During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between two fence posts. The limb exerts a downward force of 176 N on the wire. The left section of the wire makes an angle of 12.5° relative to the horizontal and sustains a tension of 413 N. Find the (a) magnitude and (b) direction (as an angle relative to horizontal) of the tension that the right section of the wire sustains.

Answers

Answer 1
Answer:

Answer:

a. 12.12°

b. 412.04 N

Explanation:

Along vertical axis, the equation can be written as

T_1 sin14 + T_2sinA = mg

T_2sinA = mg - T_1sin12.5           ....................... (a)

Along horizontal axis, the equation can be written as

T_2×cosA = T_1×cos12.5    ......................... (b)

(a)/(b) given us

Tan A = (mg - T_1sin12.5) / T_1 cos12.5

 = (176 - 413sin12.5) / 413×cos12.5

A = 12.12 °

(b) T2 cosA = T1 cos12.5

T2 = 413cos12.5/cos12.12

= 412.04 N

Answer 2
Answer:

Answer:

Magnitude - 11.83 Degree

Direction - 422.42 N

Explanation:

Given data:

Downward force on wire 176 N

Angle made by left section of wire 12.5 degree with horizontal

Tension force = 413 N

From figure

Applying quilibrium principle at point A

The vertical and horizontal force is 0

then we have

Tcos\theta = 413 N   ........1

176 = 413 sin 12.5 + Tsin\theta     .......2

Tsin\theta = 176 - 89.39  = 86.6.......3

divide equation 3 by 1

we get

\theta = tan^(-1) (0.2096)

theta = 11.83^o  ...........4

from equation 3 and 4

T = (86.6)/(sin 11.83)

T = 422.42 N


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An object moving with uniform acceleration has a velocity of 10.5 cm/s in the positive x-direction when its x-coordinate is 2.72 cm. If its x-coordinate 2.30 s later is ?5.00 cm, what is its acceleration? The object has moved to a particular coordinate in the positive x-direction with a certain velocity and constant acceleration; then it reverses its direction and moves in the negative x-direction to a particular x-coordinate in time t. We are given an initial velocity vi = 10.5 cm/s in the positive x-direction when the initial position is xi = 2.72 cm (t = 0). We are given that at t = 2.30 s, the final position is xf = ?5.00 cm. The acceleration is uniform so that we have the following equation in terms of the constant acceleration a. Xf-Xi=Vit-1/2at^2 Now we substitute the given values into this equation. (___cm)-(___cm)=(___cm/s)(__s)+1/2a(___s)

Answers

Answer:

Acceleration = 8.27 cm/s²

Explanation:

We are given;

initial velocity; v_i = 10.5 cm/s

Initial position; x_i = 2.72 cm

Time; t = 2.30 s

final position; x_f = 5.00 cm

To find the acceleration, we will make use of the formula;

x_f - x_i = (v_i * t) - (½at²)

Plugging in the relevant values, we have;

5 - 2.72 = (10.5 × 2.3) - (½ × a × 2.3²)

2.28 = 24.15 - 2.645a

24.15 - 2.28 = 2.645a

2.645a = 21.87

a = 21.87/2.645

a = 8.27 cm/s²

Using the kinematic equation, the acceleration of the object was calculated to be approximately8.27 cm/s² given its initial velocity, position, time, and final position.

We are given:

Initial velocity (vᵢ) = 10.5 cm/s

Initial position (xᵢ) = 2.72 cm

Time (t) = 2.30 seconds

Final position (x_f) = 5.00 cm

We want to find the acceleration (a) of the object using the kinematic equation:

x₋ᵢ - xᵢ = (vᵢ * t) - (1/2) * a * t²

Now, let's substitute the given values:

5.00 cm - 2.72 cm = (10.5 cm/s * 2.30 s) - (1/2) * a * (2.30 s)²

Simplify the equation:

2.28 cm = 24.15 cm - (1/2) * a * 5.29 s²

Now, isolate 'a' by rearranging the equation:

-1.09 cm = (-1/2) * a * 5.29 s²

To remove the negative sign, multiply both sides by -1:

1.09 cm = (1/2) * a * 5.29 s²

Next, solve for 'a' by multiplying both sides by (2 / 5.29):

a ≈ (1.09 cm) / (2 / 5.29) s²

a ≈ 8.27 cm/s²

So, the acceleration of the object is approximately 8.27 cm/s².

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The Richter scale is used to determine how strong the earthquake is (magnitude) of the earthquake. It touches the ground and feels the earth shaking. With the momentum of the earth shaking the device the needle on the device moves creating a wave looking line. According to the theory of plate tectonics, what happens at transform, divergent and convergent boundaries? On which of these boundary types would a volcano most likely take place, and why?

Answers

convergent, because it is where the tectonic plates shove themselves together usually resulting in a mountain or volcano

Final answer:

In plate tectonics theory, transform boundaries occur when two plates slide past each other, divergent boundaries occur when two plates move away from each other, and convergent boundaries occur when two plates collide. A volcano is most likely to occur at a divergent boundary because the plates move away from each other, allowing magma from the mantle to reach the surface and create new crust.

Explanation:

In plate tectonics theory, Transform boundaries occur when two plates slide past each other horizontally, creating earthquakes. Divergent boundaries occur when two plates move away from each other, creating volcanic activity.

Convergent boundaries occur when two plates collide, and depending on the type of plates involved, can result in volcanic activity as one plate is forced beneath the other.

A volcano is most likely to occur at a divergent boundary because the plates move away from each other, allowing magma from the mantle to reach the surface and create new crust.

The Richter scale indicates the magnitude of an earthquake. The figure drawn by the needle during shaking is an outcome of earthquake's energy. This energy is what results in seismic waves travelling through various layers of earth causing shaking on the surface.

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What is the average velocity if the initial velocity is at rest and the final velocity is 16 m/s

Answers

Answer:

8m/s

Explanation:

Vavg= 16-0/2=8m/s

Points A, B, and C are at the corners of an equilateral triangle of side 8 m. Equal positive charges of 4 mu or micro CC are at A and B. (a) What is the potential at point C? 8.990 kV * [2.5 points] 2 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 8.990 OK (b) How much work is required to bring a positive charge of 5 mu or micro CC from infinity to point C if the other charges are held fixed? .04495 J * [2.5 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] .04495 OK (c) Answer parts (a) and (b) if the charge at B is replaced by a charge of -4 mu or micro CC. Vc= kV [2.5 points] 0 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] W =

Answers

Answer:

a) 8.99*10³ V  b) 4.5*10⁻² J c) 0 d) 0

Explanation:

a)

  • The electrostatic potential V, is the work done per unit charge, by the electrostatic force, producing a displacement d from infinity (assumed to be the reference zero level).
  • For a point charge, it can be expressed as follows:

        V =(k*q)/(d)

  • As the electrostatic force is linear with the charge (it is raised to first power), we can apply superposition principle.
  • This means that the total potential at a given point, is just the sum of the individual potentials due to the different charges, as if the others were not there.
  • In our case, due to symmetry, the potential, at any corner of the triangle, is just the double of the potential due to the charge located at  any other corner, as follows:

        V = (2*q*k)/(d) = (2*8.99e9N*m2/C2*4e-6C)/(8m) =\n \n V= 8.99e3 V

  • The potential at point C is 8.99*10³ V

b)

  • The work required to bring a positive charge of 5μC from infinity to the point C, is just the product of the potential at this point times the charge, as follows:

        W = V * q = 8.99e3 V* 5e-6C = 4.5e-2 J

  • The work needed is 0.045 J.

c)

  • If we replace one of the charges creating the potential at the point  C, by one of the same magnitude, but opposite sign, we will have the following equation:

       V = (8.99e9N*m2/C2*(4e-6C))/(8m)  + ((8.99e9N*m2/C2*(-4e-6C))/(8m)) = 0

  • This means that the potential due to both charges is 0, at point C.

d)

  • If the potential at point C is 0, assuming that at infinity V=0 also, we conclude that there is no work required to bring the charge of 5μC from infinity to the point C, as no potential difference exists between both points.

The spring constant, k, for a 22cm spring is 50N/m. A force is used to stretch the spring and when it is measured again it is 32cm long. Work out the size of this force

Answers

Answer:

5N

Explanation:

Given parameters:

Original length = 22cm

Spring constant, K  = 50N/m

New length = 32cm

Unknown

Force applied  = ?

Solution:

The force applied on a spring can be derived using the expression below;

   Force  = KE

 k is the spring constant

 E is the extension

  extension = new length - original length

  extension  = 32cm  - 22cm  = 10cm

convert the extension from cm to m;  

   100cm  = 1m;

    10cm will give 0.1m

So;

  Force  = 50N/m x 0.1m  = 5N

Final answer:

To calculate the force used to stretch the spring, Hooke's Law is utilized, which leads to the conclusion that a force of 5 N was exerted to stretch the spring from its original length of 22 cm to a final length of 32 cm.

Explanation:

The force exerted by a spring is governed by Hooke's Law, which states that the force required to stretch or compress a spring by a certain distance is proportional to that distance. In this case, the spring constant, k, is given as 50 N/m and the spring is stretched from its original length of 22 cm to a final length of 32 cm. This represents a stretch, or displacement, of 10 cm (or 0.1 m when converted to the standard unit).

The force (F) can be calculated using Hooke's law: F = kx, where x is the displacement of the spring. Substituting the given values, the force amounts to F = (50 N/m) * (0.1 m) = 5 N. Therefore, the force used to stretch the spring to its final length of 32 cm is 5 N.

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An interstellar space probe is launched from Earth. After a brief period of acceleration it moves with a constant velocity, 70.0% of the speed of light. Its nuclear-powered batteries supply the energy to keep its data transmitter active continuously. The batteries have a lifetime of 15.9 years as measured in a rest frame. (a) How long do the batteries on the space probe last as measured by mission control on Earth? yr
(b) How far is the probe from Earth when its batteries fail, as measured by mission control?
ly
(c) How far is the probe from Earth, as measured by its built-in trip odometer; when its batteries fail?
ly

Answers

Answer:

22.26 years

, 15.585 light years  , 11.13 light years

Explanation:

a)

t' = t/(√(1-(v/(c*v)/c))

= 15.9/√((1-0.7*0.7))

= 22.26 years

b)

0.7*c*22.26 years

=15.585 light years  

c)

0.7*c*15.9

=11.13 light years

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