A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.
If the angle between the beam and the cable is θ = 57.0° what is the vertical component of the force exerted by the hi.nge on the beam?

The change in temperature is 20 kelvin


What is Temperature?


Temperature can simply be described as how hot or how cold an object is at a particular period in time. The unit of temperature is Kelvin.

The formula for calculating change in temperature is

Final temperature - Initial temperature

Final temperature = 300 kelvin
Initial temperature= 280 kelvin

Change in temperature= 300-280

= 20 kelvin

Thus the change in temperature for this process is 20 kelvin


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Change in potential energy = 1.02 * 10⁶ J

Minimum work required of the hiker = 1.89 * 10⁶ J

The potential energy of a body is calculated as follows:

Change in potential energy = Final PE - Initial PE

Change in potential energy = mg(H - h)

Change in potential energy = 72.5 * 9.81 * (2660 - 1230)

Change in potential energy = 1.02 * 10⁶ J

The minimum work required of the hiker is the potential energy at the highest point.

Minimum work = mgH

Minimum work required of the hiker = 1.89 * 10⁶ J

In conclusion, potential energy is energy due to state or position of a body.

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Explanation:

gravitational potential energy = mgh (must be in S.I. unit)

m= 0.4 kg ; g= 10m/s (gravitational acceleration occurs); h=9.2 m

hence mgh=0.4×10×9.2= 36.8J

unit for energy is joules and since the variables are in S.I. unit, we can use Joules as the final unit for measurement

Answer:

36.06 J if you define g = 9.8 m/s² or 36.8 J if you define g = 10 m/s²

Explanation:

Potential Energy = Mass x Gravitational Acceleration x Height

It can be expressed as [tex]\displaystyle{PE = mgh}[/tex] where PE is potential energy, m is mass, g is gravitational acceleration and h is height.

In this case, we know mass is 0.40 kg and height of 9.2 m as well as gravitational acceleration is defined to be 9.8 m/s² (You can also define g = 10 m/s²)

Therefore, substitute given information in formula:

[tex]\displaystyle{PE=0.40 \ \times \ 9.8 \ \times 9.2 }\\\\\displaystyle{PE=36.06 \ J}[/tex]

With g = 10 m/s², you'll get:

[tex]\displaystyle{PE = 0.40 \ \times 10 \ \times 9.2}\\\\\displaystyle{PE = 36.8 \ J}[/tex]

Note that J is for joule unit.

Therefore, the answer is 36.06 J if you define g = 9.8 m/s² or 36.8 J if you define g = 10 m/s² - both work.

Answer:

Object A's momentum is larger

Explanation:

as the formula for momentum goes:

P = M * V

where P is momentum, M is mass, V is velocity.

so where Va (Object A's velocity) is larger than Vb (Object B's velocity) we get:

( i ) Va > Vb

as the masses of both objects are equal, we mark:

( ii) Ma = Mb = M

we multiply both sides of ( i ) by M to get:

( iii ) Va × M > Vb × M

and we finally get:

( iv ) Pa > Pb

Answer: Object A's momentum of larger than compared to that of object B.

Explanation:

The force required to pull the two hemispheres apart 53696.25N and  the minimum number of horses required is 37 .

( Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 )

(1)The contact area between the hemispheres is (π x 0.430²) = 0.5805m²

Pressure difference = (940 - 15) = 925millibars.

(925 x 100) = 92,500N/m^2.

(92500 x 0.5805) = 53696.25N. is the force required to part the hemispheres.

(2)[tex]\frac{53696.25N}{1450N}[/tex] = 37 horses for each side .

37 + 37 = 74 horses will be required.

Force is something which can change the motion of an object, stop it or move it, change its shape or size with it. There are two types of forces, contact forces and non-contact forces. Here, it is a contact force at first, then when the horses come it becomes non-contact force.

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Sound will travel fastest in air at 15°C.

The speed of sound in air, given in the range of 100 degrees Celsius and 0 degree Celsius include;

Air at 100°C 387 m/s

Air at 0°C 331 m/s

From the date above, the speed of sound in air increases with increases in temperature. Thus, Sound will travel fastest in air at 15°C.

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288.51 N is  the magnitude of the force that the beam exerts on the hi.nge.

Given

Mass 0f beam = 40 Kg

The horizontal component of the force exerted by the hi_nge on the beam is 86.62 N

Angle between the beam and cable is = 90°

Angle between beam and the horizontal component = 31°

As the system of the beam, hi_nge and cable are in equilibrium.

The magnitude of the force that the beam exerts on the hi_nge can be calculated by -

F =The  horizontal component of force + the vertical component of force  

F = 86.62 N + 40 × 9.8 × sin 31°

F =86.62 N + 201.89 N

F = 288.51 N

Hence, the magnitude of the force that the beam exerts on the hi_nge is 288.51  N.

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The minimum necessary height of the IV bag above the position of the needle, if the density of the solution is 1.308kg/L, and the pressure in the vein is 35.7 mmHg is, 2.78m.

To find the answer, we need to know more about the pressure exerted by a liquid column.

                                                [tex]P=[/tex] ρgh

                                           [tex]density=1.308 kg/L\\\\P=35.7 mmHg.\\[/tex]

                                       [tex]h=\frac{P}{density*g} =\frac{35.7}{1.308*9.8}\\\\ h=2.78 m[/tex]

Thus, we can conclude that, the height of the Iv bag is 2.78 m above the position of needle.

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If the solution has a density of 1.308 kg/L and the vein pressure is 35.7 mmHg, the minimum height of the IV bag above the needle position is 2.78 m.

We need to learn more about the pressure that a liquid column exerts in order to determine the solution.

                            [tex]P=[/tex] ρgh

                        [tex]Density=1.308kg/L\\P=35.7mmHg[/tex]

                           [tex]h=\frac{P}{density*g}\\\\ h=2.78m[/tex]

As a result, we may say that the I.V. bag is 2.78 m above the ground.

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The velocity of a 55.0-kg person hitting the ground, is mathematically given as

vt=39.5983m/s

Generally, the equation for is  mathematically given as

mass of squirrel,

[tex]m=550 \mathrm{~g}\\\\Surface area, $A=945 \mathrm{~cm}^{2}=88 \times 10^{-3}$\\\\Height, $h-4 \mathrm{~m}$\\[/tex]

Terminal velocity is given by:

[tex]$v_{i}=\sqrt{\frac{2 m g}{\rho A C}}$[/tex]

where \rho is the density of fluid that is falling and it is given by

[tex]$\rho=\frac{m}{V}$[/tex]

since, volume =area * height

[tex]^{\rho=} \frac{0.55 \mathrm{Kg}}{0.0945 \mathrm{~m}^{2} \times 4.0 \mathrm{~m}}\\\\$\rho=0.1455 \mathrm{Kg} / \mathrm{m}^{3}$[/tex]

A is the surface area of squirrels.

C is the drag coefficient.

The surface area facing the fluid is given by:

[tex]A_{f}=\frac{0.0945 \mathrm{~m}^{2}}{2} \\\\\\ A_{f}=0.04725 \mathrm{~m}^{2}[/tex]

so, terminal velocity is :

[tex]$v_{t}=\sqrt{\frac{2 \times 0.55 \mathrm{Kg} \times 9.8 \mathrm{~m} / \mathrm{s}^{2}}{0.1455 \mathrm{Kg} / \mathrm{m}^{3} \times 0.04725 \mathrm{~m}^{2} \times 1}}$[/tex]

Vt=39.5983

In conclusion, the terminal velocity of the squirrel is 39.5983m/s

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(a) The speed of a satellite on a low lying circular orbit around this planet is  7,338.93 m/s.

(b) The minimum speed required for a satellite in order to break free permanently from the planet is 10,378.82 m/s.

(c) The radius of the synchronous orbit of a satellite is 69,801 km .

v = √GM/r

where;

v = √[(6.67 x 10⁻¹¹ x 4.74 x 10²⁴) / (5870 x 10³)]

v = 7,338.93 m/s

v = √2GM/r

v = √[( 2 x 6.67 x 10⁻¹¹ x 4.74 x 10²⁴) / (5870 x 10³)]

v = 10,378.82 m/s

v = 2πr/T

r = vT/2π

r = (7,338.93 x 16.6 x 3600 s) / (2π)

r = 69,801 km

Thus, the speed of a satellite on a low lying circular orbit around this planet is  7,338.93 m/s.

The minimum speed required for a satellite in order to break free permanently from the planet is 10,378.82 m/s.

The radius of the synchronous orbit of a satellite is 69,801 km .

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3.00 m is the magnitude of the resultant vector

137.1° is the directional angle of the resultant vector.

1) To find the magnitude of the resultant

Discover each vector's individual components first. 

We must consider the signs of the components because the angles in the figure are measured in various ways. 

In this case, both the y component of vector C and the x component of vector A are negative. 

The vectors' components are as follows, using a little trigonometry:  

Magnitude of A = 5 m

[tex]A_{x}[/tex] = - (5.00m) cos20° = -5 ×0.408 = -4.698 m

[tex]A_{y}[/tex] = + (5.00m)sin20° = +5 × 0.342 = +1.710 m

Magnitude of B = 5m

[tex]B_{x}[/tex] = +(5.00m)cos60° = 5 × 0.5 = +2.5m

[tex]B_{y}[/tex] = +(5.00m)sin60° = 5 × √3/2 = +4.33 m

Cx = 0

Cy = -4.00m

The sum of all three vectors, which we refer to as R, produces components.

Rₓ = Aₓ + Bₓ + Cₓ

   = -4.698 + 2.5 + 0

   = -2.198 m

[tex]R_{y}[/tex] = [tex]A_{y} + B_{y} + C_{y}[/tex]

    = +1.71 +4.33 - 4.00

    = 2.040 m

R = [tex]\sqrt{R_{x^{2} }+R_{y^{2} }[/tex]

  = [tex]\sqrt{(-2.198)^{2 } + (2.040)^{2} }[/tex]

  = 3.00 m

2) To find the directional angle of resultant

tan θ = 2.040/-2.198 = -0.928

θ = -42.9°

Such a vector would be in the so-called "fourth quadrant," as it is well known. However, we discovered that R has a negative x component and a positive y component, indicating that such a vector must reside in the "second quadrant.

The calculator accidentally returned an angle that is 180 degrees off, thus we must add 180 degrees to the naïve angle in order to get the right angle.

Therefore, R's actual direction is determined by-

θ = -42.9° + 180° = 137.1°

Hence, the magnitude of resultant vector is 3.00 m and directional angle is 137.1°

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K is cation by losing of electron whereas Br is anion due to accepting of electrons.

Yes, the positive ion appears on K and become cation whereas the negative ion bears on Br which make it anion because of losing and gaining of electron by these atoms. This transferring of electrons leads to formation of ionic bonds between them.

So we can conclude that K is cation by losing of electron whereas Br is anion due to accepting of electrons.

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Answer:

Explanation:

The kinetic energy of the first block just at the foot of the incline is 78.4J, the kinetic and gravitational potential energies of the first block halfway down the incline are same, and which is equal to 39.2J. The speeds of the two blocks just after their collision interchange with the values before collision.

To find the answer, we need to know about the concept of collision and kinetic energy.

                 [tex]TE=KE+PE\\T=PE=m_1gh=(0.4*9.8*20)=78.4 J[/tex]

                   [tex]TE=KE=78.4J[/tex]

                          [tex]U'=m_1gh'=mg\frac{h}{2} \\U'=39.2J[/tex]

                      [tex]TE=78.4 J\\KE'+PE'=78.4J\\KE'=78-U'=78.4-39.2=39.2J[/tex]

                            [tex]KE=\frac{1}{2} mv^2=78.4J\\v=\sqrt{\frac{2KE}{m} } =4m/s[/tex]

               [tex]\frac{1}{2}m_1u_1^2+ \frac{1}{2}m_2u_2^2=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2[/tex]

                 [tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]

                            [tex]m_1=m_2=m\\u_1=4m/s\\u_2=0\\v_1=?\\v_2=?[/tex]

                       [tex]\frac{1}{2}m*4^2=\frac{1}{2}m(v_1^2+v_2^2)\\(v_1^2+v_2^2)=16[/tex]  from resolving KE equation.

                        [tex]4m=m(v_2+v_1)\\4=v_2+v_1\\v_1=4-v_2[/tex] From resolving momentum conservation.

                            [tex]v_2=4m/s\\v_1=0[/tex]

Thus, we can conclude that, the kinetic energy of the first block just at the foot of the incline is 78.4J, the kinetic and gravitational potential energies of the first block halfway down the incline are same, and which is equal to 39.2J. The speeds of the two blocks just after their collision interchange with the values before collision.

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a)New position vector in vector form= r = 0.233404 î + 2.94056j m

b) Lies in second quadrant at 161.391°

c)Velocity =4.8675 m/s

d)It is moving in a direction making 161.391° with positive x-direction.

e)Acceleration will be centripetal acceleration (8.031 m/s²).

Given:

Mass of the object m = 3.95 kg

ω=1.65 rad/s

Radius of the circle = 2.95 m

a)

new position vector in vector form

=R cos1.65 î  + R sin 1.65 j

= 2.95 cos1.65 î  +2.95 sin1.65 j

= 2.95 x 0.07912 î + 2.95 x 0.9968 j

r = 0.233404 î + 2.94056j

b)

Angular Displacement = θ₀ = 9.10 rad

9.10 radian = 180/π× 9/10 degree

= 521.391°

=521.391°- 360°

=161.391°

This will lie in second quadrant.

Angle made with positive x-axis

=161.391°

c)

Velocity

v = ω R

= 1.65 x 2.95

=4.8675 m/s

d)

It is moving in a direction making 161.391° with positive x-direction.

e)

Acceleration will be centripetal acceleration.

= v²/R

=(4.8675)² / 2.95

=23.6925562 / 2.95

=8.031 m/s²

f) Position, Velocity and Acceleration graph:

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a) New position vector is r = 0.233404 î + 2.94056j m

b) Lies in second quadrant at 161.391°

c) Velocity =4.8675 m/s

d) It is moving in a direction making 161.391° with positive x-direction.

e) Acceleration will be centripetal acceleration (8.031 m/s²).

Given:

Mass of the object m = 3.95 kg

ω=1.65 rad/s

The radius of the circle = 2.95 m

a) new position vector in vector form

 r =R cos1.65 î  + R sin 1.65 j

 r = 2.95 cos1.65 î  +2.95 sin1.65 j

 r = 2.95 x 0.07912 î + 2.95 x 0.9968 j

 r = 0.233404 î + 2.94056j

b) Angular Displacement = θ₀ = 9.10 rad

9.10 radian = 180/π× 9/10 degree

 = 521.391°

 = 521.391°- 360°

 =161.391°

This will lie in the second quadrant.

Angle made with the positive x-axis =161.391°

c) Velocity

   v = ω R

   v = 1.65 x 2.95

   v = 4.8675 m/s

d) It is moving in a direction making 161.391° with positive x-direction.

e) Acceleration will be centripetal acceleration.= v²/R

   =(4.8675)² / 2.95

   =23.6925562 / 2.95

   =8.031 m/s²

f) Position, Velocity, and Acceleration graph:

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Hi there!

We can use Biot-Savart's Law for a moving particle:
[tex]B= \frac{\mu_0 }{4\pi}\frac{q\vec{v}\times \vec{r}}{r^2 }[/tex]

B = Magnetic field strength (T)
v = velocity of electron (0.130c = 3.9 × 10⁷ m/s)

q = charge of particle (1.6 × 10⁻¹⁹ C)

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

r = distance from particle (2.10 μm)

There is a cross product between the velocity vector and the radius vector (not a quantity, but specifies a direction). We can write this as:

[tex]B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }[/tex]

Where 'θ' is the angle between the velocity and radius vectors.

a)
To find the angle between the velocity and radius vector, we find the complementary angle:

θ = 90° - 60° = 30°

Plugging 'θ' into the equation along with our other values:

[tex]B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }\\\\B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(30)}{(2.1*10^{-5})^2 }[/tex]

[tex]B = \boxed{7.07 *10^{-10} T}[/tex]

b)
Repeat the same process. The angle between the velocity and radius vector is 150°, and its sine value is the same as that of sin(30°). So, the particle's produced field will be the same as that of part A.

c)

In this instance, the radius vector and the velocity vector are perpendicular so

'θ' = 90°.

[tex]B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(90)}{(2.1*10^{-5})^2 } = \boxed{1.415 * 10^{-9}T}[/tex]

d)
This point is ALONG the velocity vector, so there is no magnetic field produced at this point.

Aka, the radius and velocity vectors are parallel, and since sin(0) = 0, there is no magnetic field at this point.

[tex]\boxed{B = 0 T}[/tex]

The horizontal force applied is  160 N while the velocity is  2.03 m/s.

The work done by the car is obtained as the product of the force and the distance;

W = F x

F = ?

x = 30.0 m

W = 4,800 J

F = 4,800 J/30.0 m

F = 160 N

But F = ma

a = F/m

a = 160 N/2.30 ✕ 10^3-kg

a= 0.069 m/s

Now;

v^2 = u^2 + 2as

u = 0/ms because the car started from rest

v = √2as

v = √2 * 0.069 * 30

v = 2.03 m/s

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(a) The weight of the drop is 1.54 x 10⁻²⁵ N,

(b) The the charge on the drop, in terms of e is 5 x 10⁻¹²e and

(c) The excess electrons is 5 x 10⁻¹² electron.

The weight of the drop is calculated as follows;

Volume of the drop; V = ⁴/₃πr³

V = ⁴/₃π(1.64 x 10⁻⁶)³ = 1.845 x 10⁻¹⁷ m³

mass = density x volume

mass =  0.851 g/cm³ x  1.845 x 10⁻²³ cm³ = 1.572 x 10⁻²³ g =  1.57 x 10⁻²⁶ kg.

Weight = 1.57 x 10⁻²⁶ kg x 9.8 m/s² = 1.54 x 10⁻²⁵ N.

q = F/E

q = (1.54 x 10⁻²⁵ N)/(1.92 x 10⁵)

q = 8.01 x 10⁻³¹ C

1.6 x 10⁻¹⁹ C = 1e

8.01 x 10⁻³¹ C = ?

= 5 x 10⁻¹²e

1.6 x 10⁻¹⁹ C ------- 1 electron

8.01 x 10⁻³¹ C ------- ?

= 5 x 10⁻¹² electron

Thus, the weight of the drop is 1.54 x 10⁻²⁵ N, the the charge on the drop, in terms of e is 5 x 10⁻¹²e and the excess electrons is 5 x 10⁻¹² electron.

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C. The center of mass and center of gravity will coincide if the acceleration due to gravity is constant or uniform.

The center of mass of an object is the unique point where the weighted relative position of the distributed mass sums to zero.

Center of gravity is the point from which the weight of a body or system may be considered to act.

Thus, the center of mass and center of gravity will coincide if the acceleration due to gravity is constant or uniform.

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The speed of the satellite in a circular orbit around the Earth is 1.32 x 10⁵ m/s.

v = √(GM/r)

where;

v = √(6.67 x 10⁻¹¹ x 5.98 x 10²⁴/3.57 x 6.37x 10³)

v = 1.32 x 10⁵ m/s

Thus, the speed of the satellite in a circular orbit around the Earth is 1.32 x 10⁵ m/s.

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Ant is performing a work

what is work?

Work is the force applied on an individual with respect to displacement.

Work = Force × displacement

Unit is Nm

Elephant is pushing bt there is no displacement occurred so the work of elephant is zero.

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The ant works, but the elephant does not.

Work done = Force × Displacement.

If there are ants and houseflies,

Ants drag the house bug, so they use specific force to move the house bug from one point to another, so we can say they work.

In the case of the elephant and the tree,

When the elephant pushes the tree (applying a force), the tree does not move, i.e., there is no displacement, so there is work.

Work done = Force × Displacement

= Force × 0

= 0

Therefore,

The ant works, but the elephant does not.

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Answer: True

hope this helps!

Answer:

Create space

Watch your breath

Watch your thoughts

Be gentle with yourself

Affirm what you want

Explanation:

The highest point of the wheel is the position of the wheel when its potential energy is greatest.

The position of the wheel when its potential energy is greatest when it is at the highest point because potential energy depends on the height of an object. If the object is at more height then it has more potential energy and vice versa.

So we can conclude that the highest point of the wheel is the position of the wheel when its potential energy is greatest.

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(a) The proton’s potential energy change is 3.6 x 10⁻¹⁸ J.

(b) The potential difference between the negative plate and a point midway between the plates is 11.25 V.

(c) The speed of the proton just before it hits the negative plate is 6.57 x 10⁴ m/s.

U = qΔV

where;

U = q(Ed)

U = (1.6 x 10⁻¹⁹)(1500 x 1.5 x 10⁻²)

U = 3.6 x 10⁻¹⁸ J

ΔV = E(0.5d)

ΔV = 0.5Ed

ΔV = 0.5 (1500)(1.5 x 10⁻²)

ΔV = 11.25 V

U = ¹/₂mv²

U = mv²

v² = 2U/m

where;

v² = (2 x 3.6 x 10⁻¹⁸) / ( 1.67 x 10⁻²⁷)

v² = 4.311 x 10⁹

v = √(4.311 x 10⁹)

v = 6.57 x 10⁴ m/s

Thus, the proton’s potential energy change is 3.6 x 10⁻¹⁸ J.

The potential difference between the negative plate and a point midway between the plates is 11.25 V.

The speed of the proton just before it hits the negative plate is 6.57 x 10⁴ m/s.

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Answer:

Yes.

Explanation:

When you move your hand from the cold water to the “warmer” (room temp) water, one hand feels warm.

As you move your hand from the warm water to the “colder” (room temp) water, that hand feels colder.

Although both hands experience the last bowl of water at the same temperature, your brain senses two

separate sensations. So the water feels “warm” or “cold” relative to the water your hand was in previously.

The greater the difference in temperature, the easier it is to sense a difference.

Answer:

C. The atom loses 1 electron to have a total of 36

Explanation:

The number of electrons in a Rubidium atom is 37. Since the atom loses 1 electron, it has 36 left.

Answer:

D. Impulse

Explanation: Hope this helps

No. of turns in the solenoid is an option (b) 144.

The self-inductance of a long solenoid depends only on its physical properties (such as the number of turns of wire per unit length and the volume), and not on the magnetic field or the current.

Self-inductance of solenoid = 39.1 μH

                                           = 39.1 × [tex]10^-^6[/tex] H

Length of the solenoid = 4.0 cm

Cross-sectional area = 0.60 cm²

Expression for the self-inductance of a coil ;

L = µ₀N²A / [tex]l[/tex]

where,

L = Self- Inductance

N = No. of turns.

A = Cross-sectional area

[tex]l=[/tex] Length of the solenoid

L =( 4π × [tex]10^-^7[/tex] × N² × 0.60 ) / 4.0

39.1× 4.0 / 4π × [tex]10^-^7[/tex] × 0.60 = N²

N² = 2.07 × [tex]10^6[/tex]

N = 144

Therefore, the no. of turns of the solenoid is 144.

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