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Readme [11.4K]
4 years ago
8

The aircraft wing from problem 6 experiences temperature extremes that span 210 degrees Celsius. The component for the wing will

have a length of exactly 3 meters. Testing indicates that the aircraft wing will remain stable only if the component never expands to a length larger than 3.017 meters. If the component is made from the metal alloy in question, will it meet this requirement. An unknown metal alloy is being tested to discover its thermal properties to see if it is suitable for use as a component in an aircraft wing. The alloy is formed into a bar measuring 1 meter in length, and is then heated from its starting temp. of 30°C to a final temperature of 100°C. The length of the heated bar is measured to be exactly 1.002 meters in length.
Required:
What is the coefficient of thermal expansion of the alloy?
Physics
1 answer:
ExtremeBDS [4]4 years ago
5 0

Answer:

α = 2,857 10⁻⁵ ºC⁻¹

Explanation:

The thermal expansion of materials is described by the expression

          ΔL = α Lo ΔT

          α = \frac{\Delta L}{L_o \ \Delta T}

in the case of the bar the expansion is

         ΔL = L_f - L₀

         ΔL=   1.002 -1

         ΔL = 0.002 m

the temperature variation is

         ΔT = 100 - 30

         ΔT = 70º C

we calculate

         α = 0.002 / 1 70

         α = 2,857 10⁻⁵ ºC⁻¹

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3 years ago
Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates
Ksivusya [100]

1. How fast is the blue car going 1.8 seconds after it starts?

Recall this kinematic equation:

Vf = Vi + aΔt

Vf is the final velocity.

Vi is the initial velocity.

a is the acceleration.

Δt is the amount of elapsed time.


Given values:

Vi = 0 m/s (the car starts at rest)

a = 3.7 m/s² (this is the acceleration between t = 0s and t = 4.4s)

Δt = 1.8 s


Substitute the terms in the equation with the given values and solve for Vf:

Vf = 0 + 3.7×1.8

<h3>Vf = 6.66 m/s</h3><h3 />

2. How fast is the blue car going 10.0 seconds after it starts?

The car stops accelerating after t = 4.4s and continues at a constant velocity for the next 8.3 seconds. This means the car is traveling at a constant velocity between t = 4.4s and t = 12.7s. At t = 10s the car is still traveling at this constant velocity.


We must use the kinematic equation from the previous question to solve for this velocity. Use the same values except Δt = 4.4s which is the entire time interval during which the car is accelerating:

Vf = 0 + 3.7×4.4

Vf = 16.28 m/s

<h3>The constant velocity at which the car is traveling at t = 10s is 16.28 m/s</h3>

3. How far does the blue car travel before its brakes are applied to slow down?

We must break down the car's path into two parts: When it is traveling under constant acceleration and when it is traveling at constant velocity.


Traveling under constant acceleration:

Recall this kinematic equation:

d = \frac{Vi+Vf}{2}×Δt

d is the distance traveled.

Vi is the initial velocity.

Vf is the final velocity.

Δt is the amount of elapsed time.


Given values:

Vi = 0 m/s (the car starts at rest).

Vf = 16.28 m/s (determined from question 2).

Δt = 4.4 s


Substitute the terms in the equation with the given values and solve for d:

d = \frac{0+16.28}{2}×4.4

d = 35.8 m


Traveling at constant velocity:

Recall the relationship between velocity and distance:

d = vΔt

d is the distance traveled.

v is the velocity.

Δt is the amount of elapsed time.


Given values:

v = 16.28 m/s (the constant velocity from question 2).

Δt = 8.3 s (the time interval during which the car travels at constant velocity)

Substitute the terms in the equation with the given values:

d = 16.28×8.3

d = 135.1 m


Add up the distances traveled.

d = 35.8 + 135.1

<h3>d = 170.9 m</h3>

4. What is the acceleration of the blue car once the brakes are applied?

Recall this kinematic equation:

Vf²=Vi²+2ad

Vf is the final velocity.

Vi is the initial velocity.

a is the acceleration

d is the distance traveled.


Given values:

Vi = 16.28 m/s

Vf = 0 m/s

d = 216 m - 170.9 m = 45.1 m (subtracting the distance already traveled from the total path length)


Substitute the terms in the equation with the given values and solve for a:

0² = 16.28²+2a×45.1

<h3>a = -2.94 m/s²</h3>

5. What is the total time the blue car is moving?

We already know the time during which the car is traveling under constant acceleration and traveling at constant velocity. We now need to solve for the amount of time during which the car is decelerating.

Recall again:

d = \frac{Vi+Vf}{2}×Δt

Given values:

d = 45.1 m

Vi = 16.28 m/s (the velocity the car was traveling at before hitting the brakes).

Vf = 0 m/s (the car slows to a stop).


Substitute the terms in the equation with the given values and solve for Δt:

45.1 = \frac{16.28+0}{2}×Δt

Δt = 5.54s


Add up the times to get the total travel time:

t = 4.4 + 8.3 + 5.54 =

<h3>t = 18.24s</h3>

6. What is the acceleration of the yellow car?

Recall this kinematic equation:

d = ViΔt + 0.5aΔt²

d is the distance traveled.

Vi is the initial velocity.

a is the acceleration.

Δt is the amount of elapsed time.


Given values:

d = 216 m (both cars meet at 216m)

Vi = 0 m/s (the car starts at rest)

Δt = 18.24 s (take the same amount of time to reach 216m)


Substitute the terms in the equation with the given values and solve for a:

216 = 0×18.24 + 0.5a×18.24²

<h3>a = 1.3 m/s²</h3>
6 0
4 years ago
Two engineering students, John with a weight of 96 kg and Mary with a weight of 48 kg, are 30 m apart. Suppose each has a 0.04%
djyliett [7]

Answer:

6.8370869499\times 10^{20}\ N

Explanation:

N_A = Avogadro's number = 6.022\times 10^{23}

e = Charge of electron = 1.6\times 10^{-19}\ C

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

Z = Atomic number of water = 18

M = Molar mass of water = 0.018 kg/mol

m = Mass of person

The charge is given by

q=imbalance\times n\times e

Total number of protons and electrons in each sphere

n=\dfrac{mN_AZe}{M}

q=imbalance\times \dfrac{mN_AZe}{M}

q_1=0.0004\times \dfrac{96\times 6.022\times 10^{23}\times 18\times 1.6\times 10^{-19}}{0.018}\\\Rightarrow q_1=3699916.8\ C

q_2=0.0004\times \dfrac{48\times 6.022\times 10^{23}\times 18\times 1.6\times 10^{-19}}{0.018}\\\Rightarrow q_1=1849958.4\ C

Electrical force is given by

F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^{9}\times 3699916.8\times 1849958.4}{30^2}\\\Rightarrow F=6.8370869499\times 10^{20}\ N

The electrostatic force of attraction between them is 6.8370869499\times 10^{20}\ N

4 0
4 years ago
We can use the formula r=squareroot s/12.6 to relate a ball's surface area (in square inches) to its radius (in inches). Suppose
Charra [1.4K]

Answer:

The radius is 5.56 inches.

Explanation:

Given that,

Surface area = 390.6 inch²

We need to calculate the radius

Using given formula

r =\sqrt{\dfrac{S}{12.6}}

S = surface area

r = radius

Put the value of S in to the formula

r =\sqrt{\dfrac{390.6}{12.6}}

r =5.56\ inches

Hence, The radius is 5.56 inches.

3 0
3 years ago
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