

Wf = Ff x d x cosθ

where;

Ff = μmg cosθ

Ff = 0.0165 x 431 kg x 9.81 x cos (4.03)

Ff = 69.59 N

Wf = 69.59 x 163 x cos (4.03)

Wf = 11,315.12 J

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Answer:

138.6 megacalculations

Explanation:

This is a pretty straightforward one.

All it needs is to convert the degree of measurement.

Dimensions in physics are attributed names, which state the power to which they're are raised. Just as how

Kilo and Mega means the numbers are raised to the power of 3 and 6 respectively. There also exists the ones that indicates how small, such as milli and micro, which are to the powers of -3 & -6.

The question says the IBM computer calculates at an astonishing 136.8 teracalculations.

Tera in physics means it's raised to the power of 12. Thus, the IBM calculates at an astonishing rate of

136.8*10^12 calculations per second.

We're then asked how many calculations it does in 1 micro second. Like I had highlighted earlier, 1 micro second is 1 raised to the power of -6. Or succinctly put,

1 micro second = 1*10^-6.

If the IBM does

138.6*10^12 = 1 second,

Then it does

x = 1*10^-6 second.

When we cross multiply, we have

138.6*10^12 * 1*10^-6, and that is

138.6*10^6 calculations, or say, 138.6 megacalculations.

The IBM does 138.6 megacalculations in 1 micro second, which is still astonishing, by the way

Pressure builds under the surface of Earth, causing mountains to form.The correct answer is option D.

Uplift refers to the geological process that elevates the Earth's surface, resulting in the formation of mountains. This process is primarily driven by tectonic forces, including the movement and collision of Earth's lithospheric plates.

When two plates converge, immense pressure builds up beneath the surface, causing the crust to buckle and fold. This deformation leads to the formation of mountains, as rocks are pushed upward and displaced vertically.

As the uplift process continues over millions of years, mountains gradually take shape. Erosion and weathering play significant roles in shaping their features, but it is the initial uplift that initiates the formation of mountains.

As the Earth's surface is elevated, a wide range of landforms can emerge, including rugged peaks, deep valleys, and steep slopes.

Uplift has a profound impact on the Earth's surface and ecosystems. Mountains alter local climates, influencing precipitation patterns and creating variations in temperature and wind patterns.

Therefore, uplift plays a crucial role in shaping the Earth's surface and influencing various geological, biological, and climatic processes.

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The spring constant of the spring is 178.4 N/m.

In this problem, we are given a mass of 0.25 kg that is attached to a spring and set into vibration with a period of 0.22 s.

We can use the equation for the period of a mass-spring system, which relates the period of oscillation to the mass and spring constant, to calculate the spring constant.

We can use the equation for the period of a mass-spring system:

T = 2π√(m/k)

where T is the period, m is the mass, and k is the spring constant.

Rearranging this equation to solve for k, we get:

k = \((4\pi^2m) / T^2\)

Substituting the given values, we get:

k = \((4\pi^2 \times 0.25 \:kg) / (0.22 s)^2\)

k = 178.4 N/m

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Answer:

I would say color.

Explanation:

Because for me most of the time the warm color mean It hot. and the cool color most likily to mean cold  

Answer:

Color

Explanation:

The temperature of a star is determined by the color. lol :)

the frictional force on the broom has magnitude μ * m * g * cos(theta)

The frictional force on the broom can be determined using the equation:

F_friction = μ * F_norm

Where F_friction is the frictional force, μ is the coefficient of friction, and F_norm is the normal force. The normal force is equal to the force exerted on the broom perpendicular to the surface of the floor. Since the broom is being pushed across a horizontal surface, the normal force is equal to the force of gravity, which is equal to the mass of the broom (m) multiplied by the acceleration due to gravity (g).

F_norm = m * g

The force being applied to the broom is 7N and it is at angle theta with the horizontal. The component of this force perpendicular to the surface of the floor is the force that acts on the normal force.

F_norm = 7 * cos(theta)

Therefore,

F_friction = μ * m * g * cos(theta)

So, the frictional force on the broom has magnitude μ * m * g * cos(theta)

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The compressive stress in the steel column is found to be approximately 397.6 MPa.

The formula for calculating the area of a circle can be used to determine the steel column's cross-sectional area (A),

A = π*(d/2)², diameter of the column is d,

A = π*(0.4/2)²

A = 0.1257m²

The compressive stress (σ) in the column can be calculated using the formula, σ = F/A, F is the load carried by the column is F.

σ = 50 MN/0.1257m²

σ = 397.6 MPa

Therefore, the compressive stress in the steel column is approximately 397.6 MPa.

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Gender inequality in mechanical engineering can manifest through biased hiring practices, limited opportunities for women to advance to leadership roles, and a lack of representation in the industry.

Despite efforts to promote gender equality, women are still underrepresented in the field of mechanical engineering. This can be due to a variety of factors, including biased hiring practices, limited opportunities for career advancement, and a lack of representation in the industry.

To address these issues, it is important for companies and organizations to promote diversity and inclusion initiatives, such as actively recruiting women and people of diverse backgrounds, providing mentorship and networking opportunities, and advocating for policies that support work-life balance and equal pay. By creating a more inclusive environment, the field of mechanical engineering can attract and retain more talented individuals and foster innovation and growth.

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--The complete question is, How does gender inequality manifest in the field of mechanical engineering, and what are some potential solutions to address this issue?--