PHYSICS COLLEGE

The Golden Gate Bridge in San Francisco has a main span of length 1.28 km, one of the longest in the world. Imagine that a steel wire with this length and a cross-sectional area of 3.10 ✕ 10^−6 m^2 is laid on the bridge deck with its ends attached to the towers of the bridge, on a summer day when the temperature of the wire is 43.0°C. When winter arrives, the towers stay the same distance apart and the bridge deck keeps the same shape as its expansion joints open. When the temperature drops to −10.0°C, what is the tension in the wire? Take Young's modulus for steel to be 20.0 ✕ 10^10 N/m^2. (Assume the coefficient of thermal expansion of steel is 11 ✕ 10−6 (°C)−1.)

Answers

Answer 1

Answer:

Answer:

361.46 N

Explanation:

$\alpha$ = Coefficient of thermal expansion = $11* 10^(-6)\ /^(\circ)C$

Y = Young's modulus for steel = $20* 10^(10)\ Pa$

A = Area = $3.1* 10^(-6)\ m^2$

$L_0$ = Original length = 1.28 km

$\Delta T$ = Change in temperature = $45-(-10)$

Length contraction is given by

$\Delta L=\alpha L_0\Delta T$

Also,

$\Delta L=(L_0T)/(YA)$

$\alpha L_0\Delta T=(L_0T)/(YA)\n\Rightarrow T=\alpha \Delta TYA\n\Rightarrow T=11* 10^(-6)* (43-(-10))* 20* 10^(10) * 3.1* 10^(-6)\n\Rightarrow T=361.46\ N$

The tension in the wire is 361.46 N

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Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.

Answers

Answer:

115 ⁰C

Explanation:

Step 1: The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

$q_(1) +q_(2) =-q_(3)$ -----eqution 1

where,

$q_(1)$ is the heat absorbed by the solid at 0⁰C

$q_(2)$ is the heat absorbed by the liquid at 0⁰C

$q_(3)$ the heat lost by the warmer water sample

Important equations to be used in solving this problem

$q=m *c*\delta {T}$ , where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

$\delta {T}$ is change in temperature

Again,

$q=n*\delta {_f_u_s}$ -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

Step 2: calculate how many moles of water you have in the 100.0-g sample

$=237g *(1 mole H_(2) O)/(18g) = 13.167 moles of H_(2)O$

Step 3: calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

$q_(1) = 13.167 moles *6.01(KJ)/(mole) = 79.13KJ$

This means that equation (1) becomes

79.13 KJ + $q_(2) = -q_(3)$

Step 4: calculate the final temperature of the water

$79.13KJ+M_(sample) *C*\delta {T_(sample)} =-M_(water) *C*\delta {T_(water)$

Substitute in the values; we will have,

$79.13KJ + 237*4.18(J)/(g^(o)C)*(T_(f)-218}) = -350*4.18(J)/(g^(o)C)*(T_(f)-100})$

79.13 kJ + 990.66J* $(T_(f)-218})$ = -1463J* $(T_(f)-100})$

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* $(T_(f)-218})$ = -1.463KJ* $(T_(f)-100})$

$79.13 + 0.99066T_(f) -215.96388= -1.463T_(f)+146.3$

collect like terms,

2.45366 $T_(f)$ = 283.133

∴ $T_(f) =$ = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

How many joules of work are done on an object when a force of 10 N pushes it 5 m?A) 2 J
B) 5 J
C) 50 J
D) 1 J
E) 10 J

Answers

Answer:

option C

Explanation:

given,

Force on the object = 10 N

distance of push = 5 m

Work done = ?

we know,

work done is equal to Force into displacement.

W = F . s

W = 10 x 5

W = 50 J

Work done by the object when 10 N force is applied is equal to 50 J

Hence, the correct answer is option C

Final answer:

The work done on an object when a force of 10 N pushes it 5 m is 50 Joules, calculated by multiplying the force and the displacement. So, the correct option is C.

Explanation:

The question is asking about work, which in physics is the result of a force causing a displacement. The formula for work is defined as the product of the force (in Newtons) and the displacement (in meters) the force causes. If a force of 10 N pushes an object a distance of 5 m, the work done is calculated by multiplying the force and the displacement (10 N * 5 m), yielding 50 Joules of work.

Therefore, the correct answer is 50 J (C).

Learn more about Work here:

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If gear a rotates with a constant angular acceleration of aa = 90 rad>s2, starting from rest, determine the time required for gear d to attain an angular velocity of 600 rpm. Also, find the number of revolutions of gear d to attain this angular velocity. Gears a, b, c, and d have radii of 15 mm, 50 mm, 25 mm, and 75 mm, respectively.

Answers

solution:

$Given: (let a = alpha)aA = 90 rad/s^2(wd) = 600 rpm = 62.831 rad/s(w0) = 0 rpmra = .015 mrb = .05 mrc = .025 mrd = .075aB = aA*(ra/rb) = 90 * (.015/.05)aB = 27 rad/s^2aC = aB, Therefore aC = 27 rad/s^2aD = aC*(rc/rb) = 27 * (.025/.075)aD = 9 rad/s^2$

What is the frequency of a photon that has the same momentum as a neutron moving with a speed of 1.90 × 103 m/s?

Answers

The mass of a neutron is:
$m=1.67 \cdot 10^(-27)kg$
Since we know its speed, we can calculate the neutron's momentum:
$p=mv=(1.67 \cdot 10^(-27)kg)(1.90 \cdot 10^3 m/s)=3.17 \cdot 10^(-24) kg m/s$

The problem says the photon has the same momentum of the neutron, p. The photon momentum is given by
$p= (h)/(\lambda)$
where h is the Planck constant, and $\lambda$ is the photon wavelength. If we re-arrange the equation and we use the momentum we found before, we can calculate the photon's wavelength:
$\lambda= (h)/(p)= (6.6 \cdot 10^(-34)Js)/(3.17 \cdot 10^(-24) kg m/s)=2.08 \cdot 10^(-10) m$

And since we know the photon travels at speed of light c, we can now calculate the photon frequency:
$f= (c)/(\lambda)= (3 \cdot 10^8 m/s)/(2.08 \cdot 10^(-10) m)= 1.44 \cdot 10^(18) Hz$

According to American Heart Association, your target heart rate is__________. Group of answer choices

A .all of the above

B. 85% - 95% of your max heart rate

C. 20%-30% of your max heart rate

D. 50%-85% of your max heart rate

Answers

Answer:

Explanation:

A normal heart rate is from 85-95% the other heart rate is not normal because your heart is beating more than normal

A baseball is thrown horizontally with a velocity of 44 m/s. It travels a horizontal distance of 18m to the plate before it is caught. How long does the ball stay in the air?

A.0.41 sec
B.41 sec
C.4.1 sec
D.4 sec

Answers

A horizontal baseball pitch is launched at 44 m/s. The ball will stay for 4.1 sec (approx) in the air. Hence, option C is correct.

What is Velocity?

The rate at which an object's position changes when observed from a specific point of view and when measured against a specific unit of time is known as its velocity.

Its SI unit is represented as m/s, and it is a vector quantity, it means that it has both magnitude and direction.

According to the question, the given values are :

Initial Velocity, u = 44 m/s,

Distance travelled, s = 18 m and,

Final velocity, v = 0.

Use equation of motion :

v = u + at

0 = 44 + (-9.8)t

t = 44 / 9.8

t = 4.3 (approx)

Hence, the time for which the ball stay in the air is 4.1 sec (approx).

To get more information about velocity :