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brilliants [131]
3 years ago
7

Calculate the change in length of concrete sidewalk (coefficient of linear expansion for concrete is 12*10^-6/celcius) that is 1

50 meters that undergoes a temperature change of 30 degrees
Physics
1 answer:
anyanavicka [17]3 years ago
7 0

Answer:

The answer to your question is 5.4 cm

Explanation:

This problem refers to calculate the change in length in one dimension due to a change in temperature.

Data

α = 12 x 10⁻⁶

Lo = 150 meters

ΔT = 30 °C

Formula

                    ΔL/Lo = αΔT

solve for ΔL

                    ΔL = αLoΔT

Substitution

                    ΔL = (12 x 10⁻⁶)(150)(30)

Simplification

                    ΔL = 0054 m = 5.4 cm

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Precision is how close a series of measurements are to ______
Leya [2.2K]

Answer:

Accuracy is how close a measured value is to an accepted value. <u>Precision is how close measurements are to one another.</u> To make measurements, you have to evaluate both the accuracy and the precision to get a correct value.

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2 years ago
Two identical balls are thrown vertically upward. the second ball is thrown with an initial speed that is twice that of the firs
Temka [501]
The motion of the ball on the vertical axis is an accelerated motion, with acceleration 
a=g=-9.81 m/s^2
The following relationship holds for an uniformly accelerated motion:
2aS=v_f^2 - v_i^2
where S is the distance covered, vf the final velocity and vi the initial velocity.

If we take the moment the ball reaches the maximum height (let's call this height h), then at this point of the motion the vertical velocity is zero:
v_f =0
So we can rewrite the equation as
2(-9.81 m/s^2) h=-v_i^2
from which we can isolate h
h= \frac{v_i^2}{19.62} (1)

Now let's assume that v_i is the initial velocity of the first ball. The second ball has an initial velocity that is twice the one of the first ball: 2v_i. So the maximum height of the second ball is
h= \frac{(2v_i)^2}{19.62}= \frac{4v_i^2}{19.62} (2)

Which is 4 times the height we found in (1). Therefore, the maximum height of ball 2 is 4 times the maximum height of ball 1.
8 0
3 years ago
A water main pipe of diameter 10 cm enters a house 2 m below ground. A smaller diameter pipe carries water to a faucet 5 m above
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Explanation:

Given that,

Diameter = 10 cm

Distance = 2 m

Speed v_{1}= 2\ m/s

Speed v_{2}=7\ m/s

Pressure in main pipe P_{1}=2\times10^{5}\ Pa

(I). We need to calculate the diameter

Using equation of continuity

Av_{1}=Av_{2}

\pi(\dfrac{d_{1}}{2})^2\times v_{1}=\pi(\dfrac{d_{2}}{2})^2\times v_{2}

(\dfrac{10}{2})^2\times2=(\dfrac{d_{2}}{2})^2\times7

d_{2}=\sqrt{\dfrac{25\times2\times4}{7}}

d_{2}=5.345\ cm

(II). We need to calculate the pressure the gauge pressure

Using Bernoulli equation

P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}

P_{2}=P_{1}+\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)-\rho g(h_{1}-h_{2})

P_{2}=2\times10^{5}+\dfrac{1}{2}\times1000(4-49)-1000\times 9.8\times(5)

P_{2}=1.28500\times10^{5}\ Pa

(III).  If it is possible to carry water to a faucet 17 m above ground,

Using Bernoulli equation

P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh=P_{3}+\dfrac{1}{2}\rho v_{3}^2+\rho g h_{3}

P_{3}=P_{1}+\dfrac{1}{2}\rho v_{1}^2-\rho g(h_{1}-h_{3})

Here, h_{3}=0

Put the value in the equation

P_{3}=2\times10^{5}+\dfrac{1}{2}\times1000\times4-1000\times 9.8\times17

P_{3}=3.5400\times10^{5}\ Pa

Hence, This is required solution.

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Explanation:

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