The angle between the two beams inside the glass is mathematically given as
<(BR)= 0.563°
What is the angle between the two beams inside the glass?Generally, the equation for an angle for blue is mathematically given as
[tex]< B= arcsin( sin\theta / i )[/tex]
Therefore
<B= arcsin( sin30 / 1.660 )
<B= 17.53°
For angle for red
<R= arcsin(sin30 / 1.610)
<R= 18.09°
For the angle in between
<(BR)= 18.093 - 17.530
<(BR)= 0.563°
In conclusion, the angle between the two beams inside the glass
<(BR)= 0.563°
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Two planets X and Y travel counterclockwise in circular orbits about a star, as seen in the figure.
The radii of their orbits are in the ratio 4:3. At some time, they are aligned, as seen in (a), making a straight line with the star. Five years later, planet X has rotated through 88.0°, as seen in (b).
1. By what angle has planet Y rotated through during this time?
The angle of the planet is mathematically given as
dY= 704 degrees
What angle has planet Y rotated through during this time?With Kepler's third rule, which states that a planet's orbit squared is a function of cubed radius, we can prove that this is the case.
Generally, the equation for the period is mathematically given as
(periodX / periodY)^2 = (radius X / radius Y)^3
Therefore
(pX / pY)^2 = 4^3
(pX / pY)^2 = 64
\sqrt{(pX / pY )^2}= \sqrt{64}
(pX / pY=8
In conclusion, Because it takes 8 times longer to complete one orbit on planet X, planet Y travels 8 times farther than planet X does in the same time period...
planet Y travels ;
dY=8 * 88.0
dY= 704 degrees
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In the given figure, weight of stone inside water
is 9N and water displaced by stone is 2N then,
i)What is the actual weight of stone?
ii) Which principle is the
experiment based on?
The actual weight of the stone is 11 N. It is based on the Archimedes principles.
What is Archimedes principle?Archimedes principle states that the up thrust by water on an object is equal to the weight of water displaced.Upthrust by water on an object= actual weight of object - weight inside waterWhat is the actual weight of the object, if its weight inside water is 9N and weight of water displaced is 2N?Actual weight= weight inside water+ weight of water displaced
= 9N + 2N = 11N
Thus, we can conclude that the actual weight of the object is 11N.
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A person standing at the edge of a cliff throws one ball straight up and another ball straight down, each at the same initial speed. Neglecting air resistance, which ball hits the ground below the cliff with the greater speed?
a 5.5kg bowling ball has a weight on earth closest to what in N
The weight of the body is obtained as 53.9 N.
What is the weight of an object?The term weight refers to the product of the mass and the acceleration due to gravity.
Now we have the mass of the body as 5.5kg and the acceleration due to gravity as 9.8 m/s^2.
It the follows that the weight is;
W = mg = 5.5kg * 9.8 m/s^2 = 53.9 N
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Three ropes A, B and C are tied together in one single knot K.
If the tension in rope A is 65.3 N, then what is the tension in rope B?
The tension in the rope B is determined as 10.9 N.
Vertical angle of cable Btanθ = (6 - 4)/(5 - 0)
tan θ = (2)/(5)
tan θ = 0.4
θ = arc tan(0.4) = 21.8 ⁰
Angle between B and Cθ = 21.8 ⁰ + 21.8 ⁰ = 43.6⁰
Apply cosine rule to determine the tension in rope B;
A² = B² + C² - 2BC(cos A)
B = C
A² = B² + B² - (2B²)(cos A)
A² = 2B² - 2B²(cos 43.6)
A² = 0.55B²
B² = A²/0.55
B² = 65.3/0.55
B² = 118.73
B = √(118.73)
B = 10.9 N
Thus, the tension in the rope B is determined as 10.9 N.
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Hello!
This is an example of a force summation in the vertical direction.
We have the tension of rope A upward (+), and the equal vertical components of the tensions of rope B and C downward (-).
These forces sum to zero, since the knot is stationary.
[tex]\Sigma F = T_A - T_{By} - T_{Cy} \\\\0 = T_A - T_{By} - T_{Cy}[/tex]
Ropes 'B' and 'C' form equivalent angles from the vertical. (If you were to draw a line from rope A down). We can use right-triangle trig to determine the angle:
[tex]tan^{-1}(\frac{O}{A}) = \theta[/tex]
The ropes are 5 m long and 2 m tall, which are the opposite and adjacent sides respectively:
[tex]tan^{-1}(\frac{5}{2}) = 68.2^o[/tex]
The vertical components are the adjacent sides from this angle, so, we would use cosine.
[tex]0 = T_A - T_Bcos\theta - T_Ccos\theta[/tex]
Rope 'B' and 'C' have the same tensions since they form the same angle with the vertical and are the same length, so we can call them 'T'.
[tex]0 = T_A - 2Tcos\theta[/tex]
Solving for 'T':
[tex]2Tcos\theta = T_A \\\\T = \frac{T_A}{2cos\theta}\\\\T = \frac{65.3}{2cos(68.2)} = \boxed{87.92 N}[/tex]
Three equal positive charges 'q' are at the corners of an equilateral triangle of side 'a'.
a. Assuming that the three charges together create an electric field, find the location of a point other than the obvious one where the electric field is zero.
b. What is the magnitude and direction of the electric field at the top corner due to the two charges at the base?
(a) The location of a point where the electric field is zero is at the center of the triangle which is equal to ¹/₆√3a.
(b) The magnitude and direction of the electric field at the top corner due to the two charges at the base is 1.732 kq/a².
Position where the electric field is zero
The electric field is zero at the center of the equilateral triangle whose magnitude is equal to √3a/6.
Electric field at top corner due to two charges at the baseE = E₁ + E₂
where;
E₁ is electric field at the left base cornerE₂ is electric field at the right base cornerE = kq/a²[(cos 60i + sin 60j) + (-cos 60i + sin 60j)]
E = kq/a²[2(sin 60j)] = 1.732 kq/a²
Thus, the location of a point where the electric field is zero is at the center of the triangle which is equal to ¹/₆√3a.
The magnitude and direction of the electric field at the top corner due to the two charges at the base is 1.732 kq/a².
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Which best describes the results of Becquerel’s experiments?
Both forming images when placed in their respective places best describes Becquerel’s experiments.
What is Becquerel’s experiment?This was conducted by Henri Becquerel in which he sought to know how uranium salts are affected by light.
He discovered that the salts emits a penetrating radiation and formed an image in the presence of light but didn't form any in darkness.
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the temperature at which the velocity of sound in air is twice its velocity at 15°C
With the use of below formula, at 879 °C, velocity will be double the velocity at 15 °C.
What is the relationship between Velocity and sound ?The velocity of sound waves in air is proportional to the square root of Thermodynamic temperature. That is, V = K[tex]\sqrt{T}[/tex]
Given that the temperature at which the velocity of sound in air is twice its velocity at 15°C, Let us make use of the formula;
(v2/v1) = √(T2 / T1)
Where
v2 = final velocityv1 = initial velocityT2 = final absolute temperatureT1 = initial temperature.Recall that absolute temperature = °C + 273.
If v2 = 2 × v1 and temperature in degree Celsius = 15°C, then,
Temperature in Kelvin K = 15 + 273 = 288
Substitute all the parameters into the formula
(2 × v1)/v1 = √(T2/288)
2 = √ (T2 /288)
Square both sides
4 = (T2/288)
T2 = 4 × 288
T2 = 1152K
Temperature in degrees Celsius = 1152 - 273 = 879 °C.
Therefore, at 879 °C, velocity will be double the velocity at 15 °C.
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A canon ball is shot out of a cannon at an angle of 45 degrees. What is the initial velocity of the cannon ball if its initial horizontal velocity is 8 m/s?
Answer:
11.31 [m/s].
Explanation:
1. the required velocity can be calculated according to
[tex]V=\frac{V_{horizontal}}{sin45};[/tex]
2. according to the formula above:
V=8*1.41≈11.3137085 [m/s].
Calculate the de Broglie wavelength of a 0.56 kg ball moving with a constant velocity of 26 m/s (about 60 mi/h)
The de Broglie wavelength of a 0.56 kg ball moving with a constant velocity of 26 m/s is 4.55×10⁻³⁵ m.
De Broglie wavelength:The wavelength that is incorporated with the moving object and it has the relation with the momentum of that object and mass of that object. It is inversely proportional to the momentum of that moving object.
λ=h/p
Where, λ is the de Broglie wavelength, h is the Plank constant, p is the momentum of the moving object.
Whereas, p=mv, m is the mass of the object and v is the velocity of the moving object.
Therefore, λ=h/(mv)
λ=(6.63×10⁻³⁴)/(0.56×26)
λ=4.55×10⁻³⁵ m.
The de Broglie wavelength associated with the object weight 0.56 kg moving with the velocity of 26 m/s is λ=4.55×10⁻³⁵ m.
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How much work must be done to stop a 975- kg car traveling at 105 km/h ?
Express your answer to two significant figures and include the appropriate units.
The amount of work done to stop a 975- kg car traveling at 105 km/h is 414,808.34J.
How to calculate work done?The amount of work done by a moving object can be calculated using the following formula:
W (Kinetic energy) = ½ mv²
Where;
m = massv = velocityAccording to this question, a car of 975 kg is traveling at 105 km/h. This speed in m/s is 29.17m/s.
K.E = ½ × 975 × 29.17²
K.E = 414,808.34J
Therefore, the amount of work done to stop a 975- kg car traveling at 105 km/h is 414,808.34J.
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A 1300 kg steel beam is supported by two ropes. (Figure
1)
What is the tension in rope 1?
What is the tension in rope 2?
Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.
By Newton's second law,
the net horizontal force acting on the beam is[tex]R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0[/tex]
where [tex]R_1,R_2[/tex] are the magnitudes of the tensions in ropes 1 and 2, respectively;
the net vertical force acting on the beam is[tex]R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0[/tex]
where [tex]m=1300\,\rm kg[/tex] and [tex]g=9.8\frac{\rm m}{\mathrm s^2}[/tex].
Eliminating [tex]R_2[/tex], we have
[tex]\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)[/tex]
[tex]R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2[/tex]
[tex]R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2[/tex]
[tex]-R_1 \sin(50^\circ) = -\dfrac{mg}2[/tex]
[tex]R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}[/tex]
Solve for [tex]R_2[/tex].
[tex]\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0[/tex]
[tex]\dfrac{R_2}2 = -mg\cot(110^\circ)[/tex]
[tex]R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}[/tex]
A closed curve encircles several conductors. The line integral around this curve is (image attached below)
a) What is the net current in the conductors?
b) If you were to integrate around the curve in the opposite direction, what would be the value of the line integral?
The net current in the conductors and the value of the line integral
[tex]I=\frac{3.2\cdot 10^{-4}}{4\pi \cdot 10^{-7}}=254.77\, A[/tex]The resultant remains same 3.2 *10^4 TmThis is further explained below.
What is the net current in the conductors?Generally,
To put it another way, the total current In flowing across a surface S (contained by C) is proportional to the line integral of the magnetic B-field (in tesla, T).
[tex]\oint_C \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} = \mu_0 \iint_S \mathbf{J} \cdot \mathrm{d}\mathbf{S} = \mu_0I_\mathrm{enc}[/tex]
[tex]I=\frac{3.2\cdot 10^{-4}}{4\pi \cdot 10^{-7}}=254.77\, A[/tex]
B)
In conclusion, It is possible for the line integral to go around the loop in either direction (clockwise or counterclockwise), the vector area dS to point in either of the two normal directions and Ienc, which is the net current passing through the surface S, to be positive in either direction—but both directions can be chosen as positive in this example. The right-hand rule solves these ambiguities.
The resultant remains the same at 3.2 *10^4 Tm
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Discuss the aspects of either the Gemini Program or the Soyuz Program.
Answer:
The Gemini program
Explanation:
The Gemini Program was the second human spaceflight program hosted by Nasa in the year 1961. Taking place between mission Mercury and Apollo, the Gemini spacecraft carried two people to space and marked the foundation to the upcoming Apollo mission to Moon. It was a series of missions into the outer orbital which took place between 1965 and 1966. Prior to the Gemini missions, NASA had little to no information about space and space traveling. It was crucial for them to get acquainted with life outside before establishing successful Moon landings. And the series of Gemini missions helped them do just that.
A 2.30 mH toroidal solenoid has an average radius of 6.20 cm and a cross-sectional area of 2.80 cm2.
a) How many coils does it have? In calculating the flux, assume that B is uniform across a cross section, neglect the variation of B with distance from the toroidal axis.
b) At what rate must the current through it change so that a potential difference of 2.60 V is developed across its ends?
(a) The number of turns of the coil is determined as 1,596 turns.
(b) The rate of change of current is determined as 1,130.43 A/s.
Number of turns of the solenoid
L = N²μA/l
where;
L is inductance N is number of turnsA is areal is average length = 2πrN²μA = LI
N² = LI/μA
N² = (2.3 x 10⁻³ x 2π x 0.062)/(4π x 10⁻⁷ x 2.8 x 10⁻⁴)
N² = 2,546,428.6
N = √2,546,428.6
N ≈ 1,596 turns
Rate of current changeL = (emf)/I
I = (emf)/L
I = (2.6)/(2.3 x 10⁻³)
I = 1,130.43 A/s
Thus, the number of turns of the coil is determined as 1,596 turns.
The rate of change of current is determined as 1,130.43 A/s.
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I need help with my homework
C. The center of mass and center of gravity will coincide if the acceleration due to gravity is constant or uniform.
What is center of mass?The center of mass of an object is the unique point where the weighted relative position of the distributed mass sums to zero.
What is center of gravity?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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How big is this restoring force compared with the tensile force stretching the spring?
A. Bigger
B. Not enough info
C. Smaller
D. Same size
The restoring force on the spring is found to have exactly the same magnitude as the stretching force. Option D
What is the restoring force?The restoring force is the force that seeks to restore the spring to its equilibrium position. It has the same magnitude as the stretching force but acts in opposite direction.
Thus, the restoring force on the spring is found to have exactly the same magnitude as the stretching force.
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What is the x-component of a vector with a magnitude of 115 km at an angle of 22°?
The x-component of a vector are < 106.6, 43.07 >
Depending on the angle we are provided, the x-component of a vector can either be cos or sin. Cos always corresponds to the right triangle's side that contacts the specified angle.
If a vector v with magnitude ||v|| makes an angle θ with the positive x-axis then,
v = ||v|| cos θi + ||v|| sin θj
= < ||v|| cos θ , ||v|| sin θ >
Magnitude p = 115 km
Angle = 22°
p = ||p|| < cos θ, sin θ >
p = 115 < cos 22°, sin 22° >
p = 115 < 0.927, 0.3746 >
p = < 106.6, 43.07 >
Therefore, the x-component of a vector are < 106.6, 43.07 >
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Define the term work and state its unit. An ant is dragging a house-fly and the elephant is pushing a big tree which is not moving. Who is doing work, the ant or the elephant? Justify your answer. 922.5 205
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.
Who works, how do find the ant and the elephant?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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Haley is trying to pull an object upward. The below forces are acting on the object.
Fp= 5500 N
Fg= 6000 N
Which represents the net force?
The net force is represented by ↓ 500N.
What is the net force?The net force is the force that has the same effect in magnitude and direction as two or more forces acting together.
Now we have the forces;
Fp= 5500 NFg= 6000 NThus we can obtain the net force as;
5500 N - 6000 N
= - 500 N
Therefore the net force is represented by ↓ 500N.
Missing parts:
Haley is trying to pull an object upward. The below forces are acting on the object.
Fp = 5500N
Fg = 6000N
Which represents the net force?
← 500N
→ 500N
↑ 500N
↓ 500N
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When using a stream table in a classroom setting what are three factors that can be controlled?
Answer:
These include the slope of the land, the nature of the land surface, the placement of dams, and the direction of topsoil disturbance as created by farming activities. Materials: Students should work in groups of 3 or 4, or as materials allow.Explanation:
Hope this helpsA crate with a mass of 175.5 kg is suspended from the end of a uniform boom with a mass of 94.7 kg. The upper end of the boom is supported by a cable attached to the wall and the lower end by a pivot (marked X) on the same wall.
Calculate the tension in the cable.
The tension, T in the cable is equal to 323.5 N.
What is the tension?Tension is force exerted by a cable or string on another object usually a weight suspended from the cable or string
The tension in the cable is found this:
Angle of the boom with horizontal, θ = tan⁻¹(5/10) = 26.56°
The angle of cable with horizontal, B = tan⁻¹(4/10) = 21.80
Taking moments about the pivot:
175.5 * cos 26.56 + 94.7 * cos 26.56 * 0.5 = T (sin(26.56 + 21.80) * 1
T = 241.68/0.747
T = 323.5 N
In conclusion, the tension in the cable is determined by taking moments about the pivot.
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A thin flexible gold chain of uniform linear density has a mass of 17.1 g. It hangs between two 30.0 cm long vertical sticks (vertical axes) which are a distance of 30.0 cm apart horizontally (x-axis), as shown in the figure below which is drawn to scale.
Evaluate the magnitude of the force on the left hand pole.
The Force on the left hand pole, F' = 0.167N
What is the force on the left hand pole?Force is an agent which produces a change in the motion or state of an object.
Force is a vector quantity.
The general force is calculated as follows:
F = mg/sinθ
m = 17.1 g = 0.0171 kg
g = 9.81 m/s²
θ = 45°
F = 0.0171 * 9.81/sin45
F = 0.237 N
Force on the left hand pole, F' = Fcosθ
F' = 0.237 * cos 45
F' = 0.167N
In conclusion, the force on the left hand pole is the horizontal component of force.
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Earth travels around the Sun each year in sn elliptical path, as opposed to a perfect curcle. This means that the speed if earth and its dustance from the Sun change over rhe course if a year. What does this sayabout the magnitude of the centripetal acceleration if earth over the course of a year
A change in the linear speed of the Earth around the sun will cause a change in the magnitude of the centripetal acceleration.
What is centripetal acceleration ?
Centripetal acceleration is the acceleration of a body moving a circular path.
The relationship between centripetal acceleration and speed;
a = v²/r
where;
v is linear speeda is centripetal accelerationr is radius of the pathSince the centripetal acceleration is directly proportional to square of linear speed, a change in the linear speed of the Earth around the sun will cause a change in the magnitude of the centripetal acceleration.
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Based on the way living things are organized ehat level combines to form organ ststems
Answer:
Higher levels of organization are built from lower levels.
Molecules combine to form cells.
Cells combine to form tissues.
Tissues combine to form organs.
Organs combine to form organ systems,
and organ systems combine to form organisms.
Explanation:
Hope it helps.
By how many newtons does the weight of a 100-kg person decrease when he goes from sea level to mountain top at an altitude of 5000 m? The mean radius of the earth is 6.38 × 106 m.
The weight of a 100-kg person decreases by 4 N when he goes from sea level to mountain top at an altitude of 5000 m.
What determines the weight of a person?The weight of a person is determined by the mass of the body and the acceleration due to gravity.
The acceleration due to gravity, g is dependent on the mass of the earth, M the radius of the earth and the gravitational force constant , G.
Mathematically, the acceleration due to gravity at the mountain top is determined using the formula:
g = GM/r²where:
G = 6.67 × 10⁻¹¹ Nm²/kg²
M = 5.9736 x 10²⁴ kg
r = 638000 + 5000 = 6385000
g = (6.67 × 10⁻¹¹ * 5.9736 x 10²⁴ )(6385000)²
g = 9.77 m/s²
His weight at the mountain top will be:
weight = 100 * 9.77
weight = 977 N
Weight at sea level = 100 * 9.81 = 981 N
Decrease in weight = 981 - 977
Decrease in weight = 4 N
In conclusion, the weight of the man varies according to his distance from the earth.
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A hypothetical planet has a mass 2.81 times that of Earth, but the same radius.
What is g near its surface?
The acceleration due to gravity near the surface of the planet is 27.38 m/s².
Acceleration due to gravity near the surface of the planet
g = GM/R²
where;
G is universal gravitation constantM is mass of the planetR is radius of the planetg is acceleration due to gravity = ?g = (6.626 x 10⁻¹¹ x 2.81 x 5.97 x 10²⁴) / (6371 x 10³)²
g = 27.38 m/s²
Thus, the acceleration due to gravity near the surface of the planet is 27.38 m/s².
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Light of intensity I0 is polarized vertically and is incident on an analyzer rotated at an angle from the vertical. Find the angle if the transmitted light has intensity
I = (0.750)I0, I = (0.500)I0, I = (0.250)I0, and I = 0.
(Enter your answers in degrees.)
a. θ = 41. 4°
b. θ = 60°
c. θ = 75. 5°
d. θ = 90°
How to determine the angleFrom the given information, we would be using the Malus' law
It is given as;
I = I0 cos²θ
Where I0 is the intensity of the polarized light after passing through P
a. To find the angle, compare with the given equation
I = (0.750)I0
I = I0 cos θ
then
cos θ = 0. 750
θ = [tex]cos^-^1(0. 750)[/tex]
θ = 41. 4°
b. I = (0.500)I0
cos θ = 0. 500
θ = [tex]cos^-^1(0. 500)[/tex]
θ = 60°
c. I = (0.250)I0
cos θ = 0. 250
θ = [tex]cos^-^1 (0. 250)[/tex]
θ = 75. 5°
d. I = 0
cos θ = 0
θ = [tex]cos^-^1 (0)[/tex]
θ = 90°
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An object with a density of 941.0 kg/m3 and a mass of 1039.0 kg is thrown into the ocean. Find the volume that sticks out of the water. (use ρseawater = 1024 kg/m3)
The volume that sticks out of the water is 83 m³.
To find the answer, we need to know about the archimedes principle.
What's archimedes principle?It says that when an object is on a water surface, the amount of force on the object is equal to the weight of water displaced by it.Mathematically, weight of the object= weight of water displacedWhat's the volume of an object remain on the water surface, if the density and mass of the object are 941.0 kg/m³, 1039.0 kg respectively?Let V = volume of the object, v= volume of water displacedV-v = volume that sticks out of the waterWeight of the object = V× density of object × gWeight of water displaced= v× density of water × gAs per archimedes principle, V× density of object × g=v× density of water × gV-v = density of water - density of object= 1024 - 941 = 83 m³
Thus, we can conclude that the volume that sticks out of the water is 83 m³.
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A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the cable is 90°. If the beam is inclined at an angle of θ = 31.0° with respect to horizontal..
1. What is the horizontal component of the force exerted by the hi.nge on the beam? (Use the `to the right' as + for the horizontal direction.)
2. What is the magnitude of the force that the beam exerts on the hi.nge?
The tension in the cable is 380.55 N and the vertical component of the force exerted by the hi.nge on the beam is 11.45 N.
Tension in the cableApply the principle of moment and calculate the tension in the cable;
Clockwise torque = TL sinθ
Anticlockwise torque = ¹/₂WL
TL sinθ = ¹/₂WL
T sinθ = ¹/₂W
T = (W)/(2 sinθ)
T = (40 x 9.8)/(2 x sin31)
T = 380.55 N
Vertical component of the forceT + F = W
F = W - T
F = (9.8 x 40) - 380.55
F = 11.45 N
Thus, the tension in the cable is 380.55 N and the vertical component of the force exerted by the hi.nge on the beam is 11.45 N.
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