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3 years ago

if an object exposed to salt water that seeps into cracks and forms crystals what is likely to occur?

Physics

1 answer:

suter [353]3 years ago

8 0

Breakage and/or expansion I'm pretty sure
Also, larger cracks bc of the crystallization of the salt water

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A scientist is trying to determine whether a newly observed phenomenon can be described as a wave. Which of the following questi

Step2247 [10]

Answer:

B. does it transfer energy

A wave is a disturbance in a medium. Particles vibrate and TRANSFER their ENERGY to their neighbors.

7 0

3 years ago

You are camping with two friends, Joe and Karl. Since all three of you like your privacy, you don't pitch your tents close toget

aleksklad [387]

Answer:

35.7 m

Explanation:

Let

$\mid A\mid=18.5 m$

$\mid B\mid=41 m$

We have to find the distance between Joe's and Karl'e tent.

$A_x=Acos\theta$

$A_y=Asin\theta$

Substitute the values then we get

$A_x=18.5cos23^{\circ}=17 m$

$A_y=18.5sin 23^{\circ}=7.2 m$

$B_x=41cos37.5^{\circ}=32.5 m$

$B_y=41sin37.5^{\circ}=-24.96 m$

Because vertical component of B lie in IV quadrant and y-inIV quadrant is negative.

By triangle addition of vector

$B=A+C$

$C=B-A$

$C_x=B_x-A_x=32.5-17=15.5 m$

$C_y=B_y-A_y=-24.96-7.2=-32.16\approx=-32.2 m$

$\mid C\mid=\sqrt{C^2_x+C^2_y}$

$\mid C\mid=\sqrt{(15.5)^2+(-32.2)^2}=35.7 m$

Hence, the distance between Joe's and Karl's tent=35.7 m

6 0

3 years ago

The average coefficient of linear expansion of copper is 1.7 10-5 (°c)−1. the statue of liberty is 93 m tall on a summer morning

sammy [17]

Let the rise in temperature be $5^0C$

The expansion in length due to change in temperature is given by the expression lαΔt , where l is the length, α is the coefficient of linear expansion, Δt is the change in temperature.

Here l = 93 m, α = $1.7*10^{-5}$ $^0C^{-1}$ , and Δt = $5^0C$

So expansion in length = 93* $1.7*10^{-5}$ *5 = 0.007905 m = $0.79*10^{-3}m$

So order of magnitude in change in length = -3

3 0

3 years ago

Read 2 more answers

HELP ME HOW ARE BLACK HOLES FROMED

V125BC [204]

Answer:

Stellar black holes form when the center of a very massive star collapses in upon itself.

6 0

3 years ago

Read 2 more answers

A piece of aluminum has a volume of 1.50 10-3 m3. the coefficient of volume expansion for aluminum is β = 69 ✕ 10-6 (°c)-1. the

Alex17521 [72]

Answer:

W = 3.12 J

Explanation:

Given the volume is 1.50*10^-3 m^3 and the coefficient of volume for aluminum is β = 69*10^-6 (°C)^-1. The temperature rises from 22°C to 320°C. The difference in temperature is 320 - 22 = 298°C, so ΔT = 298°C. To reiterate our known values we have:

β = 69*10^-6 (°C)^-1 V = 1.50*10^-3 m^3 ΔT = 298°C

So we can plug into the thermal expansion equation to find ΔV which is how much the volume expanded (I'll use d instead of Δ because of format):

$dV = \beta V_{0} dT\\dV = (69*10^{-6})( C)^{-1} * (1.50*10^{-3})m^{3} * (298)C\\dV = 3.0843*10^-5$

So ΔV = 3.0843*10^-5 m^3

Now we have ΔV, next we have to solve for the work done by thermal expansion. The air pressure is 1.01 * 10^5 Pa

To get work, multiply the air pressure and the volume change.

$W = P * dV = (1.01 * 10^5)Pa * (3.0843*10^{-5})m^3\\W = 3.115143J$

W = 3.12 J

Hope this helps!

4 0

3 years ago