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.
Answers
Answer: 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. Then, the magnitude of the force on the left-hand pole will be, 0.167N.
Explanation: To find the correct answer, we have to know more about the Basic forces that acts upon a body.
What is force and which are the basic forces that acts upon a body?A push or a pull which changes or tends to change the state or rest, or motion of a body is called Force.Force is a polar vector as it has a point of application.Positive force represents repulsion and the negative force represented attraction.There are 3 main forces acting on a body, such as, weight mg, normal reaction N, and the Tension or pulling force.How to solve the problem?We have given that, the gold chain hangs between the vertical sticks of 30cm and the horizontal distance between then is 30cm.From the given data, we can find the angle [tex]\alpha[/tex] (in the free body diagram, it is given as θ).[tex]tan\alpha =\frac{30}{30}[/tex]
[tex]\alpha =tan^{-1}(1)=45[/tex] degree.
From the free body diagram given, we can write the balanced equations of total force along y direction as,[tex]y- direction,\\T_2sin\alpha =mg\\T_2=\frac{mg}{sin \alpha } =\frac{17.1*10^{-3}kg*9.8m/s^2}{sin 45}=0.236 N[/tex]
From the free body diagram given, we can write the balanced equations of total force along x direction as,[tex]x- direction\\T_1-T_2cos\alpha =0\\T_1=0.236*cos45=0.167N[/tex]
Thus, we can conclude that, the magnitude of force on the left-hand pole will be 0.167N.
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Answer:
0.1426 N
Explanation:
A chain of uniform mass density is suspended between two poles 30 cm apart. The geometry of the problem is such that the left support only supplies a horizontal force on the chain. The right support must both balance that horizontal force and supply a vertical force that balances the weight of the chain.
Magnitude of forcesFor some tension T in the chain at the right support, the vertical force will be ...
vertical force = T·sin(α) = W . . . . . matches the weight (W) of the chain
for some angle α between the horizontal and the chain at the right pole.
The corresponding horizontal force is ...
horizontal force = T·cos(α)
This force balances the horizontal force at the left support pole. In terms of W, this force is ...
horizontal force = W/sin(α)·cos(α) = W/tan(α)
AngleThe curve assumed by a chain of uniform mass density can be demonstrated to be a catenary. For supports 30 cm apart, its equation can be described by ...
y = 30·cosh(x/30)
The diagram shows that y=4 for x=0, so we need to subtract 26 cm from this:
y = 30·cosh(x/30) -26
The slope of the curve at any point is the derivative of this function:
y' = 30(1/30)(sinh(x/30)) = sinh(x/30)
At the right support, the slope of the curve is ...
y' = sinh(30/30) = sinh(1) ≈ 1.1752012
This is the tangent of the angle that the curve makes with the horizontal at the right support.
tan(α) = 1.1752012
Note, you can see from the grid squares on the graph that the slope at the right support is slightly more than 1.
WeightThe weight of the chain is the product of its mass and the acceleration due to gravity:
W = ma = (0.0171 kg)(9.8 m/s²) = 0.16758 N
Force on the PoleThen the force on the left-side pole is ...
horizontal force = W/tan(α) = (0.16758 N)/1.1752012
horizontal force ≈ 0.1426 N
__
Additional comment
The attached graph is a plot of the catenary curve we have assumed for the gold chain. We have attempted to match the vertical height on the left side, but we note that there seems to be a small discrepancy at the right side. The graph in the problem statement seems to show the right attach point at about y=21, not 20.3.
Related Questions
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?
Answers
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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Based on the way living things are organized ehat level combines to form organ ststems
Answers
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.
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
Answers
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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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?
Answers
(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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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)
Answers
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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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.
Answers
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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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?
Answers
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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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?
Answers
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 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?
Answers
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].
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.
Answers
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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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?
Answers
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]
Which best describes the results of Becquerel’s experiments?
Answers
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 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.)
Answers
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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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.
Answers
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 hypothetical planet has a mass 2.81 times that of Earth, but the same radius.
What is g near its surface?
Answers
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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the temperature at which the velocity of sound in air is twice its velocity at 15°C
Answers
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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What is the fastest possible speed called in our universe and what is the equation for it?
Answers
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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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?
Answers
(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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Calculate the de Broglie wavelength of a 0.56 kg ball moving with a constant velocity of 26 m/s (about 60 mi/h)
Answers
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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Discuss the aspects of either the Gemini Program or the Soyuz Program.
Answers
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.
please help me with this physics question ASAP
Answers
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
What is the resistance of a carbon rod at 25.8 ∘C if its resistance is 0.0200 Ω at 0.0 ∘C ?
Answers
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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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?
Answers
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]
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.)
Answers
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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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?
Answers
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 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?
Answers
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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a 5.5kg bowling ball has a weight on earth closest to what in N
Answers
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
Answers
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 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?
Answers
When using a stream table in a classroom setting what are three factors that can be controlled?
Answers
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:
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