The value of Fmax = 10-³/2 N
Fmax = 500N
Explanation:
Impulse is the area under force curve
The ramp up and down each have area 1/2*base*height
= 1/2 *2ms*Fmax areas for the combined area (2ms)Fmax
The middle rectangle has area = base * height
= (4ms)Fmax
So, total (6ms)Fmax is equal to 3Ns
Fmax = 3Ns/6ms = 10^3 /2 N
Fmax = 1000/2 = 500N
The value of Fmax gives an impulse of 9.0 Ns is found to be 2.25N
How are the force F, the impulse J, and the time t related?The term "impulse" describes how much of an influence a force has overall over the course of time. J = F t for a constant force. The continuous net force applied on an object throughout time and its duration are both factors that affect impulse.
For a constant force, J = F ⋅ Δ t
Impulse J => 9 Ns
F max at given impulse = 9/Δ t => 9/(6-2)
=> 2.25N
What does F in impulse stand for?F stands for the applied resultant force, t1 and t2 for the start and end of the impulse, respectively, m for the object's mass, v2 for the object's ultimate velocity at the conclusion of the time interval, and. When the time interval starts, the object's starting velocity is v1.
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The light beam shown in the figure below makes an angle of = 24.8° with the normal line NN' in the linseed oil. Determine the angles and '. (The refractive index for linseed oil is 1.48.)
The angle of refraction of the light beam is determined as 16.46 ⁰.
Angle of refraction of the light beamn = sin i / sin r
where;'
n is refractive indexi is angle of incidencer is angle of refractionAngle between the ray and the normal = incident angle = 24.8⁰
1.48 = sin (24.8) / (sin r)
sin r = sin (24.8) / (1.48)
sin r = 0.283
r = sin ⁻¹(0.283)
r = 16.46 ⁰
Thus, the angle of refraction of the light beam is determined as 16.46 ⁰.
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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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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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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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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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It's a snowy day and you're pulling a friend along a
level road on a sled. You've both been taking physics,
so she asks what you think the coefficient of friction
between the sled and the snow is. You've been
walking at a steady 1.5 m/s, and the rope pulls up
on the sled at a 35 ° angle. You estimate that the
mass of the sled, with your friend on it, is 57 kg and
that you're pulling with a force of 75 N
The coefficient of friction between the sled and the snow is 0.119.
To find the answer, we need to know about the friction.
How to find the coefficient of friction between the sled and the snow?Whenever a body moves over the surface of another body, a force come into play, which acts parallel to the surface of contact and oppose the relative motion. This opposing force is called friction.To solve the problem, we have to draw the free body diagram of the given system.We have given with the following values,[tex]a=0\\\alpha =35^0\\T=75N\\m=57kg[/tex]
Here, acceleration will be equal to zero, because the velocity is given as constant.
Thus, from the diagram, we can write the balancing equations as follows,[tex]ma=Tcos\alpha -f\\\where\\f=kN\\\N+Tsin\alpha=mg\\Thus,\\N=mg-Tsin\alpha[/tex]
Substituting N in f and f in the equation of ma, then we get,[tex]ma= Tcos\alpha -k(mg-Tsin\alpha )[/tex]
Substituting values, we get the coefficient of friction as,[tex]0=(75*cos35)-k((57*9.8)-(75sin35))\\\\k((57*9.8)-(75sin35))=(75*cos35)\\\\515.6k=61.44\\\\k=\frac{61.44}{515.6}=0.119[/tex]
Thus, we can conclude that, the coefficient of friction between the sled and the snow is 0.119.
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The sled's coefficient of friction with the snow is 0.119.
We must understand the friction in order to choose the solution.
How can I determine the sled and snow's coefficient of friction?A force that works parallel to the surface of contact and opposes the relative motion is present whenever one body moves over the surface of another body. Friction is the name for this opposing force.We must create the given system's free body diagram in order to solve the issue.The values that we have provided are[tex]\alpha =35\\T=75N\\m=57kg\\a=0[/tex]
Because the velocity is specified as constant in this case, the acceleration will be equal to zero.
Consequently, we can express the balancing equations as follows using the diagram:[tex]ma=Tcos\alpha -f\\ where,f=kN\\N+Tsin\alpha =mg\\ thus,\\N=mg-Tsin\alpha[/tex]
When we substitute N for f and f in the equation for ma, we obtain,[tex]ma=Tcos\alpha -k(mg-Tsin\alpha )[/tex]
By substituting values, we obtain the friction coefficient as.[tex]k=0.119[/tex]
As a result, we may say that there is 0.119 coefficient of friction between the sled and the snow.
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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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A truck is carrying a refrigerator as shown in the figure. The height of the refrigerator is 158.0 cm, the width is 60.0 cm. The center of gravity of the refrigerator is 67.0 cm above the midpoint of the base.
What is the maximum acceleration the truck can have so that the refrigerator doesn't tip over? (The static friction between the truck and the refrigerator is very large.)
The maximum acceleration the truck can have so that the refrigerator does not tip over is 1.86m/s².
What is acceleration?Acceleration is the change in velocity with time.
The maximum acceleration is obtained by taking moments about the tipping point of rotation.
Moment = force * perpendicular distanceClockwise moment = Anticlockwise momentF₂ * 1.58 m = F₁ * 0.67 m
where
F₂ is tipping force = mass * acceleration, a
F₁ is weight = mass * acceleration due to gravity, g
The weight acts at a distance half the width of the refrigerator = 30 cm or 0.3 m
Height of refrigerator is 158 cm 0r 1.58 cm
m * a * 1.58 = m * 9.81 * 0.30
a = 1.86 m/s²
The maximum acceleration the truck can have so that the refrigerator does not tip over is 1.86m/s².
In conclusion, if the maximum acceleration is exceeded, the refrigerator will tip over.
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please help me with this physics question ASAP
Answer:
See below
Explanation:
With switches open, the circuit is a simple series circuit ....the ammeters will have the same readings
V = IR
I = V/R = 5 / (10+5+5) = .25 A
b) With S1 closed 5 ohm and 10 ohm in parallel become = 5 *10 / (5+10) = 3.33 ohm
then the series circuit current becomes
5 v / ( 10 + 3.33 + 5 ) = ammeter 1 = .273 amps
ammeter 2 will get a portion of this ...the smaller resistor will get 2/3 ...the 10 ohm resistor will get 1/3 .273 * 10 / 15 =.182 amps
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 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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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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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].
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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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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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 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]
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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A boat moves through the water with two forces acting on it. One is a 1,575-N forward push by the water on the propeller, and the other is a 1,200-N resistive force due to the water around the bow. (Review attachment)
(a) What is the acceleration of the 1,100-kg boat?
______ m/s2
(b) If it starts from rest, how far will the boat move in 20.0 s?
______ m
(c) If it starts from rest, how far will the boat move in 20.0 s?
______ m/s
(a) The acceleration of the 1,100-kg boat is 0.341 m/s².
(b) The distance covered by the boat is 68.2 m.
(c) The speed of the boat is 6.82 m/s.
Acceleration of the boat
Net force on the boat = 1,575 N - 1,200 N = 375 N
F(net) = ma
a = F(net)/m
a = 375/1100
a = 0.341 m/s²
Distance moved in 20 ss = ut + ¹/₂at²
s = 0 + ¹/₂(0.341)(20)²
s = 68.2 m
Speed of the boat in 20 sv = u + at
v = 0 + 0.341(20)
v = 6.82 m/s
Thus, the acceleration of the 1,100-kg boat is 0.341 m/s², the distance covered by the boat is 68.2 m and the speed of the boat is 6.82 m/s.
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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.
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.
Checking the what will reduce the possibility of having to rebuild or replaced the engine?
Checking the oil will reduce the possibility of having to rebuild or replaced the engine.
what is an engine ?
A device created to transform one or more sources of energy into mechanical energy is known as an engine or motor. Potential energy, heat energy, chemical energy, electric potential, and nuclear energy are all forms of energy that are readily available.
The lubricating function of engine oil is crucial. It shields and stops all the moving parts from rubbing against one another. Metal-on-metal wear would quickly kill your engine without lubrication.
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What is the resistance of a carbon rod at 25.8 ∘C if its resistance is 0.0200 Ω at 0.0 ∘C ?
0.02020 ohm is the resistance of a carbon rod at 25.8 ∘C if its resistance is 0.0200 Ω at 0.0 ∘C.
What is a resistor?A resistor is an electrical component that controls or restricts how much electrical current can pass across a circuit in an electronic device. A specified voltage can be supplied via resistors to an active device like a transistor.
The temperature of the resistor varies based on the variation in the temperature. The equation that describes the relationship between the two of them is:
R = R0[1+ alpha(T-T0)] where:
R is the new resistance we are looking for
alpha is the temperature coefficient of resistance. For carbon rod, alpha = ₋ 4.8 x [tex]10^{-4}[/tex](1/°c)
T0 is the standard temperature =25.8°C
R0 is the resistance at T0 = 0.0200 ohms
T is the temperature at which we want to get R = 0
Substitute in the equation to get R as follows:
R = 0.0200 [1+( ₋ 4.8 x [tex]10^{-4}[/tex]) (0-25.8)] = 0.02020 ohm
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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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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 helpsWhat is the fastest possible speed called in our universe and what is the equation for it?
Answer:
The speed of light traveling through a vacuum is exactly 299,792,458 meters (983,571,056 feet) per second. That's about 186,282 miles per second — a universal constant known in equations as "c," or light speed.
[tex]s\frac{d}{t}[/tex]
Explanation:
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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). By what angle has planet Y rotated through during this time?
Answer: 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). Then, at an angle 135.48°, the planet Y rotated through during this time.
Explanation: To find the answer, we need to know about the Kepler's third law of planetary motion.
What is Kepler's third law of planetary motion?Kepler's third law of planetary motion states that, the square of the period of revolution is proportional to the cube of the orbital radius of the elliptical path.It can be expressed as,T² ∝ r³
How to solve the problem?We have given with the ratio of the radii of their orbits as,4:3.planet X rotated through an angle of 88°.thus,[tex]\frac{r_1}{r_2}=\frac{4}{5} \\\frac{T_1}{T_2} =(\frac{r_1}{r_2})^{3/2}\\[/tex]
As we know that,[tex]T=\frac{2\pi }{w}[/tex] where, w is the angular velocity.
Angular displacement is the angle swept by the position vector of a particle in a given interval of time.[tex]\alpha[/tex] =wt.
We can rewrite our equation as,[tex]\frac{T_x}{T_y}=\frac{w_y}{w_x}\\thus,\\\frac{w_y}{w_x}=(\frac{r_1}{r_2})^{3/2}[/tex]
We have to find the angle rotated by planet Y during 5 yrs. So, we can rewrite the above equation in terms of angular displacement.[tex]\frac{\alpha _y}{\alpha _x} = (\frac{r_1}{r_2})^{3/2}\\where,\\\alpha _x=\frac{88^0}{5 yrs} because,\\here, angle \beta_x =88^0.\\[/tex]
Thus, the angle rotated by planet Y during 5 yrs will be [tex]\beta _y[/tex] =[tex]\alpha _y=\frac{88}{5yrs} *(\frac{4}{3} )^{3/2}=\frac{135.48^0}{5yrs} .\\thus,\\\beta _y=135.48^0.[/tex]
Thus, we can conclude that the angle rotated by planet Y during 5 yrs will be 135.48 degrees.
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The planet Y then rotated through at this time at an angle of 135.48°.
In order to understand the solution, we must be familiar with Kepler's third law of planetary motion.
What does the third law of planetary motion by Kepler say?According to Kepler's third law of planetary motion, the elliptical path's orbital radius is proportional to the cube of the square of the revolution's period.It can be stated as follows:T² ∝ r³
How can the issue be resolved?The ratio of their orbital radii that we have provided is 4:3.Planet X rotated at an 88° angle. thus,[tex]\frac{R_1}{R_2}=\frac{4}{5} \\\frac{T_1}{T_2}=(\frac{4}{5} ) ^{\frac{3}{2} }[/tex]
As we are aware,[tex]T=\frac{2\pi }{w}[/tex]
where w is the angle of rotation per time.
The angle that a particle's position vector sweeps over in a specific amount of time is known as the angular displacement.[tex]\alpha[/tex]=wt.
Our equation can be rewritten as,[tex]\frac{w_y}{w_x} =(\frac{4}{5} ) ^{\frac{3}{2} }[/tex]
We have to find the angle that planet Y rotated at over the course of five years. Consequently, we can express the equation above in terms of angular displacement.[tex]\frac{\alpha _y}{\alpha _x}=(\frac{4}{3} ) ^{\frac{3}{2} } , where\\\alpha _x=\frac{88}{5yrs} \\[/tex]
So, during a period of five years, planet Y will rotate at an angle,[tex]\alpha _y=\frac{88}{5yrs} *(\frac{4}{3} ) ^{\frac{3}{2} }=\frac{135.48}{5yrs}[/tex]
Thus, we may infer that planet Y will revolve at an angle of 135.48 degrees during the course of five years.
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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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if f=30 m=4 calculate value of m in equation
m=f/a
Answer:
a = 7.5
Explanation:
m = f/a
f = 30
m = 4
Thus;
a = f/m
a = 30/4
a = 7.5