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The density of the planet is determined as 1,974.26 kg/m³.

√(⁴/₃πGρ) = 2π/(2.35 x 3600)

where;

√(⁴/₃πGρ) = 2π/(2.35 x 3600)

√(⁴/₃πGρ) = 2π/8460

(⁴/₃πGρ)  = (2π/8460)²

⁴/₃πGρ = 4π²/(8460)²

ρ = 12π/(8460² x 4G)

ρ = (12π) / (8460² x 4 x 6.67 x 10⁻¹¹)

ρ = 1,974.26 kg/m³

Thus, the density of the planet is determined as 1,974.26 kg/m³.

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Some of the techniques used by astronomers to determine the type of galaxy in which we live are:

This refers to the study of heavenly bodies and space and other things that space is made up of.

Hence, we can see that the reason why it is so difficult to determine the shape of our galaxy is that astronomers can only infer its presence from the motions of stars in the galaxy, and a precise shape is difficult to be determined.

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The minimum diameter of the steel wire is 7.58 mm.

Young's modulus = stress/strain

strain = e/l = (7.32 x 10⁻³ m) / (19.3 m) = 3.79 x 10⁻⁴

stress = Young's modulus  x  strain

stress = 200 x 10⁹ N/m²   x 3.79 x 10⁻⁴  = 7.59 x 10⁷ N/m²

stress = Force/Area

Area = Force/stress

Area = mg/stress

Area = (350 x 9.8) / (7.59 x 10⁷)

Area = 4.519 x 10⁻⁵ m²

Area = πd²/₄

πd²/₄ = 4.519 x 10⁻⁵ m²

πd² = 4(4.519 x 10⁻⁵)

d² = (4 x 4.519 x 10⁻⁵)/π

d² = 5.75 x 10⁻⁵

d = √(5.75 x 10⁻⁵)

d = 7.58 x 10⁻³ m

d = 7.58 mm

Thus, the minimum diameter of the steel wire is 7.58 mm.

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Radiation from the sun hits earth unequally and is absorbed by different materials in varying amounts. This is called differential heating.

Differential heating is the property that causes different surfaces to heat up and cool down at different rates. The earth's surface receives different magnitudes of solar radiation and also the earth's surfaces absorb thermal energy in different magnitudes.

The color, shape, texture, surface and presence of constructions can influence the heating or cooling of the earth.

The earth in the equator line is heated more by solar action than that of the poles, since it receives more amount of radiation per unit area.

In general, dry surfaces heat up and cool down faster than wet ones.

Therefore, we can confirm that when radiation from the sun hits earth unequally and is absorbed by different materials in varying amounts is called differential heating.

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A psychologist will need to have good linguistic intelligence in other to be successful.

This is referred to as a professional who specializes in the handling of mental health challenges in individuals.

It is best for such professional to have a good linguistic intelligence as the right words being said to the patient will solve the problem thereby bringing in more success.

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The two units for measuring the diameter of nucleus atom are femtometre and metre.

How do you measure the size of the nucleus ?

Nucleus size is expressed in fermi, often known as femtometers. between a lighter and a heavier nucleus. Despite its modest size, the nucleus contains the majority of an atom's mass. The weight or mass of the atom's nucleus and neutrons are determined by neutrons.

femtometre (fm), which equals [tex]10^{-15}[/tex] metre.

A nucleus' diameter largely depends as to how many particles it contains, from about 4 fm for a light nucleus like carbon to 15 fm for a heavy nucleus as lead.

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The magnitude of the tension in the string marked A is 39.5 N.

The tension in A is determined thus:

The angle at A, θ = tan⁻¹(3/8) = 20.56

When extrapolated below negative x, the angle at B, α = tan⁻¹(5/4) = 51.34

When extrapolated below negative x, the angle at C, β = tan⁻¹(1/6) = 9.46

Taking the horizontal components of tension;

56.3cos(9.46) = A * cos(20.56) + B * cos(51.34)

0.6247B= 55.53 - 0.936A

B = (55.53 - 0.936A)/0.6247 ----(1)

Taking the vertical components of tension;

56.3 * sin(9.46) + A * sin(20.6) = B * sin(51.3)

9.25 + 0.35A = 0.78B  ---- (2)

substitute the value (1)  in (2)

9.25 + 0.35A = 0.78{(55.53 - 0.936A)/0.6247}

(9.25 + 0.35A) * 0.6247 = 43.31 - 0.73A

0.22A + 0.73A = 43.31 - 5.78

0.93A = 37.53

A = 39.5 N

In conclusion, the tension in A is determined by solving for the vertical and horizontal components of tension.

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The correct option is C.   Air resistance friction is not useful for an airplane because it slows it down.

Air resistance is the opposition to motion of an object caused by air flow.

High air resistance may cause turbulence during the motion of an air plane. This can lead to engine failure or some mishap during flight.

Air resistance is also a type of friction between air and another material such as airplane.

Friction between the air and the plane slows the airplane down. This is known as air resistance, or drag.

The faster an object travels through the air, the more it has to fight against drag.

Thus, air resistance friction is not useful for an airplane because it slows it down.

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The tension in the upper rope is 50.53 N.

The tension in the upper rope is calculated as follows;

T(up) = T(dn) + mg

where;

T(up) = 12.8 N + (3.85 x 9.8) N

T(up) = 50.53 N

Thus, the tension in the upper rope is 50.53 N.

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The correct option is D.   The initial momentum of the blue and green train combined during the collision is 200 kgm/s.

Apply the principle of conservation of linear momentum as follows;

Pi = m1v1 + m2v2

where;

Pi = (50 x 4) + 30(0)

Pi = 200 kgm/s

Thus, the initial momentum of the blue and green train combined is 200 kgm/s.

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The final temperature of the system will be equal to the initial temperature, and which is 373K. The work done by the system is 409.8R Joules.

To find the answer, we need to know about the thermodynamic processes.

                                [tex]dU=dQ-dW[/tex]

                                  [tex]dU=0[/tex],

                             [tex]dU=C_pdT=0\\\\thus, dT=0\\\\or , T=constant\\\\i.e, T_1=T_2[/tex]

                          [tex]T_1=T_2=100^0C=373K[/tex]

                               [tex]W=RT*ln(\frac{V_2}{V_1} )=373R*ln(\frac{4}{\frac{4}{3} })\\ \\W=409.8R J[/tex]

Where, R is the universal gas constant.

Thus, we can conclude that, the final temperature of the system will be equal to the initial temperature, and which is 373K. The work done by the system is 409.8R Joules.

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The system's final temperature will be 373K, which is the same as its starting temperature. The system exerts 409.8R Joules of work.

We need to understand the thermodynamic processes in order to locate the solution.

                    [tex]dU=dQ-dW[/tex]     , It is 0 here.

                  [tex]dU=C_pdT=0\\dT=0\\T=constant\\T_1=T_2=373K[/tex]

As a result, the system's final temperature will be equal to its starting temperature.

                  [tex]W=RT ln(\frac{V_2}{V_1} )=373R*ln(3 )\\W=409.8R Joules[/tex]

R is the gaseous universal constant.

Thus, we can draw the conclusion that the system's end temperature will be 373K, the same as its starting temperature. The system exerts 409.8R Joules of work.

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The kind of equation that can be used to differentiate the kind of separatrix that shows change on motion is

H = 2g/l.

A simple pendulum can be defined as the equipment that displays an oscillatory motion when a mass is tied on a rope and is suspended from it.

The various movements that occur using a simple pendulum is translational ( side to side) or continuous circle (oscillatory motion).

The equation that show that a change from one type of motion to another is H = 2g/l.

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(a) The  acceleration of the ball while it is in flight has a magnitude of 9.81 m/s2 in downward direction.

(b) The velocity of the ball when it reaches its maximum height is zero.

(c) The initial velocity of the ball is 17.36 m/s.

(d) The maximum height it reaches is 15.36 m.

The acceleration of the ball while it is in flight has a magnitude of 9.81 m/s2 in downward direction.

The velocity of the ball decreases as the ball moves upwards and eventually becomes zero at maximum height.

v = u - gt

at maximum height, final velocity, v = 0

0 = u - gt

u = gt

u = 9.81 x 1.77

u = 17.36 m/s

h = ut - ¹/₂gt

h = 17.36(1.77) - ¹/₂(9.81)(1.77²)

h = 15.36 m

Thus, the  acceleration of the ball while it is in flight has a magnitude of 9.81 m/s2 in downward direction.

The velocity of the ball when it reaches its maximum height is zero.

The initial velocity of the ball is 17.36 m/s.

The maximum height it reaches is 15.36 m.

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Hi there!

Part A.

To solve this part, all we need to do is a summation of vertical forces.

We have the following acting on the beam :
- Force of gravity (Fg, down)
- Force of tension from the rope holding the box (T, down)
- Force exerted by pivot (Py, up)

These sum to zero because the beam is not accelerating vertically.

[tex]\Sigma F = -F_g - T + P_y = 0[/tex]

[tex]P_y = F_g + T[/tex]

The tension force is equal to the box's weight because the forces on the box are balanced. Let's use values and solve.

[tex]P_y = 49(9.8) + 49(9.8) = \boxed{960.4N}[/tex]

Part B.

We must begin by doing a summation of torques. Placing the pivot point at the pivot, we have the following present:
- Force of gravity acting at the center of mass of the rod (CC, at L/2)

- The tension of the horizontal cable acting at the end of the rod (CCW, at L)

- The force of tension in the rope holding the box (CC, at 3L/4)

Since the rod is not rotating, these torques sum to zero.

The equation for torque is:
[tex]\tau = r \times F[/tex]

This is a cross-product, and you must find the lever arm (perpendicular distance between pivot and line of action of force). We will need to use trigonometry for this.

Now, let's find the torque from all three of these forces.

- Force of gravity at center:

The perpendicular distance between the force of gravity and the pivot point is the cosine with respect to the angle made with the floor.

[tex]\tau = Mg\frac{L}{2}cos(\theta) = 49(9.8)*\frac{2.2}{2} cos(18) = 502.367 Nm[/tex]

- Tension of horizontal cable:

The lever arm is the sine with respect to the angle. We will still have to solve for the value of 'T'.

[tex]\tau = TLsin\theta = T(2.2)sin(18) = 0.68T[/tex]

- Tension of rope holding box:
The tension is equal to the weight of the box since the box isn't accelerating. Thus, the torque would be:
[tex]\tau = Mg(\frac{3L}{4}) = 49(9.8)*\frac{3(2.2)}{4}cos(18) = 753.55 Nm[/tex]

Summing with clockwise torques + and counterclockwise -:
[tex]\Sigma \tau = 502.367 + 753.55 - 0.68 T = 0 \\\\1255.917 = 0.68T\\T = \boxed{1847.38 N}[/tex]

Part C.

This part is a lot easier than it seems. All we need to do is a summation of horizontal forces.

We only have two:
- The horizontal tension in the cable to the left (1847.38 N)

- The horizontal force exerted by the pivot on the beam to the right

These two balance out because there is no acceleration of the beam horizontally, so:
[tex]\Sigma F = P_x - T = 0 \\\\P_x = T\\\\P_x = \boxed{1847.38 N}[/tex]

**to the right

Answer:

a) 960.4 N

b) T= 5/4 Mg CotanΘ

c) 1847. 38

Explanation:

a) Py= 2Mg

=2(49 x 9.8)

= 960.4

b) T= (Mg x 1/2 x cos Θ + Mg x 3/4 x cos Θ) / sin Θ

T= 5/4 X Mg cotanΘ

c) T= (5/4) x (49 x 9.8) cotan (18)

T= 1847.37954

= 1847.38

(a) The magnitude of the force that the magnetic field exerts on the loop is 0.042 N.

(b) The direction of the force that the magnetic field exerts on the loop will be out of the plane.

F = BIL

where;

emf = BVb

where;

emf = 2.6 x 3 x 0.018 = 0.1404 V

Current in the loop, I = emf/R = 0.1404/0.8 = 0.18 A

F = BIL

where;

F = 2.6 x 0.18 x 0.09 = 0.042 N

The direction of the force that the magnetic field exerts on the loop will be out of the plane.

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The solution for the three questions  is mathematically given as

Parts A and C are both zeros.

Part B B = 0.001855Tesla

Parts A and C are both zeros.

For component A, the magnetic field is zero since 12.6 cm is still inside the toroidal solenoid. Part C has no magnetic field since it is 20.7 cm outside of the toroidal solenoid.

Generally, the equation for  magnetic field is  mathematically given as

B = (mu_0*N*I)/(2*pi*r)

Therefore

B = ((4*pi*10^{-7})*180*8.40)/(2*pi*0.163)

B = 0.001855Tesla

In conclusion, the magnitude of the magnetic field at 16.3 cm from the center of the torus is

B = 0.001855Tesla

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Answer:

Part A & C: 0 T

Part B: 1.855*10^-3 T

Explanation:

The formula that models the magnetic field of a toroid is B=μ0*N*I/2π*r .

Note: Toroids keep their magnetic field rotating within the coils that carry current.

Part A & C: Thus the B field magnitude 12.6 cm & 20.7 cm away from the center is 0 T.

Part B: [tex]B=\frac{(4\pi *10x^{-7} )(180)(8.4 A)}{(2\pi )(16.3*10x^{-2} m)}[/tex]

B=1.855*10^-3 T

Maximum work done by the hammer on the is 10 J.

Force is added to the hammer in order to increase the power of the hammer; option B.

Work done is defined as the product of force and the distance travelled by force.

Energy is used to do work.

The potential energy of the hammer is converted to work done.

The maximum amount of work it could do on the nail is given below:

Maximum work = 1.5 * 9.81 * 0.7

Maximum work = 10 J

Force is added to the hammer in order to increase the power of the hammer.

In conclusion, the work done by the hammer is obtained from the potential energy of the hammer.

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The orbiting speed of the satellite orbiting around the planet Glob is 60.8m/s.

To find the answer, we need to know about the orbital velocity a satellite.

= 60.8m/s

Thus, we can conclude that the speed of the satellite orbiting the planet Glob is 60.8m/s.

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(1) The critical angles for the air is 90⁰,

(2) The critical angles for the polystyrene is 39.1 ⁰ and

(3) The critical angles for the flint glass is 37.2 ⁰.

θc = sin⁻¹( 1/η)

where;

Refractive index of air = 1

Refractive index of polystyrene = 1.5865

Refractive index of flint glass = 1.655

θc = sin⁻¹( 1/1)

θc = 90⁰

θc = sin⁻¹( 1/1.5865)

θc =  39.1 ⁰

θc = sin⁻¹( 1/1.655)

θc = 37.2 ⁰

Thus, the critical angles for the air is 90⁰, the critical angles for the polystyrene is 39.1 ⁰ and the critical angles for the flint glass is 37.2 ⁰.

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According to mathematics, the planet's angle is stated as

dY=704 degrees.

We may demonstrate this by using Kepler's third law, which asserts that a planet's orbit squared is a function of cubed radius.

The equation for the period is often expressed numerically as

[tex](periodX / periodY)^2 = (radius X / radius Y)^3[/tex]

Therefore

(pX / pY)^2 = 4^3

(pX / pY)^2 = 64

[tex]\sqrt{(pX / pY )^2}= \sqrt{64}[/tex]

pX / pY=8

In conclusion, planet Y travels 8 times further than planet X does in the same amount of time since one orbit on planet X takes 8 times longer to complete.

planet Y travels ;

dY=8 * 88.0

dY= 704 degrees

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The wavelength of A2 with frequency (110hz) is 3.01 m.

Wavelength of a sound wave is the distance between successive similar points in the wave such as rarefactions or compressions.

A tuba is a musical instrument that produces sound waves.

Wavelength is related to frequency and velocity by the formula below:

The wavelength of A2 with frequency (110hz) is 3.01 m.

In conclusion, the wavelength of a wave is inversely proportional to frequency.

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Answer: A treasure chest full of silver and gold coins is being lifted from a pirate ship to the shore using two ropes as shown in the figure. The mass of the treasure chest is 75.6 kg. The tension in Rope A is 7.42x10^2 N, and the tension Rope B carries is 7.52x10^2 N. Then, the tension in rope C is 376N

Explanation: To find the correct answer, we have to know more about the Basic forces that acts upon a body.

                 [tex]T_A=7.42*10^2N\\T_B=7.54*10^2N[/tex]

                 [tex]T_C=T_B sin30\\thus,\\T_C=7.52*10^2*0.5=376N[/tex].

Thus, we can conclude that the tension in the rope c will be 376N.

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376N is the tension in rope C.

In order to determine the right response, we must have a better understanding of the fundamental forces that affect a body.

                              [tex]T_c=T_Bsin30\\T_c=7.52*10*2*0.5=376N[/tex]

Thus, we can infer that the rope's c tension will be 376N.

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The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

To find the answer, we need to know about the time of flight and range of projectile motion.

U= √{2.20×9.8/sin(73)} = 22.5m/s

= 2.73 s

Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

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Answer:

Average density of the crown: approximately [tex]8\; {\rm g \cdot mL^{-1}}[/tex].

Hence, if this crown contains no empty space, this crown is not made of pure gold.

Explanation:

Let [tex]m(\text{crown})[/tex] and [tex]V(\text{crown})[/tex] denote the mass and volume of this crown. Let [tex]g[/tex] denote the gravitational field strength.

Since this crown is fully immersed in water, the volume of water displaced [tex]V(\text{water, displaced})[/tex] is equal to the volume of this crown:

[tex]V(\text{water, displaced}) = V(\text{crown})[/tex].

The mass of water displaced would be:

[tex]\begin{aligned}m(\text{water, displaced}) &= \rho(\text{water}) \, V(\text{water, displaced}) \\ &= \rho(\text{water}) \, V(\text{crown})\end{aligned}[/tex].

The weight of water displaced would be [tex]m(\text{water, displaced})\, g = \rho(\text{water}) \, V(\text{crown})\, g\end{aligned}[/tex].

The buoyancy force on this crown is equal to the weight of water that this crown displaced:

[tex]F(\text{buoyancy}) = \rho(\text{water}) \, V(\text{crown})\, g[/tex].

The magnitude of this buoyancy force is [tex]7.84\; {\rm N} - 6.86\; {\rm N} = 0.98\; {\rm N}[/tex]. Rearrange the equation for buoyancy to find [tex]V(\text{crown})[/tex]:

[tex]\begin{aligned} V(\text{crown}) &= \frac{F(\text{buoyancy}) }{\rho(\text{water}) \, g}\end{aligned}[/tex].

Since the weight of this crown is [tex]\text{weight}(\text{crown}) = m(\text{crown})\, g[/tex], the mass of this crown would be [tex]m(\text{crown})= \text{weight}(\text{crown}) / g[/tex].

The average density of this crown would be:

[tex]\begin{aligned}\rho(\text{crown}) &= \frac{m(\text{crown})}{V(\text{crown})} \\ &= \frac{\text{weight}(\text{crown}) / g}{F(\text{buoyancy}) / (\rho(\text{water})\, g)} \\ &= \frac{\text{weight}(\text{crown})}{F(\text{buoyancy})}\, \rho(\text{water}) \\ &= \frac{7.84\; {\rm N}}{0.98\; {\rm N}}\times 1.000 \; {\rm g\cdot mL^{-1}} \\ &= 8.0\; {\rm g \cdot mL^{-1}}\end{aligned}[/tex].

The density of pure gold is significantly higher than [tex]8.0\; {\rm g\cdot mL^{-1}}[/tex]. Hence, if this crown contains no empty space (i.e., no air bubble within the crown), the crown would not be made of pure gold.

The total current through the battery is 4.77 A.

The current through resistor R1 is 2.4 A.

The current through resistor R2 is 1.2 A

The current through resistor R3 is 0.8 A

The current through resistor R4 is 0.36 A.

The total resistance of the circuit is calculated as follows;

1/Rt = 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄

1/Rt = 1/5 + 1/10 + 1/15 + 1/33

1/Rt = 0.397

Rt = 1/0.397

Rt = 2.518 ohms

I = V/Rt

I = 12/2.518

I = 4.77 A

I₁ = V/R₁

I₁ = 12/5

I₁ = 2.4 A

I₂ = 12/10

I₂ = 1.2 A

I₃ = 12/15

I₃ = 0.8 A

I₄ = 12/33

I₄ = 0.36 A

Thus, the total current through the battery is 4.77 A.

The current through resistor R1 is 2.4 A.

The current through resistor R2 is 1.2 A

The current through resistor R3 is 0.8 A

The current through resistor R4 is 0.36 A.

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The period of M2, in years. is mathematically given as

T2= 134.3968years

M1 = 4.00 kg

M2 = 1.00 kg

T1 = 34.0 years.

Generally, the equation for is  mathematically given as

T2 = T1 (a2/a1)3/2

T2=  34.0  * (5/2)^{3/2}

T2= 134.3968years

In conclusion, the period of M2, in years

T2= 134.3968years

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The required horsepower engine is 32 horsepower.

The force of the 30 skiers is calculated as follows:

Mass of the skiers = 30 * 67kg = 2010 kg

Net force acting on the skiers along the x-axis

where f is the frictional force

The kinetic frictional force, f = μN

where

μ = The coefficient of the kinetic friction

N = normal reaction

Net force acting on the skiers along y axis, the

Fy = ma

N = mg cos θ

Substituting for N above

f = μk mg cos θ

Substituting for f in  (1)

F = mg sin θ + μk mg cos θ

F = mg(sinθ + μk cos θ)

Work done by the engine in pilling up the skiers, W = Fx

W = mg ( sinθ + μk cos θ)x

x = 300 m

W = (2010 kg) (9.81 m/s²) (sin 23° + (0.10) cos 23°) (300 m)

Work done, W = 2.86 * 10⁶ J

Time taken, t = 2.0 * 60sec = 120 s

1 horsepower = 746 W

Power = 2.86 * 10⁶/ 120 * 1/746

Power = 31.9 horsepower

In conclusion, the power of the engine is the ratio of the work done and time taken.

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Answer: A train traveling initially at 16 m/s is under constant acceleration of 2 m/2. At a distance of 720m it will travel in 20 s, and the final velocity will be 56m/s.

Explanation: To find the answer, we need to know about uniformly accelerated motion.

                     [tex]u=16m/s\\a=2 m/s^2\\t=20s[/tex]

                      [tex]S=ut+\frac{1}{2}at^2[/tex]

                      [tex]S=(16*20)+\frac{2*20^2}{2} =720 m.[/tex]

                    [tex]v^2=u^2+2aS\\thus,\\v=\sqrt{u^2+2aS} =\sqrt{16^2+(2*2*720)} =56 m/s.[/tex]

Thus, we can conclude that, a train traveling initially at 16 m/s is under constant acceleration of 2 m/2. At a distance of 720m it will travel in 20 s, and the final velocity will be 56m/s.

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A train experiencing constant acceleration of 2 m/s^2 is moving at an initial speed of 16 m/s. It will go 720 meters in 20 seconds, with a final velocity of 56 m/s.

Understanding uniformly accelerated motion is necessary in order to determine the solution.

                         [tex]S=ut+\frac{1}{2}at^2 \\S=720m\\where,\\u=16m/s, a=2m/s^2,t=20s[/tex]

                          [tex]v^2-u^2=2aS\\v=\sqrt{u^2+2aS} \\v=56m/s[/tex]

Thus, we may say that a train moving at 16 m/s initially experiences constant acceleration of 2 m/2. It will go 720 meters in 20 seconds, with a final velocity of 56 meters per second.

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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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