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Lerok [7]
3 years ago
6

The concrete sections of a certain superhighway are designed to have a length of 23.0 m. The sections are poured and cured at 10

.0°C. What minimum spacing should the engineer leave between the sections to eliminate buckling if the concrete is to reach a temperature of 42.0°C?
Physics
1 answer:
loris [4]3 years ago
8 0

Answer:

0.88 cm

Explanation:

Initial length at 10°c, L = 23 m

Rise of temperature, Δt = 42 - 10 = 32°c

Coefficient of linear expansion of concrete, α = 12 x 10^(-6) per°c

Minimum spacing to be left = α L Δt

=12\times10^{-6}\times23\times32

=8.832\times10^{-3}m

= 8.832 x 10^{-1} cm

= 0.88 cm

Note: I have a chosen a general value of 12 x 10^-6 per deg c for coefficient of expansion of  concrete. However, please refer to the value given in your text book and substitute it for an accurate answer.

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What are tectonic plates, and which of Earht's layers are they composed of?
motikmotik

The tectonic plates are made up of Earth's crust and the upper part of the mantle layer underneath. Together the crust and upper mantle are called the lithosphere. hope this helps :)

3 0
3 years ago
A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the ca
svetoff [14.1K]

Answer:

q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Explanation:

Given that L=1H, R=5000\Omega, \ C=0.25\times10^{-6}F, \ \ E(t)=12V, we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:

E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000  \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}

#To find the particular solution:

Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Hence the charge at any time, t is q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

6 0
3 years ago
At time t1 = 14 s, a car is located at 99, 80, 27 m and has velocity 4, 0, −3 m/s. At time t2 = 18 s, what is the position of th
Korvikt [17]

Answer:

115, 80, 15m

Explanation

t1 = 14s

t2 = 18s

change in time = 4s (18-14)

r(final) = r(initial) + (average velocity) x (change in time)

multiply the average velocity with the change in time

= (4, 0, -3) x 4 = 16, 0, -12

now we'll add this value to the initial position of the car

(99, 80, 27)m + (16, 0, -12)m = (115, 80, 15)m

8 0
4 years ago
Which best explains how an object at rest deep in space and far from any massive body behaves compared to an object in free fall
7nadin3 [17]

Answer: They behave the same because, according to the principle of equivalence, the laws of physics work the same in all frames of reference.

Explanation:

According to the equivalence principle postulated by Einstein's Theory of General Relativity, acceleration in space and gravity on Earth have the same effects on objects.

To understand it better, regarding to the equivalence principle, Einstein formulated the following:  

A gravitational force and an acceleration in the opposite direction are equivalent, both have indistinguishable effects.  Because the laws of physics must be accomplished in all frames of reference.

Hence, according to general relativity, gravitational force and acceleration in the opposite direction (an object in free fall, for example) have the same effect.  This makes sense if we deal with gravity not as a mysterious atractive force but as a geometric effect of matter on spacetime that causes its deformation.

3 0
3 years ago
If you can swim in still water at 0.5m/s, the shortest time it would take you to swim from bank to bank across a 20m wide river,
expeople1 [14]

Answer:

t=40s,

Explanation:

If you can swim in still water at 0.5m/s, the shortest time it would take you to swim from bank to bank across a 20m wide river, if the water flows downstream at a rate of 1.5m/s, is most nearly:

from the question the swimmer will have a velocity which is equal to the sum of the speed of the water and the velocity to swi across the bank

Vt=v1+v2

the time is takes to swim across the bank will be

DY=Dv*t

DY=distance across the bank

Dv=ther velocity of the swimmer across the bank

t=20/ 0.5m/s,

t=40s, time it takes to swim across the bank

velocity is the rate of displacement

displacement is distance covered in a specific direction

4 0
3 years ago
Read 2 more answers
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