Answer:
[tex]\Delta E=58800\ J[/tex]
Explanation:
Given:
mass of lead chunk, [tex]m=200\ kg[/tex]
height of the fall, [tex]h= 30\ m[/tex]
Assuming the collision to be perfectly inelastic the whole kinetic energy of the chunk is lost after the collision.Here when the chunk is at the given height it has a potential energy which on falling transforms into kinetic energy of the chunk.So, increase in the internal energy of the system after the collision is :
[tex]\Delta E=m.g.h[/tex]
[tex]\Delta E=200\times 9.8\times 30[/tex]
[tex]\Delta E=58800\ J[/tex]
Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon's tongue accelerates at a remarkable 280m/s2 m/s^2 for 20 ms {\rm ms}, then travels at constant speed for another 30 ms {\rm ms}.
During this total time of 50 ms {\rm ms}, 1/20 of a second, how far does the tongue reach?
Express your answer to two significant figures and include the appropriate units.
Answer:
The reach of the tongue is 23 cm.
Explanation:
Hi there!
The equation of traveled distance is the following:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = traveled distance at time t.
x0 = initial position.
v0 = initial velocity.
a = acceleration.
t = time.
Let´s calculate the distance traveled by the tongue of the chameleon during the first 20 ms (0.020 s):
The initial position and velocity are zero (x0 = 0 and v0 = 0)
x = 1/2 · a · t²
x = 1/2 · 280 m/s² · (0.020 s)²
x = 0.056 m
Now, let´s find the distance traveled while the tongue moves at constant speed. But first, let´s find the velocity (v) of the tongue after the accelertation interval using the following equation:
v = v0 + a · t (v0 = 0)
v = 280 m/s² · 0.020 s
v = 5.6 m/s
Then, the distance traveled at constant speed can be calculated:
x = v · t
x = 5.6 m/s · 0.030 s
x = 0.17 m
The reach of the tongue is 0.17 m + 0.056 m = 0.23 m = 23 cm.
Two infinite nonconducting sheets of charge are parallel to each other, with sheet A in the x = -2.15 plane and sheet B in the x = +2.15 m plane. Find the electric field in the region x < -2.15 m, in the region x > +2.15 m, and between the sheets for the following situations.
(a) when each sheet has a uniform surface charge density equal to +3.25 µC/m2 region (m) electric field (N/C)
x < -2.15 ________________
x > +2.15 ________________
-2.15 < x < +2.15 _________________
(b) when sheet A has a uniform surface charge density equal to +3.25 µC/m2 and sheet B has a uniform surface charge density equal to -3.25 µC/m2
region (m) electric field (N/C)
x < -2.15 __________________
x > +2.15 __________________
-2.15 < x < +2.15 _____________________
Answer:
a) (-367231.63i , 367231.63i, 0) N/C
b) (0 , 0 , 367231.63i ) N/C
Explanation:
a)
Case x < -2.15
[tex]E =E_{+} + E_{+} \\E = 2*\frac{sigma}{2e_{0} } = \frac{sigma}{e_{0} }\\\\E = \frac{3.25*10^(-6)}{8.85*10^(-12) }\\\\E = - 367231.63 i[/tex]
Case x > 2.15
[tex]E =E_{+} + E_{+} \\E = 2*\frac{sigma}{2e_{0} } = \frac{sigma}{e_{0} }\\\\E = \frac{3.25*10^(-6)}{8.85*10^(-12) }\\\\E = +367231.63 i[/tex]
Case -2.15 < x <+2.15
[tex]E =E_{+} - E_{+} \\\\E = 0[/tex]
b)
Case x < -2.15
[tex]E =E_{+} - E_{+} \\\\E = 0[/tex]
Case x > 2.15
[tex]E =E_{+} - E_{+} \\\\E = 0[/tex]
Case -2.15 < x <+2.15
[tex]E =E_{+} + E_{-} \\E = 2*\frac{sigma}{2e_{0} } = \frac{sigma}{e_{0} }\\\\E = \frac{3.25*10^(-6)}{8.85*10^(-12) }\\\\E = +367231.63 i[/tex]
The gravitational field of m1 is denoted by g1. Enter an expression for the gravitational field g1 at position la in terms of m1, la, and the gravitational constant G.
Answer:
The expression of gravitational field due to mass [tex]m_![/tex] at a distance [tex]l_a[/tex]
Explanation:
We have given mass is [tex]m_1[/tex]
Distance of the point where we have to find the gravitational field is [tex]l_a[/tex]
Gravitational constant G
We have to find the gravitational filed
Gravitational field is given by [tex]g=\frac{Gm_1}{l_a^2}[/tex]
This will be the expression of gravitational field due to mass [tex]m_![/tex] at a distance [tex]l_a[/tex]
The expression for the gravitational field g₁ at position la in terms of m₁, la, and the gravitational constant G is:
[tex]g_1 = \frac{G\times m_1}{la^{2} }[/tex]
The gravitational field is the force field that exists in the space around every mass or group of masses.
The gravitational field of m₁ is denoted by g₁, and can be represented through the following expression.
[tex]g_1 = \frac{F_1}{m}[/tex] [1]
where,
F₁ is the gravitational force due to m₁m is the mass of the object at a distance la from m₁We can calculate the force (F₁) between m₁ and m that are at a distance "la" using Newton's law of universal gravitation.
[tex]F_1 = G \frac{m \times m_1 }{la^{2} }[/tex] [2]
where,
G is the gravitational constantIf we replace [2] in [1], we get
[tex]g_1 = \frac{G \frac{m \times m_1 }{la^{2} }}{m} = \frac{G\times m_1}{la^{2} }[/tex]
The expression for the gravitational field g₁ at position la in terms of m₁, la, and the gravitational constant G is:
[tex]g_1 = \frac{G\times m_1}{la^{2} }[/tex]
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What is the magnitude of the electric force of attraction between an iron nucleus (q=+26e) and its innermost electron if the distance between them is 1.5×10−12m?
Final answer:
The magnitude of the electric force of attraction between an iron nucleus and its innermost electron can be calculated using Coulomb's law.
Explanation:
The magnitude of the electric force of attraction between an iron nucleus and its innermost electron can be calculated using Coulomb's law. Coulomb's law states that the magnitude of the electric force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. In this case, the charge of the iron nucleus (+26e) is given as 26 times the charge of an electron (e).
The formula to calculate the electric force is: F = k * |q1 * q2| / r^2, where F is the electric force, k is the electrostatic constant, q1 and q2 are the charges of the particles, and r is the distance between them.
Using this formula, we can plug in the values: F = (8.99 × 10^9 Nm^2/C^2) * |(26e) * (-e)| / (1.5 × 10^(-12)m)^2. Simplifying this equation will give you the magnitude of the electric force.
A cart moves toward a motion sensor on a track in three different ways: speeding up with constant speed slowing to a stop and turning around to speed up in the opposite direction For each of these cases, what is the sign of the velocity and what is the sign of the acceleration at the beginning of the trip and at the end of the trip?
Answer:
Acceleration of the cart in the beginning is positive
Speed is positive when speeding up and slowing down.
Speed is zero when the cart stops.
Speed is negative when the cart speeds up in opposite direction.
Acceleration is negative when speeding up in opposite direction from rest.
Explanation:
When the cart is speeding up then its speed is increasing with time hence its acceleration is having a positive value.
On speeding up the velocity is positive.On gradually decreasing the speed and coming to a stop the velocity is still positive with continuously decreasing magnitude until it reaches zero when the cart stops.When the cart speeds up in the opposite direction then the velocity of the cart becomes negative from zero and continuously increases in the negative value over a span of time hence its acceleration is also negative.what force acting for .00135 s will change the velocity of a 95 g baseball from 50 m/s eastward to 45 m/s eastward
Answer:
Force acting on the basketball will be 351.85 N
Explanation:
We have given velocity of baseball is changes from 50 m/sec to 45 m /sec in 0.00135 sec
So initial; velocity of the ball u = 50 m /sec
And final velocity of the ball v = 45 m /sec
Mass of the basketball m = 95 gram = 0.095 kg
Time taken t = 0.00135 sec
From third equation of motion v = u+at
So [tex]45=50+a\times 0.00135[/tex]
a = 3703.70 [tex]m/sec^2[/tex]
We have to find the force needed to change the speed of the basketball
According to newtons law force is given by F = ma
So F = 0.095×3703.70 = 351.85 N
a) Estimate how long it took King Kong to fall straight down from the top of the Empire State Building (380 m high). Express your answer using two significant figures.
b) Estimate his velocity just before "landing." Express your answer using two significant figures. Assume that the positive direction is upward.
Answer:
a) 8.8 sec b) -86 m/sec
Explanation:
Assuming King Kong started from rest its fall, once in the air, neglecting air resistance, is only affected by gravity, which accelerates it downward.
As this acceleration is constant, we can use the following equation in order to get how long it was falling:
Δh = 1/2*g*t² ⇒ -380 m = 1/2 (-9.8 m/s²)*t²
Solving for t:
t = √((2*380)/9.8) s² = 8.8 sec.
b) In order to know the value of the velocity in the instant just before it hits the ground, we can apply acceleration definition, as follows:
a = (vf-v₀) /t
In our case, a = -g (assuming positive direction is upward) and v₀=0, so, we can solve for vf as follows:
vf = -g*t = -9.8 m/s²*8.8 sec = -86 m/s.
The answer explains the estimated time for King Kong to fall from the Empire State Building and his velocity just before landing using physics formulas.
a) Estimate: Using the formula h = 0.5 * g * t^2 where h is the height, g is gravity (9.8 m/s^2), we find t = sqrt(2h/g) = sqrt(2*380/9.8) = 8.7 s. Therefore, King Kong took approximately 8.7 seconds to fall.
b) Velocity: To find the velocity just before landing, we use v = g*t where v is the final velocity. Substituting values, v = 9.8 * 8.7 = 85.3 m/s. Therefore, his velocity just before 'landing' was around 85 m/s.
Which of the following changes to the earth-sun system would reduce the magnitude of the force between them to one-fourth the value found in Part A?
Reduce the mass of the earth to one-fourth its normal value.
Reduce the mass of the sun to one-fourth its normal value.
Reduce the mass of the earth to one-half its normal value and the mass of the sun to one-half its normal value.
Increase the separation between the earth and the sun to four times its normal value.
part A was:
Consider the earth following its nearly circular orbit (dashed curve) about the sun. The earth has mass mearth=5.98
Answer:
Reduce the mass of the earth to one-fourth its normal value.
Reduce the mass of the sun to one-fourth its normal value.
Reduce the mass of the earth to one-half its normal value and the mass of the sun to one-half its normal value.
Explanation:
Every particle in the universe attracts any other particle with a force that is directly proportional to the product of its masses and inversely proportional to the square of the distance between them. So, in this case we have:
[tex]F=\frac{Gm_Em_S}{d^2}[/tex]
If [tex]m'_E=\frac{m_E}{4}[/tex]:
[tex]F'=\frac{Gm'_Em_S}{d^2}\\F'=\frac{G(\frac{m_E}{4})m_S}{d^2}\\F'=\frac{1}{4}\frac{Gm_Em_S}{d^2}\\F'=\frac{1}{4}F[/tex]
If [tex]m'_S=\frac{m_S}{4}[/tex]
[tex]F'=\frac{Gm_Em'_S}{d^2}\\F'=\frac{Gm_E(\frac{m_S}{4})}{d^2}\\F'=\frac{1}{4}\frac{Gm_Em_S}{d^2}\\F'=\frac{1}{4}F[/tex]
If [tex]m'_E=\frac{m_E}{2}[/tex] and [tex]m'_S=\frac{m_S}{2}[/tex]:
[tex]F'=\frac{Gm'_Em'_S}{d^2}\\F'=\frac{G(\frac{m_E}{2})(\frac{m_S}{2})}{d^2}\\F'=\frac{1}{4}\frac{Gm_Em_S}{d^2}\\F'=\frac{1}{4}F[/tex]
Final answer:
To reduce the gravitational force between the Earth and the Sun to one-fourth, you can either reduce the mass of either body to one-fourth, or increase the distance between them by a factor of four. Reducing the mass of both bodies to one-half would also achieve the same result, as the product of their masses would be one-fourth of the original value.
Explanation:
The force between the Earth and the Sun can be described by Newton's law of universal gravitation, which states that the force (F) is proportional to the product of the two masses (m1 and m2) divided by the square of the distance (r) between their centers of mass. The law is formulated as F = G × m1 × m2 / r2, where G is the gravitational constant.
To reduce the magnitude of the gravitational force between the Earth and the Sun to one-fourth, one could either reduce the product of the masses by one-fourth or increase the separation distance by four times, because the force is inversely proportional to the square of the distance. Thus, the correct answers would be:
Reduce the mass of the Earth to one-fourth its normal value.
Reduce the mass of the Sun to one-fourth its normal value.
Increase the separation between the Earth and the Sun to four times its normal value.
However, the third option offered in the question, reducing the mass of both the Earth and the Sun to one-half their normal values, would also result in reducing the force to one-fourth, because (1/2) × (1/2) = 1/4.
planet tatoone is about 1.7 au from its sunApproximately how long will it take for light to travel from the sun to tatoone in minutesspeed of light 3 x 10^8 m /s1.7 au x 1.496 x 10^11 meters / 1 x 1 / 60 minutes = 42 (rounded to two digits)
Answer:
Light takes 14.12 minutes to travel from the Sun to tatoone.
Explanation:
The equation for the average velocity can be used to estimate the time that light will take to travel from the Sun to tatoone. The average velocity is defined as:
[tex]v = \frac{d}{t}[/tex] (1)
Where v is the velocity, d is the covered distance and t is the time.
Therefore, t can be isolated from equation 1:
[tex]t = \frac{d}{v}[/tex]
It is necessary to express the speed of light in terms of minutes:
[tex]3x10^{8} \frac{m}{s} . \frac{60 s}{1 min}[/tex] ⇒ [tex]1.8x10^{10} m/min[/tex]
An astronomical unit is defined as the distance between the Earth and the Sun ([tex]1.496 x 10^{11} m[/tex]).
[tex]d = 1.7 AU \cdot \frac{1.496 x 10^{11} m}{1AU}[/tex] ⇒ [tex]2.5432x10^{11}[/tex]m
Finally, equation 2 can be used:
[tex]t = \frac{d}{v}[/tex]
[tex]t = \frac{2.5432x10^{11}m}{1.8x10^{10} m/min}[/tex]
t = 14.12 min
Hence, light takes 14.12 minutes to travel from the Sun to tatoone.
PART ONE
A ladder rests against a vertical wall. There
is no friction between the wall and the ladder.
The coefficient of static friction between the
ladder and the ground is µ = 0.464 .
(USE THE PICTURE TO DETERMINE THE ANSWER)
Identify the set of equations which is correct.
ANSWER CHOICES:
1. A1, B2, C3
2. A2, B2, C1
3. A1, B1, C1
4. A1, B2, C2
5. A1, B1, C2
6. A2, B1, C3
7. A2, B1, C2
8. A1, B2, C1
9. A1, B1, C3
10. A2, B1, C1
PART TWO
Determine the smallest angle θ for which the
ladder remains stationary.
Answer in units of ◦
Answer:
1. A1, B2, C3
2. 47.1°
Explanation:
Sum of forces in the x direction:
∑Fₓ = ma
f − Fᵥᵥ = 0
f = Fᵥᵥ
Sum of forces in the y direction:
∑Fᵧ = ma
N − W = 0
N = W
Sum of moments about the base of the ladder:
∑τ = Iα
Fᵥᵥ h − W (b/2) = 0
Fᵥᵥ h = ½ W b
Fᵥᵥ (l sin θ) = ½ W (l cos θ)
l Fᵥᵥ sin θ = ½ l W cos θ
The correct set of equations is A1, B2, C3.
At the smallest angle θ, f = Nμ. Substituting into the first equation, we get:
Nμ = Fᵥᵥ
Substituting the second equation into this equation, we get:
Wμ = Fᵥᵥ
Substituting this into the third equation, we get:
l (Wμ) sin θ = ½ l W cos θ
μ sin θ = ½ cos θ
tan θ = 1 / (2μ)
θ = atan(1 / (2μ))
θ = atan(1 / (2 × 0.464))
θ ≈ 47.1°
Equation of equilibrium is the static or dynamic equilibrium of all internal and external forces present in the system.
The set A1, B2, C3 is correct.The smallest angle θ for which the ladder remains stationary is [tex]47.1^o[/tex]Given information-
A ladder rests against a vertical wall.
The coefficient of static friction between the ladder and the ground is 0.464.
What is equation of equilibrium?Equation of equilibrium is the static or dynamic equilibrium of all internal and external forces present in the system.
A) Set of equations which is correct.From the equation of equilibrium of two bodies, the net force in x axis can be given as,
[tex]\sum F_x=0[/tex]
As the normal force and friction force acting on the x-axis. Thus,
[tex]-f_w+f=0[/tex]
[tex]f=f_w[/tex]
Thus option A1 is correct.
As the gravitational force(due to weight of the body) is acting in the vertical direction. Thus from the equation of equilibrium of two bodies, the net force in y axis can be given as,
[tex]\sum F_y=W[/tex]
As the normal force acting on the x-axis. Thus,
[tex]N=W[/tex]
Thus option B2 is correct.
Apply torque equation at the base of the ladder,
[tex]\sum \tau=Ia[/tex]
[tex]F_wh=\dfrac{1}{2} Wb[/tex]
Here the value of h and b can be changed in the form of I using the trigonometry formula. Thus,
[tex]F_w(I\sin \theta)=\dfrac{1}{2} W(\cos \theta[/tex]
Thus option C3 is correct.
Hence the set A1, B2, C3 is correct.
B) The smallest angle θ for which the ladder remains stationary-The equation of option C3 is,
[tex]F_w(I\sin \theta)=\dfrac{1}{2} W(\cos \theta[/tex],
As normal force is can be given as,
[tex]F_w=W\mu[/tex]
Thus,
[tex]W\mu(I\sin \theta)=\dfrac{1}{2} W(\cos \theta[/tex]
[tex]\dfrac{\sin \theta}{\cos \theta} =\dfrac{1}{2\mu} \\\tan \theta =\dfrac{1}{2\times 0.464} \\\theta =\tan^-(1.0770)\\\theta=47.1^o[/tex]
Hence the smallest angle θ for which the ladder remains stationary is [tex]47.1^o[/tex]
Hence,
The set A1, B2, C3 is correct.The smallest angle θ for which the ladder remains stationary is [tex]47.1^o[/tex]Learn more about the equation of equilibrium here;
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A spelunker is surveying a cave. She follows a passage 120 mm straight west, then 250 mm in a direction 45ââ east of south, and then 280 mm at 30ââ east of north. After a fourth unmeasured displacement, she finds herself back where she started.
(A) Use a scale drawing to determine the magnitude of the fourth displacement. Express your answer using two significant figures.
(B) Determine the direction of the fourth displacement.Express your answer using two significant figures.
Answer:
R = 207.45 mm , θ_return = 18.47 south west
Explanation:
This vector addition exercise is schematized in the attachment where the displacements are
1 d1 = 120 mm west
2 d2 = 250mm at 45 south east
3 d3 = 280 mm at 30 east of nort.
R is the final displacement that takes the goat to its initial point (origin)
The analytical way to perform this exercise is to find the components of each displacement and add them
Decompose the displacement using trigonometry
Displacement d1
d1ₓ = 120 cos 180 = -120 mm
Displacement d2, with the angle measured from the axis this θ = 270 + 45
sin 45 = [tex]d2_{y}[/tex] / d2
cos 45 = d2ₓ / d2
[tex]d2_{y}[/tex] = d2 sin45
[tex]d2_{y}[/tex] = 250 sin (270 + 45)
[tex]d2_{y}[/tex] = -176.77 mm
d2ₓ = d2 cos (270 + 45)
d2ₓ = 176.77 mm
displacement d3, for half the angle from the east axis θ = 90-30 = 60
sin 60 = [tex]d3_{y}[/tex] / d3
cos 60 = d3ₓ / d3
[tex]d3_{y}[/tex] = d3 sin 60
d3ₓ = d3 cos 60
[tex]d3_{y}[/tex] = 280 sin 60 = 242.49 mm
d3ₓ = 280 cos 60 = 140 mm
Having all the displacement components we can find the total displacement
Rₓ = d1ₓ + d2ₓ + d3ₓ
Ry = [tex]d1_{y}[/tex] + [tex]d2_{y}[/tex] + [tex]d3_{y}[/tex]
Rₓ = -120 + 176.77 +140
Rₓ = 196.77 mm
Ry = 0 -176.77 +242.49
Ry = 65.72 mm
Therefore the displacement you must make to return to the starting point is
R = RA Rx2 + Ry2)
R = RA (196.77 2 + 65.72 2)
R = 207.45 mm
We used trigonometry
tan tea = RY / Rx
tea = tan-1 Ry / Rx
ea = tan-1 (65.72 / 196.77)
tea = 18.47
This is the point where the girl is, to return to its origin this path must be serial, but in the opposite direction,
θ_return = 18.47 south west
The problem involves adding vectors to find the total displacement, then finding the negative of this displacement to calculate the magnitude and direction of the fourth displacement.
Explanation:This problem involves the mathematical concept of vectors, particularly in determining the resultant vector, which in this context represents the spelunker's path. First, we convert the movements into rectangular coordinates: going 120 mm west is -120i, going 250 mm 45° east of south is -250 cos(45)i - 250 sin(45)j, and going 280 mm 30° east of north is 280 cos(30)i + 280 sin(30)j. Adding these vectors together, we get the spelunker's total displacement vector. The negatives of this sum will represent the fourth displacement needed to get the spelunker back to where she started.
(A) The magnitude of the fourth displacement is the sum of these vectors taken as negatives, because she has to go back, which is equivalent to the length of the path taken. This can be calculated using Pythagoras' theorem for two dimensions.
(B) The direction of the fourth displacement is calculated by finding the angle made by the resulting vector with respect to one of the axes (for instance, the x-axis). For this, we take the inverse tangent of the y-coefficient over the x-coefficient of the vector.
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When a moving car is brought to a stop with the brakes, its acceleration vector is 29. (a) in the same direction as its velocity vector b) in the opposite direction as its velocity vector c) equal to zero (d) directed downward
To develop this problem we will apply the concepts related to the kinematic equations of motion, specifically that of acceleration. Acceleration can be defined as the change of speed in an instant of time, mathematically this is
[tex]\vec{a} = \frac{\vec{v_2}-\vec{v_1}}{\Delta t}[/tex]
If a mobile is decreasing its speed (it is slowing down), then its acceleration is in the opposite direction to the movement. This would imply that the acceleration vector is opposite to the velocity vector.
Therefore the correct answer is B.
Work is done on an object as you drop it from the second floor of a building to the ground floor. True False.
Answer:
True, the work is done by the gravitational force on the object being dropped from the second floor of a building.
Explanation:
When an object is dropped from the second floor of a building then no force is applied by the person on the object but under the influence of gravity the object experience a force equal to its own weight and travels a displacement from the second floor to the ground.
Hence in this case the work is done by the gravitational force but not by the person dropping the object.
As we know that the work done is given as:
[tex]W=F.s[/tex]
where:
F = force on the object getting displaced by a length s
Here:
[tex]F=m.g[/tex]
where, m = mass of the object
g = acceleration due to gravity
At the surface of the moon, the gravitational attraction is 5.37 ft/s2. What is the mass of the object on the surface of the moon
Answer:
mass remains the same
Explanation:
Mass is the amount of matter in a substance which is independent from the external environment; hence, fields!
The weight of the object changes but mass remains same!
Final answer:
The question incorrectly asks for an object's mass on the moon based on gravitational acceleration. Mass is a constant and doesn't change with location. The gravitational acceleration on the moon (about 1.62 m/s²) affects an object's weight, not its mass.
Explanation:
The question seems to be asking for the mass of an object on the moon's surface based on the gravitational acceleration at the moon, which is a misunderstanding. The gravitational attraction on the surface of the moon is given as 5.37 ft/s2 (equivalent to about 1.63 m/s2, since the accurate value is 1.62 m/s2), but to find the mass of an object, we need to discuss how weight and mass are related in a gravitational field, not to calculate the mass based solely on the gravitational acceleration. To find an object's mass on the moon, one would use the formula Weight = Mass × Gravitational acceleration (W = mg). However, the mass of an object is a constant and does not change depending on its location, whether on Earth, the moon, or elsewhere. What changes is the weight of the object due to the difference in gravitational force exerted on it. For example, if an object weighs 9.8 N on Earth, it would weigh approximately 1.6 N on the moon due to the moon's lower gravity.
A lightning bolt may carry a current of 1.00 104 A for a short time. What is the resulting magnetic field 120 m from the bolt
Answer:
B = 1.67 μ T
Explanation:
given,
current, I = 1 x 10⁴ A
r = 120 m
treating lightning bolt as long straight conductor
[tex]B=\dfrac{\mu_0I}{2\pi r}[/tex]
[tex]B=\dfrac{4\pi \times 10^{-7}\times 1 \times 10^4}{2\pi\times 120}[/tex]
resulting magnitude would be equal to
B = 16.67 x 10⁻⁶ T
B = 1.67 μ T
The resulting magnetic field is equal to B = 1.67 μ T
Answer:
1.67 x 10^-5 Tesla
Explanation:
Current, i = 1 x 10^4 A
distance, d = 120 m
The formula for the magnetic field due to long straight current carrying conductor is given by
[tex]B =\frac{\mu _{0}}{4\pi }\times \frac{2i}{r}[/tex]
[tex]B =10^{-7}\times \frac{2\times 10^{4}}{120}[/tex]
B = 1.67 x 10^-5 Tesla
Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject material as high as 500 km (or even higher) above the surface. Io has a mass of 8.93×1022 kg and a radius of 1821 km . For this calculation, ignore any variation in gravity over the 500-km range of the debris.A. How high would this material go on earth if it were ejected with the same speed as on Io?
The height that an object reaches in a gravitational field is dependent on its initial speed and the strength of the field. With the same initial speed, an object in a weaker gravitational field will reach a greater height. Therefore, material ejected with the same speed on Io and Earth will reach a greater height on Io due to its weaker gravitational field.
Explanation:To understand how high material would go on Earth if ejected at the same speed as on Io, we first need to consider that the height that an object reaches in a gravitational field (ignoring air resistance) depends on its initial speed and the strength of the gravitational field. Specifically, the maximum height is given by the equation H = (v^2)/(2g), where v is the initial speed, and g is the acceleration due to gravity.
When an object is thrown upwards, it slows down under the force of the Earth's gravity until its speed decreases to zero. At that point, it begins to fall back down. Therefore, if two objects are thrown upwards with the same speed but under the influence of different gravitational fields (one stronger and one weaker), the object in the weaker gravitational field will reach a greater height.
In the case of Io, Jupiter's moon, the strength of the gravitational field is much less than that on Earth because its mass is much smaller. That's why volcanic material can be ejected to such high altitudes. Now, if we were to eject material with the same speed on Earth, which has a stronger gravitational field compared to Io, it would not reach the same height.
To calculate the specific height that material would reach on Earth assuming the same ejection speed as on Io, we would need to know that ejection speed. Unfortunately, the question does not provide this information.
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material ejected from Io with the same speed on Earth would reach a height of approximately [tex]\( 3.41 \times 10^8 \, \text{m} \) or \( 341 \, \text{km} \)[/tex] above the surface of Earth.
To determine how high material ejected from Io would go on Earth if it were ejected with the same speed, we can use the concept of escape velocity and the energy conservation principle.
The escape velocity [tex](\( v_e \))[/tex] of a celestial body is the minimum speed an object must have to break free from its gravitational pull and escape into space. It is given by the formula:
[tex]\[ v_e = \sqrt{\frac{2GM}{r}} \][/tex]
Where:
- [tex]\( v_e \)[/tex] is the escape velocity,
- ( G ) is the universal gravitational constant[tex](\( 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)),[/tex]
- ( M \ is the mass of the celestial body,
- ( r ) is the distance from the center of the celestial body.
Let's calculate the escape velocity of Io first:
[tex]\[ v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \times 8.93 \times 10^{22} \, \text{kg}}{1821 \times 10^3 \, \text{m}}} \]\[ v_e = \sqrt{\frac{2 \times 6.67 \times 8.93}{1821}} \times 10^{11} \, \text{m/s} \]\[ v_e = \sqrt{2 \times 3.34} \times 10^{11} \, \text{m/s} \]\[ v_e \approx 8.19 \times 10^4 \, \text{m/s} \][/tex]
Now, let's assume that material ejected from Io is given this speed of [tex]\( 8.19 \times 10^4 \, \text{m/s} \)[/tex] on Earth. We can calculate how high it would go using energy conservation principles.
The kinetic energy ( KE ) of the material is given by:
[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]
Where:
- \( m \) is the mass of the material, and
- \( v \) is the velocity.
The potential energy (\( PE \)) gained by the material at height \( h \) above the surface of Earth is given by:
[tex]\[ PE = mgh \][/tex]
Where:
- ( g ) is the acceleration due to gravity on Earth[tex](\( 9.81 \, \text{m/s}^2 \))[/tex],
- ( h ) is the height above the surface of Earth.
At the maximum height, the kinetic energy is converted entirely into potential energy. So we have:
[tex]\[ \frac{1}{2} m v^2 = mgh \]\[ \frac{1}{2} v^2 = gh \]\[ h = \frac{v^2}{2g} \][/tex]
Substitute the values:
[tex]\[ h = \frac{(8.19 \times 10^4 \, \text{m/s})^2}{2 \times 9.81 \, \text{m/s}^2} \]\[ h = \frac{6696.61 \times 10^8 \, \text{m}^2/\text{s}^2}{19.62 \, \text{m/s}^2} \]\[ h \approx 3.41 \times 10^8 \, \text{m} \][/tex]
So, material ejected from Io with the same speed on Earth would reach a height of approximately [tex]\( 3.41 \times 10^8 \, \text{m} \) or \( 341 \, \text{km} \)[/tex] above the surface of Earth.
A particle's position along the x-axis is described by: x(t)=A t + B t2
where t is in seconds, x is in meters, and the constants A and B are given below.
Randomized Variables
A = -4.3 m/s
B = 4.9 m/s2
(a) What is the velocity, in meters per second, of the particle at t1 = 3.0 s?
(b) What is the velocity, in meters per second, of the particle when it is at the origin (x=0) at t0 > 0?
Answer:
a)V = 25.1 m/s
b)V = 4.226 m/s
Explanation:
Given that
x(t)=A t + B t²
A = -4.3 m/s
B = 4.9 m/s²
x(t)= - 4.3 t +4.9 t²
The velocity of the particle is given as
[tex]V=\dfrac{dx}{dt}[/tex]
V=-4.3 + 4.9 x 2 t
V= - 4.3 + 9.8 t m/s
Velocity at point t= 3 s
V= - 4. 3 + 9.8 x 3 m/s
V= - 4.3 + 29 .4 m/s
V = 25.1 m/s
At origin :
x= 0 m
0 = - 4.3 t +4.9 t²
0 = - 4.3 + 4.9 t
[tex]t=\dfrac{4.3}{4.9}\ s[/tex]
t=0.87 s
The velocity at t= 0.87 s
V= - 4.3 + 9.8 t m/s
V= - 4. 3 + 9.8 x 0.87 m/s
V= - 4.3 + 8.526 m/s
V = 4.226 m/s
a)V = 25.1 m/s
b)V = 4.226 m/s
The velocity of the particle at t = 3.0s is 25.1 m/s.
The velocity of the particle when it is at the origin is 4.324 m/s.
The given parameters:
Position of the particle, x(t) = At + Bt²A = -4.3 m/s and B = 4.9 m/s²The velocity of the particle at t = 3.0s is calculated as follows;
[tex]v = \frac{dx}{dt} = A + 2Bt\\\\v = -4.3 + 2(4.9\times 3)\\\\v = 25.1 \ m/s[/tex]
The velocity of the particle when it is at the origin (x = 0)
[tex]0 = -4.3t + 4.9t^2\\\\0 = t(-4.3 + 4.9t)\\\\t = 0 \ \ \ or \ \ -4.3 + 4.9t = 0\\\\4.9t = 4.3\\\\t = \frac{4.3}{4.9} \\\\t = 0.88 \ s\\\\v = A + 2Bt\\\\ v = -4.3 + (2\times 4.9 \times 0.88)\\\\v = 4.324 \ m/s[/tex]
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For projectile motion on level ground assuming negligible air resistance (the initial angle being neither 0o nor 90o): Can the speed ever be the same as the initial speed at a time other than at t = 0 ?
a. No
b. Yes, at the end of the arch (end of trajectory)
c. Yes, at its maximum height
d. Yes, at half of its maximum height
Answer:
B. Yes, at the end of the trajectory.
Explanation:
By the conservation of energy, in order the object to have the same speed, it must have the same potential and kinetic energy, therefore, it must be at the same height as its initial height.
Answer:
b. Yes, at the end of the arch (end of trajectory)
Explanation:
according to conservation of energy and considering no air resistance:
initially the object has some kinetic energy due to speed which gets converted to gravitational potential energy till its maximum height.
since there is no energy loss, the object again gets to same amount of kinetic energy and hence the same speed.
The back wall of an auditorium is 23.0 m from the stage.
If you are seated in the middle row, how much time elapses between a sound from the stage reaching your ear directly and the same sound reaching your ear after reflecting from the back wall? The speed of sound in the air is 343 m/s .
_________seconds
Answer:
0.067 seconds
Explanation:
The time needed for a person seated in the middle row to receive a sound from the stage, can be calculated assuming that the sound is moving in a straight line at constant speed, taking into account that the distance traveled will be half of the distance between the stage and the back wall:
t₁ = x₁ / v = 11.5 m / 343 m/s = 0.034 sec
The same sound, after reflecting from the back wall, will travel a distance equal to one and a half the distance between the stage and the back wall:
t₂ = 34.5 m /343 m/s = 0.101 sec
The time elapsed between both sounds will be equal to the difference between t₂ and t₁, as follows:
t₂ - t₁ = 0.101 sec - 0.034 sec = 0.067 sec.
A tiny ball (mass = 0.010 kg) carries a charge of -24C. What is the electric field magnitude that would cause the ball to float above the ground?
Use the relation between electric force and electric field and the concept of gravitational force to calculcate the electric field. The electric force is given by
[tex]F_e = qE[/tex]
And the gravitational force is
[tex]F_g = mg[/tex]
For the ball to float above the ground, the magnitude of electric force on the ball must be equal to the magnitude of the gravitational force. That is must be a equilibrium condition, so,
[tex]F_e = F_g[/tex]
[tex]qE = mg[/tex]
[tex]E = \frac{mg}{q}[/tex]
Replacing the values we have that,
[tex]E = \frac{(0.010)(9.8)}{-24*10^{-6}}[/tex]
[tex]E = -4.083*10^{3} N/C[/tex]
Therefore the electric field magnitude that would cause the ball to float above the ground is [tex]-4.083*10^{3} N/C[/tex]
Wave motion is characterized by two velocities: the velocity with which the wave moves in the medium (e.g., air or a string) and the velocity of the medium (the air or the string itself).Consider a transverse wave traveling in a string. The mathematical form of the wave is: y(x,t) = A sin(kx - ωt)Part AFind the velocity of propagation v_p of this wave.Express the velocity of propagation in terms of some or all of the variables A, k, and ω.Part BFind the y velocity v_y(x,t) of a point on the string as a function of x and t.Express the y velocity in terms of ω, A, k, x, and t.Part CWhich of the following statements about v_x(x,t), the x component of the velocity of the string, is true?A) v_x(x;t) = v_pB) v_x(x;t) = v_y(x;t)C) v_x(x;t) has the same mathematical form as v_y(x;t) but is 180° out of phase.D) v_x(x;t)=0Part DFind the slope of the string ∂_y(x,t) / ∂_x as a function of position x and time t.Express your answer in terms of A,k, ω, x, and t.
Answer:
Explanation:
ocities: the velocity with which the wave moves in the medium (e.g., air or a string) and the velocity of the medium (the air or the string itNBself).Consider a transverse wave traveling in a string. The mathematical form of the wave is: y(x,t) = A sin(kx - ωt)Part AFind the velocity of propagation v_p of this wave.Express the velocity of propagation in terms ofNGHJGHHG some or all of the variables A, k, and ω.Part BFind the y velocity v_y(x,t) of a point on the string as a function of x and t.Express the y velocity in terms of ω, A, k, x, and t.Part CWhich of the following statements about v_x(x,t), the x component of the velocity of the string, is true?A) v_x(x;t) = v_pB) v_x(x;t) = v_y(x;t)C) v_x(x;t) has the same mathematical form as v_y(x;t) but is 180° out of phase.D) v_x(x;t)=0Part DFind the slope of the string ∂_y(x,t) / ∂_x as a function of position x and time t.Express your answer in terms of A,k, ω, x, and t.NNNNN
a) The velocity of propogation of the wave V=w/k
b) The y velocity v_y(x,t) of a point on the string as a function of x v=-wAcos(kx-wt)
What is wave?A wave can be described as a disturbance that travels through a medium from one location to another location
y(x,t)=Asin(kx−ωt) defines the wave equation.
a)The velocity of propogation of the wave
We are asked to find wave speed (v)
Recall that v = fλ
From the wave equation above,
k = 2π/ λ where k is the wave number and λ is the wavelength, λ = 2π /k
ω = 2πf where f is the frequency and ω is the angular frequency.
f = ω/ 2π.
By substituting for λ and ω into the wave speed formulae, we have that
v =( ω/ 2π) × (2π /k)
v = ω/k
b)The y velocity v_y(x,t) of a point on the string as a function of x
y(x,t)=Asin(kx−ωt)
The first derivative of y with respect to x give the velocity (vy)
By using chain rule, we have that
v = dy/dt = A cos( kx −ωt) × (−ω)
v = - ωAcos( kx −ωt)
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The plates of a parallel-plate capacitor are maintained with a constant voltage by a battery as they are pushed together, without touching. How is the amount of charge on the plates affected during this process?
The plates of a parallel-plate capacitor are maintained with a constant voltage by a battery as they are pushed together, without touching. How is the amount of charge on the plates affected during this process?
The amount of charge remains constant.
The amount of charge on the plates decreases during this process.
The amount of charge on the plates becomes zero.
The amount of charge on the plates increases during this process.
The amount of charge on the plates increases during this process.
Explanation:
The relationship between charge and potential difference through a capacitor is
[tex]C=\frac{Q}{\Delta V}[/tex]
where
C is the capacitance
Q is the charge stored
[tex]\Delta V[/tex] is the potential difference
The capacitance of a parallel-plate capacitor is given by
[tex]C=\frac{\epsilon_0 A}{d}[/tex]
where
[tex]\epsilon_0[/tex] is the vacuum permittivity
A is the area of the plates
d is the distance between the plates
Combining the two equations, we get:
[tex]Q=\frac{\epsilon_0 A \Delta V}{d}[/tex]
In this problem:
- The potential difference between the plates, [tex]\Delta V[/tex], is kept constant
- The area of the plates, A, remains constant
- The distance between the plates, d, is decreased
Since Q is inversely proportional to d, this means that as the plates are pushed together, the amount of charge on the plates increases during the process.
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Final Answer:
By sustaining a constant voltage across two conducting plates, the parallel-plate capacitor creates an electric field, gathers positive and negative charges, and preserves the battery's electric potential difference.
Explanation:
The amount of charge on the plates remains constant during the process of pushing the plates of a parallel-plate capacitor together. When a constant voltage (potential difference) is applied across the plates by a battery, it establishes an electric field between the plates. When you push the plates together without touching, you are essentially changing the distance between them while keeping the voltage constant.
The key point to understand is that the amount of charge on the plates is determined by the voltage (potential difference) and the capacitance of the capacitor, the charge on the plates is denoted by the equation Q = C * V.
In summary, when the plates of a parallel-plate capacitor are maintained with a constant voltage by a battery and pushed together without touching, the amount of charge on the plates remains constant throughout the process.
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A 1.33 kg object is attached to a horizontal spring of force constant 2.50 N/cm and is started oscillating by pulling it 6.40 cm from its equilibrium position and releasing it so that it is free to oscillate on a frictionless horizontal air track. You observe that after eight cycles its maximum displacement from equilibrium is only 3.70 cm .
(a) How much energy has this system lost to damping during these eight cycles?
(b) Where did the "lost" energy go? Explain physically how the system could have lost energy.
Answer:
Explanation:
a ) Energy of spring = 1/2 k A² where A is amplitude of oscillation and k is force constant .
So initial energy = 1/2 x 2.5 x (6.4 x 10⁻²)²
= 51.2 x 10⁻⁴ J
So final energy = 1/2 x 2.5 x (3.7 x 10⁻²)²
= 17.11 x 10⁻⁴ J
energy lost
= 34.1 J .
This energy is dissipated in the form of heat, sound etc.
Answer:
a. [tex]\Delta U=3.375\ N.cm=3.375\times 10^{-2}\ J[/tex]
b. The energy lost during the damping is converted into kinetic energy of the molecules which is heat primarily.
Explanation:
Given:
mass of the attached object, [tex]m=1.33\ kg[/tex]spring constant, [tex]k=2.5\ N.cm^{-1}[/tex]maximum displacement, [tex]A=6.4\ cm[/tex]maximum displacement after damping, [tex]x=3.7\ cm[/tex]a)
the energy lost in damping:
[tex]\Delta U=\frac{1}{2} \times k\times A^2-\frac{1}{2} \times k\times x^2[/tex]
[tex]\Delta U=\frac{1}{2} \times 2.5\times (6.4-3.7)[/tex]
[tex]\Delta U=3.375\ N.cm=3.375\times 10^{-2}\ J[/tex]
b)
The energy lost during the damping is converted into kinetic energy of the molecules which is heat primarily.
The 2-kg spool S fits loosely on the inclined rod for which the coefficient of static friction is μs=0.2. If the spool is located 0.25 m from A, determine the maximum constant speed the spool can have so that it does not slip up the rod
Answer:
v = 0.969 m/s
Explanation:
See attachment for FBD
p = 0.25*(4/5) = 0.2 m
Sum of normal forces
Ns (3/5) - 0.2*Ns*(4/5) = 2 * (v^2 / 0.2)
Sum of vertical forces
Ns (4/5) - 0.2*Ns*(3/5) = 2*9.81
Solve both equations simultaneously to get Ns and v
Ns = 21.3 N
v = 0.969 m/s
The frequency sweep is 383-438 Mhz with a period of 15 microseconds. The first return occurs at 429 MHz. What is the range? (answer in meters to one decimal point)
Answer:
The range is 1881.8 m.
Explanation:
Given that,
Time [tex]t=15\times10^{-6}\sec[/tex]
Frequency range [tex]\Delta f= f_{f}-f_{i}[/tex]
[tex]\Delta f= 429-383[/tex]
[tex]\Delta f=46\ MHz[/tex]
The value of [tex]\dfrac{df}{dt}[/tex]
[tex]\dfrac{df}{dt}=\dfrac{438-383}{15\times10^{-6}}[/tex]
We need to calculate the range
Using formula of range
[tex]R=\dfrac{c\Delta f}{2\times\dfrac{df}{dt}}[/tex]
Put the value into the formula
[tex]R=\dfrac{3\times10^{8}\times46}{2\times\dfrac{438-383}{15\times10^{-6}}}[/tex]
[tex]R=1881.8\ m[/tex]
Hence, The range is 1881.8 m.
Two small spheres with charges q1 = +9 nC and q2 = -3 nC are placed at marks x1 = -2 cm and x2 = 13 cm. Where along the x-axis should you place a third sphere with charge q3 so that the sum of the forces exerted by the first two spheres on the third one is zero?
To find the position along the x-axis where the sum of the forces exerted by the first two spheres on the third one is zero, we can use Coulomb's law. The forces between the charges are given by F = k * (|q1*q2| / r^2). By setting the forces equal to zero and solving for x, we can determine the desired position.
Explanation:To find the location along the x-axis where a third sphere with charge q3 should be placed so that the sum of the forces exerted by the first two spheres on the third one is zero, we need to consider the forces between the charges. According to Coulomb's law, the force between two point charges is given by:
F = k * (|q1*q2| / r^2)
Where k is the Coulomb's constant, q1 and q2 are the charges of the two spheres, and r is the distance between them. In this case, the force exerted by q1 on q3 is:
F1 = k * (|q1*q3| / (x - x1)^2)
And the force exerted by q2 on q3 is:
F2 = k * (|q2*q3| / (x - x2)^2)
For the sum of the forces to be zero, we must have:
F1 + F2 = 0
Substituting the values of q1, q2, x1, and x2, we can solve for x to find the position where the sum of the forces is zero.
Here, a=2, b=−82, and
c=437. Plug these values into the formula to find x_{3} . You should find two possible positions for x_{3}
A block on the end of a spring is pulled to position x 5 A and released from rest. In one full cycle of its motion, through what total distance does it travel
Answer:
20 A
Explanation:
Given
Spring pulled from position
x = 5A
we need to calculate total distance of one full cycle of spring motion
if you see image below, you understand easily
When cycle complete
Its total distance become 20 A
Two 6-in wide wooden boards are to be joined by splice plates that will be fully glued on the contact surfaces.The glue to be used can safely provide a shear strength of 120 psi. Determine the smallest allowable length L that can be used for the splice plates for an applied load of P = 15,000 lb. Note that a gap of 0.5 in is required between boards 1 and 2.which one of these is the right answer and show work to prove why.a) 19.7 inb) 15.7 inc) 24.3 ind) 21.3 ine) 11.6 in
Answer:
D
Explanation:
The smallest possible allowable length that can be used:
Calculate area of contact between plates A
A = 2 * 6 * a
A = 12a in^2
Where a is an arbitrary constant length along the direction L.
Shear stress = Shear force / Area
Area = 15000 lb / 120 psi = 12*a
Evaluate a
a = 10.411 in
Total Length = 2*a + gap = 2*10.411 + 0.5 = 21.33 in
Answer: L = 21.33
To find the smallest allowable length for the splice plates, the required shear area was calculated using the provided load and the shear strength of the glue, then divided by the effective board width. The resulting length per plate was approximately 11.35 inches, making the smallest length that can safely be used 11.6 inches.
To determine the smallest allowable length L that can be used for the splice plates, we must first calculate the shear area required to resist the applied load P of 15,000 lb with a glue shear strength of 120 psi.
The total shear force that the splice plates must resist is equal to the applied load P. The shear strength per unit area provided by the glue is 120 psi. Therefore, the required shear area A can be found using the formula:
A = P / Shear Strength
A = 15,000 lb / 120 psi = 125 square inches
Each board has a width of 6 inches, but we account for a 0.5-inch gap between the boards, so the effective width w for the splice is (6 - 0.5) inches = 5.5 inches. Then, we can calculate the length L needed for each splice plate as follows:
L = A / w
L = 125 square inches / 5.5 inches ≈ 22.7 inches
But since the glue is applied to two plates, we divide this length by 2 to get the length required for one splice plate:
L = 22.7 inches / 2 ≈ 11.35 inches
To ensure safety, the smallest allowable length should be the next highest option available, which is 11.6 inches (Option E).
Describe the relationship between the direction of the velocity vector and the direction of the acceleration or a body moving in a circle at constant speed.
In the centripetal movement, what happens with velocity is that it will remain constant, always pointing in its tangential direction of the trajectory. Said speed, although constant, will have a constant direction that will generate an acceleration that will always point towards the center of the circle radius. Both vectors as the turn is performed will always be perpendicular to each other.
The direction of the acceleration is longitudinal to the direction of velocity in for a body moving in a circle at constant speed.
Circular motion is the motion of an object in a circle at a constant speed. As the object is constantly changing its direction as it moving in a circle at constant speed.
At all instances, the object is moving tangent to the circle. The direction of velocity vector is tangent to the circle, as the direction of velocity vector is towards the direction of object's motion.
An object moving in a circle is accelerating due to its change in direction. The direction of the acceleration is inwards.
Therefore, the direction of the acceleration is longitudinal to the direction of velocity in for a body moving in a circle at constant speed.
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A telephone cable is clamped at A to the pole AB. Knowing that the tension in the right-hand portion of the cable is T2 = 1000 lb, determine by trigonometry (a) the required tension T1 in the left-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical, (b) the corresponding magnitude of R.
Answer:
(a) T1 = 938.3lb
(b) R = 665.5lb
The detail solution to this problem can be found in the attachment below.
This problem was solved by resolving the forces along the vertical and horizontal and equating to Rx and Ry respectively. Rx = 0 because the resultant is directed along the vertical.
Explanation:
See attachment below for the full solution.
Thank you for reading.
Final answer:
To find the tension T1 and the vertical resultant R, use trigonometry, considering that T1 must have an equal and opposite horizontal force component to T2 and a vertical component that combines with T2's vertical component.
Explanation:
Understanding the Tension in Two Portions of a Cable
The student's question pertains to the tensions in a telephone cable clamped at point A to a pole. We are given that the tension in the right-hand portion of the cable (T2) is 1000 lb and are asked to determine two things:
The required tension T1 in the left-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical.
The corresponding magnitude of R.
To solve for the required tension T1 in the left-hand portion so that the resultant force R is vertical, we need to use trigonometry. Given that the resultant R must be vertical and that we already have a horizontal force component from T2, T1 must provide an equal and opposite horizontal force component to cancel it out, as well as contribute to the resultant vertical force. The magnitude of the horizontal component of T1 must equal that of T2, and the vertical component of T1 will combine with T2 to form the vertical resultant R.
To determine the magnitude of the resultant force R, which is vertical, we will sum up the vertical components of T1 and T2. This magnitude can be found using the Pythagorean theorem if necessary, or simply as the sum of the vertical components if T1 is already calculated and has the proper direction to ensure a vertical R.
For this specific scenario, without additional information such as angles or the tension division between T1 and T2, we cannot provide numerical answers but can describe the method to find the solution.