A bridge is made with segments of concrete 91 m long (at the original temperature). If the linear expansion coefficient for concrete is 1.2 × 10−5 ( ◦C)−1 , how much spacing is needed to allow for expansion for an increase in temperature of 56◦F? Answer in units of cm.

Answers

Answer 1
Answer:

Answer:

change in length is 3.397 cm

Explanation:

Given data

long = 91 m = 9100 cm

coefficient for concrete (a) =  1.2 × 10−5 ( ◦C)−1

temperature = 56 F = (56× 5/9) ◦C

to find out

how much spacing is needed to allow

solution

we know allow space is given by this formula

change in length = coefficient for concrete × given length × temperature     .............1

put all value in equation 1

change in length = 1.2 × 10−5  × 9100 × (56× 5/9)

change in length = 3.397 cm

so change in length is 3.397 cm


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What is the difference between V(peak voltage) and Vrms (root-mean-square) of AC voltage source?

Answers

Answer:

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Proper design of automobile braking systems must account for heat buildup under heavy braking. Part A Calculate the thermal energy dissipated from brakes in a 1600 kg car that descends a 15 ∘ hill. The car begins braking when its speed is 95 km/h and slows to a speed of 40 km/h in a distance of 0.34 km measured along the road.

Answers

Answer:

1838216 J

Explanation:

95 km/h = 26.39 m/s

40 km/h = 11.11 m/s

Initial kinetic energy

= .5 x 1600 x(26.39)²

= 557145.67 J

Final kinetic energy

= .5 x 1600 x ( 11.11)²

= 98745.68 J

Loss of kinetic energy

= 458400 J

Loss of potential energy

= mg x loss of height

= 1600 x 9.8 x 340 sin 15

= 1379816 J

Sum of Loss of potential energy and Loss of kinetic energy

=  1379816 + 458400

= 1838216 J

This is the work done by the friction . So this is heat generated.

Final answer:

To calculate the thermal energy dissipated from the brakes of a car, use the equation Q = Mgh/10, where Q is the energy transferred to the brakes, M is the mass of the car, g is the acceleration due to gravity, and h is the height of the hill. The temperature change of the brakes can then be calculated using the equation Q = mc∆T, where m is the mass of the brakes and c is its specific heat capacity.

Explanation:

The thermal energy dissipated from the brakes of a car can be calculated by converting the gravitational potential energy lost by the car into internal energy of the brakes. By using the equation Q = Mgh/10, where Q is the energy transferred to the brakes, M is the mass of the car, g is the acceleration due to gravity, and h is the height of the hill, we can calculate the thermal energy dissipated. From there, the temperature change of the brakes can be calculated using the equation Q = mc∆T, where m is the mass of the brakes and c is its specific heat capacity.

Learn more about Thermal energy dissipation here:

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Answers

Answer:

D. all regions of the spectrum

Explanation:

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Answers

The individual calcium atom has a positive and not negative, 2 charge

Answer:

The individual calcium atom has a positive, not negative, 2 charge.

Explanation:

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A 50n brick is suspended by a light string from a 30kg pulley which may be considered a solid disk with radius 2.0m. the brick is released from rest and falls to the floor below as the pulley rotates. it takes 4 seconds for the brick to hit the floor. i) what is the tension in newtons in the string well the brick is falling? ii) what is the magnitude of the angular momentum in kg*m^2/s of the pulley at the instant the brick hits the floor?

Answers

Brick is held at a position which is at height 2 m from the floor

Now it is released from rest and hit the floor after t = 4 s

Now the acceleration of the brick is given by

y = v_i* t + 0.5 at^2

2 = 0 + 0.5 * a * 4^2

a = 0.25 m/s^2

a)

Now in order to find the tension in the string

we can use Newton's law

F_(net) = ma

mg - T = ma

50 - T = (50)/(9.8)*0.25

T = 48.72 N

part b)

Now for the pulley

moment of inertia= (1)/(2)mr^2

m = 30 kg

R = 2 m

I = (1)/(2)*30*2^2

I = 60 kg m^2

Now the angular speed just before brick collide with the floor

w = (v)/(r)[\tex]</p><p>here we have</p><p>[tex]v = v_i + a* t

v = 0 + 0.25 * 4

v = 1 m/s

Now we will have

L = angular momentum = I w = I*(v)/(R)

L = 60 *(1)/(2)

L = 30 kg m^2/s

The brakes of a car moving at 14m/s are applied, and the car comes to a stop in 4s. (a) What was the cars acceleration? (b) How long would the car take to come to a stop starting from 20m/s with the same acceleration? (c) How long would the car take to slow down from 20m/s to 10m/s with the same acceleration?

Answers

(1) The acceleration of the car will be a=-3.5(m)/(s^2)

(2) The time taken t=5.7s

(3)  The time is taken by the car  to slow down from 20m/s to 10m/s t=2.9s

What will be the acceleration and time of the car?

(1) The acceleration of the car will be calculated as

a=(v-u)/(t)

Here

u= 14 (m)/(s)

a=(0-14)/(4) =-3.5(m)/(s^2)

(2) The time is taken for the same acceleration to 20(m)/(s)

a=(v-u)/(t)

t=(v-u)/(a)

u=20(m)/(s)

t=(0-20)/(-3.5) =5.7s

(3) The time is taken to slow down from 20m/s to 10m/s with the same acceleration

From same formula

t=(v-u)/(a)

v=10(m)/(s)

u=20(m)/(s)

t=(10-20)/(-3.5) =2.9s

Thus

(1) The acceleration of the car will be a=-3.5(m)/(s^2)

(2) The time taken t=5.7s

(3)  The time is taken by the car  to slow down from 20m/s to 10m/s t=2.9s

To know more about the Equation of the motion follow

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(a) -3.5 m/s^2

The car's acceleration is given by

a=(v-u)/(t)

where

v = 0 is the final velocity

u = 14 m/s is the initial velocity

t = 4 s is the time elapsed

Substituting,

a=(0-14)/(4)=-3.5 m/s^2

where the negative sign means the car is slowing down.

(b) 5.7 s

We can use again the same equation

a=(v-u)/(t)

where in this case we have

a=-3.5 m/s^2 is again the acceleration of the car

v = 0 is the final velocity

u = 20 m/s is the initial velocity

Re-arranging the equation and solving for t, we find the time the car takes to come to a stop:

t=(v-u)/(a)=(0-20)/(-3.5)=5.7 s

(c) 2.9 s

As before, we can use the equation

a=(v-u)/(t)

Here we have

a=-3.5 m/s^2 is again the acceleration of the car

v = 10 is the final velocity

u = 20 m/s is the initial velocity

Re-arranging the equation and solving for t, we find

t=(v-u)/(a)=(10-20)/(-3.5)=2.9 s

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