A force of 140 140 newtons is required to hold a spring that has been stretched from its natural length of 40 cm to a length of 60 cm. Find the work done in stretching the spring from 60 cm to 65 cm. First, setup an integral and find a a, b b, and f ( x ) f(x) which would compute the amount of work done.

Answer: the earth

Explanation: Earth exerts a gravitational pull on the moon 80 times stronger than the moon's pull on the Earth. Over a very long time, the moon's rotations created fiction with the Earth's tugging back, until the moon's orbit and rotational locked with Earth.

and that's why the earth pulls the moon

Final answer:

The Earth pulling on the moon and the moon pulling on the Earth exert the same amount of force on each other due to Newton's third law of motion.

Explanation:

In terms of force, the Earth pulling on the Moon and the Moon pulling on the Earth exert the same amount of force on each other. This is because of Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. So, while the Earth's gravitational force pulls the Moon towards it, the Moon's gravitational force also pulls the Earth towards it with an equal amount of force.

Newton's third law of motion states that for every action, there is an equal and opposite reaction. In the context of the gravitational interaction between the Earth and the Moon, the forces they exert on each other are equal in magnitude and opposite in direction.

The Earth pulls on the Moon with a gravitational force, and, according to Newton's third law, the Moon simultaneously pulls on the Earth with an equal gravitational force. These forces are sometimes referred to as "action and reaction pairs." The force that the Earth exerts on the Moon is often called the gravitational attraction of the Earth on the Moon, and vice versa.

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Answer:

The net downward force on the tank is

Explanation:

Given that,

Area = 1.60 m²

Suppose the design of a cylindrical, pressurized water tank for a future colony on Mars, where the acceleration due to gravity is 3.71 meters per second per second. The pressure at the surface of the water will be 150 K Pa , and the depth of the water will be 14.4 m . The pressure of the air in the building outside the tank will be 88.0 K Pa.

We need to calculate the net downward force on the tank

Using formula of formula

Where, P = pressure

g = gravity at mars

h = height

A = area

Put the value into the formula

Hence, The net downward force on the tank is

Final answer:

The net downward force on the tank's flat bottom can be found by calculating the pressure at the bottom of the container.

Explanation:

Since the density is constant, the weight can be calculated using the density:

w = mg = pVg = pAhg.

The pressure at the bottom of the container is therefore equal to atmospheric pressure added to the weight of the fluid divided by the area.

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Image is missing, so i have attached it

Answer:

19.04 × 10⁻⁴ T in the +x direction

Explanation:

We are told that the point P which is equidistant from the wires. (R = 5.00 cm). Thus distance from each wire to O is R.

Hence, the magnetic field at P from each wire would be; B = μ₀I/(2πR)

We are given;

I = 2.4 A

R = 5 cm = 0.05 m

μ₀ is a constant = 4π × 10⁻⁷ H/m

B = (4π × 10⁻⁷ × 2.4)/(2π × 0.05)

B = 9.6 × 10⁻⁴ T

To get the direction of the field from each wire, we will use Flemings right hand rule.

From the diagram attached:

We can say the field at P from the top wire will point up/right

Also, the field at P from the bottom wire will point down/right

Thus, by symmetry, the y components will cancel out leaving the two equal x components to act to the right.

If the mid-point between the wires is M, the the angle this mid point line to P makes with either A or B should be same since P is equidistant from both wires.

Let the angle be θ

Thus;

sin(θ) = (1.3/2)/5

θ = sin⁻¹(0.13) = 7.47⁰

The x component of each field would be:

9.6 × 10⁻⁴cos(7.47) = 9.52 × 10⁻⁴ T

Thus, total field = 2 × 9.52 × 10⁻⁴ = 19.04 × 10⁻⁴ T in the +x direction

Final answer:

The magnetic field at point P, which is equidistant from two long parallel wires with equal anti-parallel currents, is calculated using Ampere's law. The net magnetic field is zero because the fields due to each wire cancel each other at that point.

Explanation:

The question concerns the calculation of the magnetic field at a point equidistant from two long parallel wires that carry equal anti-parallel currents. According to the right-hand rule and Ampere's law, each wire generates a magnetic field that circles the wire. For two wires carrying currents in opposite directions, the magnetic fields at the midpoint between the wires will point in opposite directions, thus they will subtract from each other when calculating the total magnetic field at point P.

To find the magnetic field at point P, we use the formula for the magnetic field due to a long straight current-carrying wire: B = (μ₀I)/(2πd), where B is the magnetic field, μ₀ is the permeability of free space (4π x 10-7 T·m/A), I is the current, and d is the distance to the point of interest from the wire. In this case, the distance d will be the radius R = 5.00 cm since point P is equidistant from both wires.

Substituting the values into the formula, the magnetic field due to each wire at point P can be calculated. However, since the currents are anti-parallel, the net magnetic field at P would be the difference between the two fields, which is zero.

Complete question:

An electron is a subatomic particle (m = 9.11 x 10-31 kg) that is subject to electric forces. An electron moving in the +x direction accelerates from an initial velocity of +6.18 x 105 m/s to a final velocity of 2.59 x 106 m/s while traveling a distance of 0.0708 m. The electron's acceleration is due to two electric forces parallel to the x axis: F₁ = 8.87 x 10-17 N, and , which points in the -x direction. Find the magnitudes of (a) the net force acting on the electron and (b) the electric force F₂.

Answer:

(a) The net force of the electron, ∑F = 4.07 x 10⁻¹⁷ N

(b) the electric force, F₂ = 4.8 x 10⁻¹⁷ N

Explanation:

Given;

initial velocity of the electron, = +6.18 x 10⁵ m/s

final velocity of the electron, = 2.59 x 10⁶ m/s

the distance traveled by the electron, d = 0.0708 m

The first electric force,

(a) The net force of the electron is given as;

∑F = F₁ - F₂ = ma

where;

a is the acceleration of the electron

∑F = ma = (9.11 x 10⁻³¹ kg)(4.468 x 10¹³)

∑F = 4.07 x 10⁻¹⁷ N

(b) the electric force, F₂ is given as;

∑F = F₁ - F₂

F₂ = F₁ - ∑F

F₂ = 8.87 x 10⁻¹⁷ - 4.07 x 10⁻¹⁷

F₂ = 4.8 x 10⁻¹⁷ N

Final answer:

The problem involves calculating the acceleration of an electron, then using Newton's second law to find the net force on the electron. This is used to find the magnitude of a second electric force acting on the electron.

Explanation:

First, we can calculate the acceleration of the electron using the formula a = Δv/Δt, where 'a' is acceleration, 'Δv' is the change in velocity, and 'Δt' is the change in time. In this case, Δv = vf - vi = 2.59 x 106 m/s - 6.18 x 105 m/s = 1.972 x 106 m/s. The time taken by the electron to travel 0.0708 m can be found using the equation d = vi t + 0.5 a t₂. We use these values to get Δt which we use to find 'a'.

Next, let's use Newton's second law F = ma to find the net force acting on the electron. The only forces acting on the electron are electric forces, and we know one them is 8.87 x 10-17 N. If we designate this known force as F₁ then the total force F total = F₁ + F₂ where F₂ is the unknown electric force.

Finally, we can find F₂ = F total - F₁. This gives the magnitude of the second electric force.

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The anchoring force needed to hold the elbow in place is 839.5 N.

How to find anchoring force?

To determine the anchoring force needed to hold the elbow in place, use the following steps:

Apply the conservation of momentum equation to the elbow:

∑F = ˙m(V₂ - V₁)

where:

˙m = mass flow rate

V₁ and V₂ = velocities at the inlet and outlet of the elbow, respectively

F = anchoring force

The mass flow rate is given by:

˙m = ρAV

where:

ρ = density of water (1000 kg/m³)

A = cross-sectional area of the elbow

V = velocity

The velocities at the inlet and outlet of the elbow can be calculated using the following equations:

V₁ = A₁V₁

V₂ = A₂V₂

where:

A₁ and A₂ = cross-sectional areas at the inlet and outlet of the elbow, respectively

Calculate the momentum flux correction factor at the inlet and outlet of the elbow:

β₁ = 1.03

β₂ = 1.03

Substitute all of the above equations into the conservation of momentum equation:

F = ˙m(V₂ - V₁)

F = ρA₁V₁²β₁ - ρA₂V₂²β₂

Calculate the velocity at the inlet of the elbow:

V₁ = A₂V₂/A₁

V₁ = (25 cm²/150 cm²)(V₂)

V₁ = 1/6 V₂

Substitute the above equation into the conservation of momentum equation:

F = ρA₁V₁²β₁ - ρA₂V₂²β₂

F = ρA₁[(1/6 V₂)²](1.03) - ρA₂V₂²(1.03)

F = ρV₂²(1.03)(1/36 A₁ - A₂)

Calculate the anchoring force:

F = (1000 kg/m³)(V₂²)(1.03)(1/36 × 150 cm² - 25 cm²)

F = 839.5 N

Therefore, the anchoring force needed to hold the elbow in place is 839.5 N.

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The anchoring force needed to hold the elbow in place is 932 N.

The anchoring force needed to hold the elbow in place is the sum of the following forces:

The force due to the change in momentum of the water as it flows through the elbow.

The force due to the weight of the elbow and the water in it.

The force due to the buoyancy of the water in the elbow.

The force due to the change in momentum of the water can be calculated using the momentum equation:

F = mΔv

where:

F is the force

m is the mass of the fluid

Δv is the change in velocity of the fluid

In this case, the mass of the fluid is the mass of the water that flows through the elbow per second. This can be calculated using the mass flow rate equation:

m = ρAv

where:

ρ is the density of the fluid

A is the cross-sectional area of the pipe

v is the velocity of the fluid

The velocity of the fluid at the inlet can be calculated using the Bernoulli equation:

P1 + 0.5ρv1^2 = P2 + 0.5ρv2^2

where:

P1 is the pressure at the inlet

v1 is the velocity at the inlet

P2 is the pressure at the outlet

v2 is the velocity at the outlet

In this case, the pressure at the outlet is atmospheric pressure. The velocity at the outlet can be calculated using the continuity equation:

A1v1 = A2v2

where:

A1 is the cross-sectional area at the inlet

A2 is the cross-sectional area at the outlet

The force due to the weight of the elbow and the water in it is simply the weight of the elbow and the water in it. The weight can be calculated using the following equation:

W = mg

where:

W is the weight

m is the mass

g is the acceleration due to gravity

The force due to the buoyancy of the water in the elbow is equal to the weight of the water displaced by the elbow. The weight of the water displaced by the elbow can be calculated using the following equation:

B = ρVg

where:

B is the buoyancy

ρ is the density of the fluid

V is the volume of the fluid displaced

g is the acceleration due to gravity

The volume of the fluid displaced by the elbow is equal to the volume of the elbow.

Now that we have all of the forces, we can calculate the anchoring force needed to hold the elbow in place. The anchoring force is equal to the sum of the forces in the negative x-direction. The negative x-direction is the direction in which the water is flowing.

F_anchor = F_momentum + F_weight - F_buoyancy

where:

F_anchor is the anchoring force

F_momentum is the force due to the change in momentum of the water

F_weight is the force due to the weight of the elbow and the water in it

F_buoyancy is the force due to the buoyancy of the water in the elbow

Plugging in the values for each force, we get:

F_anchor = 1030 N - 490 N + 392 N = 932 N

Therefore, the anchoring force needed to hold the elbow in place is 932 N.

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