A cylindrical insulated wire of diameter 5.0 mm is tightly wound 200 times around a cylindrical core to form a solenoid with adjacent coils touching each other. When a 0.10 A current is sent through the wire, what is the magnitude of the magnetic field on the axis of the solenoid near its center

Answer:

43 m/s

Explanation:

Mass, m = 5 kg

Force, F(t) = 6t² - 4t + 3

To find the speed, we first need to get the acceleration, a.

Force is the product of mass and acceleration. It is given as:

F = ma

Therefore, acceleration is:

a(t) = F(t)/m

a(t) = (6t² - 4t + 3) / 5

a(t) = 1.2t² - 0.8t + 0.6

Acceleration is the differentiation of velocity with respect to time, t. Therefore, to get velocity, v, we integrate a(t):

a(t) = dv(t) / dt

=> v(t) = (1.2/3)t³ - (0.8/2)t² + 0.6t

v(t) = 0.4t³ - 0.4t² + 0.6t

Therefore, at time t = 5secs, velocity is:

v(5) = 0.4 * (5³) - 0.4 * (5²) + 0.6 * 5

v(5) = 50 - 10 + 3

v(5) = 43 m/s

The velocity at time, t = 5 secs is 43 m/s

Answer:

a. 60 V b. 6 A c. 360 W

Explanation:

a. Voltage in secondary

For an ideal transformer,

N₁/N₂ = I₂/I₁ = V₁/V₂ where N₁ = turns in primary = 50 turns, N₂ = turns in secondary = 250 turns, V₁ = voltage in primary = 12 V, V₂ = voltage in secondary = ? V

N₁/N₂ = V₁/V₂

V₂ = V₁N₂/N₁ = 250 × 12/50 = 60 V

b. Since V₂ = I₂R,

I₂ = V₂/R  R = 10 Ω

I₂ = 60 V/10 Ω

I₂ = 6 A

c. We first calculate the current in the primary from

N₁/N₂ = I₂/I₁  where I₁ = primary current

I₁ = N₂I₂/N₁ = 250 × 6 A/50 = 30 A

The power supplied to the primary is thus

P = I₁V₁ = 30 A × 12 V = 360 W

(a) The voltage supply at the secondary coil is of 60 V.

(b) The current flow through the 10- ohm device is 6 A.

(c)  The power supply to the primary coil is 360 W.

Given Data:

Number of turns of primary coil is, n1 = 50.

Number of turns of secondary coil is, n2 = 250.

The voltage supply to primary coil is, V1 = 12 V.

The resistance of device is, R = 10 ohm.

(a)

The first part of the given problem is based on the Transformer equation relating the number of turns of coils at primary and secondary and voltage supply at each coils.

Therefore,

n1 / n2 = V1 / V2

Here,

V2 is the voltage supply to the secondary coil.

Solving as,

50 / 250 = 12 / V2

V2 = 250 / 50 × 12

V2 = 60 V.

Thus, we can conclude that the voltage supply at the secondary coil is of 60 V.

(b)

Now in order to find the current flow through the 10 - ohm device, we can use the Ohm's law as,

V2 = I' × R

Solving as,

60 = I' × 10

I ' = 60 / 10

I' = 6 A

Thus, we can conclude that the current flow through the 10- ohm device is 6 A.

(c)

Now, we first calculate the current in the primary from

n1 / n2 = I' / I  

here,

I is the primary current

Solving as,

I = n2 × I' / n1

I = 250 × 6 /50

I = 30 A

The power supplied to the primary is thus

P1 = I × V1

P1 = 30  × 12  

P1 = 360 W

Thus, we can conclude that the power supply to the primary coil is 360 W.

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Final answer:

Newton's law of universal gravitation states that all objects in the universe attract each other with a force proportional to the product of their masses and inversely proportional to the square of the distance between them.

Explanation:

Newton's Law of Universal Gravitation:

Sir Isaac Newton's law of universal gravitation states that every object in the universe attracts every other object with a force. This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The gravitational force is universally attractive and only depends on mass and distance, following the equation F = G × (m1 × m2) / r², where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the objects, and r is the distance between the centers of the two masses.

Answer:

Hot air is less dense than cool air; the heated air causes the balloon to rise simply because it is lighter than an equal volume of cold air.So hot air balloon rises up in air.

Large ships float in water as their density is lesser than the water so.

Answer:9.5 seconds(approx.)

Explanation:time=distance/speed

here,

distance=3.3 km= 3300 m

standard speed of sound=344 m/s

so, time=3300/344

time=9.5(approx.)

Final answer:

It would take approximately 9.71 seconds for the sound of a shot to travel 3.3 km and reach you, using the speed of sound of 340 m/s.

Explanation:

To calculate how long it would take for you to hear the sound of a shot fired from 3.3 km away, we need to use the speed of sound. In general, the speed of sound is approximately 340 meters per second (m/s) at sea level under normal conditions. Therefore, we can use the formula:

Time = Distance / Speed of Sound.

Now we simply plugin the values to get:

Time = 3300 meters / 340 m/s = 9.71 seconds

Therefore, it would take approximately 9.71 seconds for the sound of the shot to travel 3.3 km and reach you.

Answer:

a) the oscillation of this field is in phase, when the magnetic field goes in the negative direction of y, the elective field goes in the positive direction of the z axis

b) the direction of the magnetic field perpendicular to this electric field and the speed in the negative x the magnetic field goes in the x direction and in the direction (1, - 1.1)

Explanation:

a) the polarization the determined wave oscillates the electric field, which is the z axis

 As the wave travels on the negative x-axis and the magnetic field is perpendicular, this field goes on the positive y-axis

the oscillation of this field is in phase, when the magnetic field goes in the negative direction of y, the elective field goes in the positive direction of the z axis

be) in the case of a polarization in the xi plane the magnetic field must go in the direction of the magnetic field perpendicular to this electric field and the speed in the negative x the magnetic field goes in the x direction and in the direction (1, - 1.1)

Answer:

469.6KJ

Explanation:

Heat energy required can be calculated using the formula

H = mc∆t where

m is the mass of the water

c is the specific heat capacity of the water

∆t is the change in temperature of the water

Given m = 9.78kg

c = 4186j/kg-c.

∆t = 52.07°C - 40.82°C

∆t = 11.25°C

H = 9.78 × 4186 × 11.25

H = 460,564.65Joules

= 460.6KJ

Answer:

the amount of energy required to raise the temperature of the water is 460564.65 J

Explanation:

The energy required to raise the temperature of water can be calculated as follows;

Q = mcΔθ

where;

Q is the quantity of heat or energy required to raise the temperature of water

m is mass of water

c is specific heat capacity of water

Δθ is change in temperature = T₂ - T₁

Given;

m =  9.78kg

c = 4186j/kg-c

Δθ = T₂ - T₁ = 52.07°C - 40.82°C  = 11.25°C

Q = mcΔθ

Q = (9.78)(4186)(11.25)

Q = 460564.65 J

Therefore, the amount of energy required to raise the temperature of the water is 460564.65 J

Answer:

For steam, heat released E = 15.26KJ

For water, heat released E2 = 6.91KJ

Explanation:

Given;

Mass(steam) ms = 25g

Mass (water) mw = 25g

Change in temperature of both steam and water ∆T = 100-34= 66°C

Specific heat of water C = 4.186 J/g.°C

Specific Latent heat L = 334J/g

For steam;

Heat released E = msL + msC∆T

E = (25×334) + (25×4.186×66)

E = 15256.9J

E = 15.26KJ

For water;

Heat released E2 = mwC∆T

E2 = 25×4.186×66

E2 = 6906.9J

E2 = 6.91KJ

Answer:

1.) A

2.) B

3.) B

4.) A

5.) A

6.) C

7.) B

Explanation:

Answer:

A directional selection

B disruptive selection

B disruptive selection

A directional selection

A directional selection

C stabilizing selection

B disruptive selection

Explanation:

Answer:

Electrical charge

Explanation:

Electrical Charge is a form of charge, designated negative, positive, or neutral (without charge) that is found on the subatomic particles that make up all atoms

Answer:

Explanation:

Given that,

A vector A has x component to be 2.7cm and y component to be 2.25cm

Then,

A = 2.7•i + 2.25•j

A vector B has x component of 0.30cm and y component of 1.75cm

B = 0.3•i + 1.75•j

So, we want to find A+B

Addition of vectors

Generally

(a•i + b•j) + (c•i + d•j) = (a+c)•i +(b+d)•j

Vectors are added component wise.

So,

A + B = (2.7•i + 2.25•j) + (0.3•i + 1.75•j)

A + B = (2.7 + 0.3)•i + (2.25 + 1.75)•j

A + B = 3•i + 4•j

We can also find it magnitude and direction

Generally,

A = a•i + b•j

|A| = √(a²+b²)

<A = arctan(b/a)

So,

|A+B| = √(3²+4²) = √9+16 = √25

|A+B| = 5

And it's direction

< = arctan(y/x)

< = arctan(4/3)

< = 53.13°

Answer:

When the sound wave enters the water its wavelength increase.

Explanation:

Given:

Speed of sound in air [tex]v = 343 \frac{m}{s}[/tex]

Speed of sound in fresh water [tex]v' = 1482 \frac{m}{s}[/tex]

We know that when wave travel into one medium to another medium frequency doesn't change.

Velocity is given by,

  [tex]v = f \lambda[/tex]

In above equation frequency is constant,

So when sound wave enters the water wavelength increase because speed sound is increase in water as compare to air.

Therefore, when the sound wave enters the water its wavelength increase.

Answer: New moon

Explanation: A solar eclipse is only possible during a new moon phase

Answer:

The second kinetic energy is 162 J.

Explanation:

Given that,

Mass, [tex]m_1=15\ g[/tex]

Velocity, [tex]v_1=7\ cm/s[/tex]

Kinetic energy, [tex]K_1=147\ ergs[/tex]

Mass, [tex]m_2=10\ g[/tex]

Velocity, [tex]v_2=9\ cm/s[/tex]

We need to find kinetic energy [tex]K_2[/tex]. Kinetic energy is given by :

[tex]K=\dfrac{1}{2}mv^2[/tex]

So,

[tex]\dfrac{K_1}{K_2}=\dfrac{m_1}{m_2}\times \dfrac{v_1^2}{v_2^2}\\\\K_2=\dfrac{K_1}{\dfrac{m_1}{m_2}\times \dfrac{v_1^2}{v_2^2}}\\\\K_2=\dfrac{147}{\dfrac{15}{10}\times \dfrac{7^2}{9^2}}\\\\K_2=162\ J[/tex]

So, the second kinetic energy is 162 J.

The work done on the gas during an adiabatic compression process is calculated by integrating the pressure as a function of volume from the initial volume to the final volume. In this case, the integral ∫from V(i) to V(f) p₀ V^(-6/5) dV provides the value for the work done.

In an adiabatic process, the work done W on the gas when it's compressed from a volume V(i) to a volume V(f) can be calculated by integrating the pressure-volume equation over the given volume range.

Given the pressure equation p = p₀ V^(-6/5), the work done on the gas (W) can be computed by the integral formula for work: W = ∫p dV = ∫from Vi to Vf p₀ V^(-6/5) dV. Evaluating this integral gives the work done on the gas during the compression process in the adiabatic process.

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Answer:

Complete question

Luis is trying to push a box of new soccer balls across the floor.

If the box is not moving, which of the following must be true?

A. The box is exerting a larger force on Luis than he is exerting on the box

B. There is another force acting on the box that balances Luis's force.

C. Luis is applying a force that acts at a distance.

D. There is no force of friction acting on the box.

Explanation:

Using newton second law of motion

ΣF = ma

Now, for a body not to move when a force is acting on it means that the body is in equilibrium and acceleration a = 0

Therefore,

Fnet = 0

So, if he is applying a force F to push the box and the box is not moving then, there is an external force that is pushing the force back opposite the direction he his pushing and this force counterbalance is own force.

F—F' = 0

F' = F

So, F' is the counter balance force and it is equal to the force applied by Luis

Or it might be frictional force, because if the static friction is not overcome, then, the body will not leave it's state of rest. So if the fictional force is very high, then the box will not leave it rest position and we also know that frictional force opposes motion,

F—Fr=0

F = Fr

So using this explanation,.

The answer is B

B. There is another force acting on the box that balances Luis's force.

Answer: The correct and is

D. Extrusion

Explanation:

What is extrusion?

Extrusion is a manufacturing process used to make basically pipes and hoses.

The pvc granules are melt into a liquid which is forced through a die, forming a long 'tube like' shape. The shape of the die determines the shape of the tube.

The extrusion is then cooled and forms a solid shape

Extrusion moulding is used to create products with a consistent cross-section.

Answer:

Explanation:

Given that,

Force is downward I.e negative y-axis

F = -2 × 10^-14 •j N

Magnetic field is westward, +x direction

B = 8.3 × 10^-2 •i T

Charge of an electron

q = 1.6 × 10^-19C

Velocity and it direction?

Force in a magnetic field is given as

F = q(V×B)

Angle between V and B is 270, check attachment

The cross product of velocity and magnetic field

F =qVB•Sin270

2 × 10^-14 = 1.6 × 10^-19 × V × 8.3 × 10^-2

Then,

v = 2 × 10^-14 / (1.6 × 10^-19 × 8.3 × 10^-2)

v = 1.51 × 10^6 m/s

Direction of the force

Let x be the direction of v

-F•j = v•x × B•i

From cross product

We know that

i×j = k, j×i = -k

j×k =i, k×j = -i

k×i = j, i×k = -j OR -k×i = -j

Comparing -k×i = -j to given problem

We notice that

-F•j = q ( -V•k × B×i)

So, the direction of V is negative z- direction

V = -1.51 × 10^6 •k m/s

Answer:

Shear modulus is equal to [tex]9.82\times 10^{10}N/m^2[/tex]

Explanation:

We have given area [tex]A=0.030m^2[/tex]

Force is given [tex]F_1=F_2=30\times 10^6N[/tex]

Height of the block d = 0.11 m

Change in height of the block [tex]\Delta d=1.12\times 10^{-3}m[/tex]

Stress is given by

[tex]stress=\frac{force}{area}[/tex]

[tex]stress=\frac{30\times 10^6}{0.030}=10^9N/m^2[/tex]

Strain is equal to

[tex]strain=\frac{\Delta d}{d}[/tex]

[tex]strain=\frac{1.12\times 10^{-3}}{0.11}=10.18\times 10^{-3}[/tex]

Shear modulus is equal to

Shear modulus [tex]=\frac{stress}{strain}[/tex]

[tex]=\frac{10^9}{10.18\times 10^{-3}}=9.82\times 10^{10}N/m^2[/tex]

Answer:

Change in momentum of the stone is 3.673 kg.m/s.

Explanation:

Given:

Mass of the ball on the horizontal the surface, m = 0.10 kg

Velocity of the ball with which it hits the stone, v = 20 m/s

According to the question it rebounds with 70% of the initial kinetic energy.

We have to find the change in momentum i.e Δp

Before that:

We have to calculate the rebound velocity with which the object rebounds.

Lets say that the rebound velocity be "v1" and KE remaining after the object rebounds be "KE1".

⇒ [tex]KE_1=0.7\times \frac{mv^2}{2}[/tex]    

⇒ [tex]KE_1=0.7\times \frac{0.10\times (20)^2}{2}[/tex]

⇒ [tex]KE_1=0.7\times \frac{0.10\times 400}{2}[/tex]

⇒ [tex]KE_1=14[/tex] Joules (J).

Rebound velocity "v1".

⇒ [tex]KE_1=\frac{m(v_1)^2}{2}[/tex]

⇒ [tex]v_1 = \sqrt{\frac{2KE_1}{m} }[/tex]

⇒ [tex]v_1 = \sqrt{\frac{2\times 14}{0.10} }[/tex]

⇒ [tex]v_1=16.73[/tex]

⇒ [tex]v_1=-16.73[/tex] m/s ...as it rebounds.

Change in momentum Δp.

⇒ [tex]\triangle p= m\triangle v[/tex]

⇒ [tex]\triangle p= 0.10\times (20-(-16.73)[/tex]

⇒ [tex]\triangle p= 0.10\times (20+16.73)[/tex]

⇒ [tex]\triangle p= 0.10\times (36.73)[/tex]

⇒ [tex]\triangle p = 3.673[/tex] Kg.m/s

The magnitude of the change in momentum of the stone is 3.673 kg.m/s.

Answer:

Speed at which the bat have to fly is 90.895 m/s away from the human (listener) in positive direction.

Explanation:

Given:

Frequency of the bat that is source here, [tex]f_S[/tex] = 25.3 kHz = 25.3 * 10^3 Hz

Frequency of the listener (human), [tex]f_L[/tex] = 20 kHz = 20*10^3 Hz

We have to identify how fast the bat have to fly in order for a person to hear these chirps .

Let the velocity of bat that is source is "Vs" and "Vs" = "Vbat".

Doppler effects formulae :

Using the above formula and considering that the bat is moving away so that the human can listen the chirps also [tex]V_L=0[/tex] as listener is stationary.

⇒ [tex]f_L=(\frac{V+V_L}{V+V_S}) f_S[/tex]   ⇒ [tex]f_L=(\frac{V+0}{V+V_S}) f_S[/tex]

Re-arranging in terms of Vs.

⇒ [tex]V+V_S =\frac{V\times f_S}{f_L}[/tex]

⇒ [tex]V_S =\frac{V\times f_S}{f_L}-V[/tex]

⇒ [tex]V_S =\frac{343\times 25.3\times 10^3}{20\times 10^3}-343[/tex]

⇒ [tex]V_S=90.895[/tex] m/s

The speed at which the bat have to fly is 90.895 m/s away from the human (listener) in positive direction.

Answer:

a.auditory ossicles

b.oval window

c.Round window

d.tympanic membrane

Answer is tympanic membrane

Explanation:

The tympanic membrane otherwise called the ear drum is a membrane shaped like a cone,it connects the outside to the inner ear,it serves to convert vibration from air into fluid membrane vibration a good example of mechanical waves for onwatd transmission into the cochlea of the inner ear through the oval window

Sound waves are converted into mechanical movements by the ear, specifically through structures in the middle and inner ear. The eardrum, ossicles, and cochlea play significant roles in this process, with the final conversion to electrical signals occurring in the cochlea.

Sound waves are converted into mechanical movements by the ear, specifically the structures within the middle and inner ear.

When sound waves enter the ear, they strike the eardrum or tympanic membrane in the middle ear, causing it to vibrate. These vibrations are then transferred to three tiny bones in the middle ear called the ossicles, consisting of the malleus, incus, and stapes. The vibrations move these bones, with the stapes pushing into the oval window of the cochlea in the inner ear.

The cochlea is filled with fluid and lined with tiny hair-like structures called stereocilia. The mechanical movement from the ossicles creates waves in this fluid, which cause the stereocilia to move. This movement is converted into electrical signals that the brain interprets as sound.

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An increase in the level of nitrogen dioxide and a decrease in the level of ground-level ozone occurs.

Explanation:

Ozone gas is normally found in stratosphere, it protects us from solar radiation. Under certain circumstances, it can be formed on the ground level.

Conditions that have to be met in order for this to happen are existing of nitrogen oxides and volatile organic compounds and their reaction catalyzed by heat and light.

During dark, cold and cloudy days, due to the lack of heat and light from the Sun, formation of ozone will be decreased.

Also, this will enable building up of nitrogen dioxide, due to the same reason, leading to increased concentration.

In a redox (reduction-oxidation) reaction, the molecule that gains an electron is reduced. Reduction and oxidation always occur together in such reactions. Reduced molecules often act as energy carriers in metabolic pathways.

During a redox (reduction-oxidation) reaction, the molecule that gains an electron is referred to as being reduced. This process is accompanied by an energy transfer. When a molecule loses an electron (a process termed as oxidation), energy is released; this energy and the electron are then transferred to another molecule, resulting in reduction of that molecule. These two processes, oxidation and reduction, always occur together in a redox reaction.

For example, considering a metabolic pathway involving Nicotinamide adenine dinucleotide (NAD+), when it accepts a hydride ion (H-) from a hydrogen atom, it gets reduced to form NADH. Here, NAD+ is the molecule that gets reduced in this redox reaction.

In summary, in redox reactions, the molecule that gains an electron and is reduced can act as an energy carrier, an important function in metabolism, facilitating the controlled transfer of energy within the cell. The concepts of oxidation and reduction are integral to understanding energy extraction and utilization in cells.

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Answer:

The phase constant

Explanation:

The phase constant tells how much a signal is shifted along the x-axis. A phase constant of ϕ means that each value of the signal happens ϕ amount of time earlier. If the signal has a beginning, then a phase constant of ϕ means the signal occurs that much sooner.

The starting conditions of an oscillator are characterized by the phase constant. Option B is correct.

Phase Constant:

This represents the number of oscillation per cycle of wave. It is denoted by [tex]\bold {\phi}[/tex]. It is also known as Propagation constant. The phase constant can be calculated using the formula.

[tex]\bold {\phi = \dfrac {2\pi }{\lambda}}[/tex]

Where, lambda is wavelength.

Therefore, option B is correct. The starting conditions of an oscillator are characterized by the phase constant.

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Answer:

118 N

Explanation:

Given mass of the block, m = 4.00kg.

The acceleration of the elevator, a = 3.0 m/s^2.

As elevotar attaced with spring scale and accelerating upward

(block and elevator), so total force

[tex]F_N-mg=ma[/tex]

Here, mg is the weight of the block downward direction.

or

[tex]F_N=ma+mg=m(g+a)[/tex]

substitute the given value, we get

[tex]F_N=4kg(9.8m/s^2+3m/s^2)[/tex]

     = 117.6 N = 118 N.

Thus, the reading on the spring scale to 3 significant figures is 118 N.

Answer:

Explanation:

The combined wave only end up been more powerful than the Longitudinal wave. This means, the transverse wave is more powerful than the combined wave. In transverse wave, the oscillation is perpendicular to the direction of the wave, while in longitudinal wave, the motion of the movement of the object is parallel to the movement of the wave. And in combined wave, the movement of the medium is in a circular manner,

The magnitude of the tension T in the cable is approximately 468.164 N.

To find the tension T in the cable supporting the beam, we can analyze the forces acting on the beam in equilibrium.

First, let's consider the forces acting on the beam:

1. The weight of the beam, acting downward at its center (2.0 m from the wall).

2. The tension T in the cable, acting upward and at an angle of 53 degrees with the horizontal.

3. The reaction force at the pin connection, acting horizontally to the left.

4. The vertical force exerted by the person standing on the beam, which contributes to the vertical component of the tension in the cable.

The beam is in equilibrium, so the sum of the torques about any point must be zero. Let's take moments about the pin connection at the wall.

Clockwise torques:

- Weight of the beam: [tex]\(200 \, \text{N} \times 2.0 \, \text{m}\)[/tex]

- Vertical component of the tension: [tex]\(T \times \cos(53^\circ) \times 1.5 \, \text{m}\)[/tex]

Counterclockwise torques:

- Tension in the cable: [tex]\(T \times \sin(53^\circ) \times 1.5 \, \text{m}\)[/tex]

- Vertical force exerted by the person: [tex]\(350 \, \text{N} \times 1.5 \, \text{m}\)[/tex]

Since the beam is in equilibrium, these torques must balance:

[tex]\[ 200 \, \text{N} \times 2.0 \, \text{m} + T \times \cos(53^\circ) \times 1.5 \, \text{m} = T \times \sin(53^\circ) \times 1.5 \, \text{m} + 350 \, \text{N} \times 1.5 \, \text{m} \][/tex]

Now, we can solve for T:

[tex]\[ 400 \, \text{N} + T \times \cos(53^\circ) \times 1.5 \, \text{m} = T \times \sin(53^\circ) \times 1.5 \, \text{m} + 525 \, \text{N} \][/tex]

[tex]\[ 400 \, \text{N} = T \times \sin(53^\circ) \times 1.5 \, \text{m} - T \times \cos(53^\circ) \times 1.5 \, \text{m} + 525 \, \text{N} \][/tex]

[tex]\[ 400 \, \text{N} = T \times (\sin(53^\circ) - \cos(53^\circ)) \times 1.5 \, \text{m} + 525 \, \text{N} \][/tex]

[tex]\[ T \times (\sin(53^\circ) - \cos(53^\circ)) \times 1.5 \, \text{m} = 125 \, \text{N} \][/tex]

[tex]\[ T = \frac{125 \, \text{N}}{(\sin(53^\circ) - \cos(53^\circ)) \times 1.5 \, \text{m}} \][/tex]

Now, we can calculate T:

[tex]\[ T = \frac{125 \, \text{N}}{(\sin(53^\circ) - \cos(53^\circ)) \times 1.5 \, \text{m}} \][/tex]

[tex]\[ T = \frac{125 \, \text{N}}{(\sin(53^\circ) - \cos(53^\circ)) \times 1.5 \, \text{m}} \][/tex]

[tex]\[ T \approx \frac{125 \, \text{N}}{(0.7986 - 0.6206) \times 1.5 \, \text{m}} \][/tex]

[tex]\[ T \approx \frac{125 \, \text{N}}{0.178 \times 1.5 \, \text{m}} \][/tex]

[tex]\[ T \approx \frac{125 \, \text{N}}{0.267 \, \text{m}} \][/tex]

[tex]\[ T \approx 468.164 \, \text{N} \][/tex]

So, the magnitude of the tension T in the cable is approximately 468.164 N.

Answer:

Explained below.

Explanation:

The water will flow more easily through a wide pipe, because as we know that if the diameter of the pipe is wider than the flow of water will be more.

And same case will be applied with the electric current, if the is wire thicker then the flow of the current will be more easy and the charge will also flow easily.