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irga5000 [103]
3 years ago
13

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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How much work must be done on a 20-kg go-cart to increase its speed from 5 m/s to 10 m/s?
Rudiy27
The answer is the change in kinetic energy

K = 1/2 • mv^2

Change in K = (1/2 • 20 • 10^2) - (1/2 • 20 • 5^2)

Change in K = 1000 - 250 = 750J
5 0
3 years ago
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Two streams merge to form a river. One stream has a width of 8.3 m, depth of 3.2 m, and current speed of 2.2 m/s. The other stre
katovenus [111]

Answer:

The depth of the resulting stream is 3.8 meters.

Explanation:

Under the assumption that streams are formed by incompressible fluids, so that volume flow can observed conservation:

\dot V_{1} + \dot V_{2} = \dot V_{3} (1)

All volume flows are measured in cubic meters per second.

Dimensionally speaking, we can determine the depth of the resulting stream (h_{3}), in meters, by expanding (1) in this manner:

w_{1}\cdot h_{1}\cdot v_{1} + w_{2}\cdot h_{2}\cdot v_{2} = w_{3}\cdot h_{3}\cdot v_{3}

h_{3} = \frac{w_{1}\cdot h_{1}\cdot v_{1}+w_{2}\cdot h_{2}\cdot v_{2}}{w_{3}\cdot v_{3}} (2)

v_{1}, v_{2} - Speed of the merging streams, in meters per second.

h_{1}, h_{2} - Depth of the merging streams, in meters.

w_{1}, w_{2} - Width of the merging streams, in meters.

w_{3} - Width of the resulting stream, in meters.

v_{3} - Speed of the resulting stream, in meters per second.

If we know that w_{1} = 8.3\,m, h_{1} = 3.2\,m, v_{1} = 2.2\,\frac{m}{s}, w_{2} = 6.8\,m, h_{2} = 3.2\,m, v_{2} = 2.4\,\frac{m}{s}, w_{3} = 10.4\,m and v_{3} = 2.8\,\frac{m}{s}, then the depth of the resulting stream is:

h_{3} = \frac{(8.3\,m)\cdot (3.2\,m)\cdot \left(2.2\,\frac{m}{s} \right) + (6.8\,m)\cdot (3.2\,m)\cdot \left(2.4\,\frac{m}{s} \right)}{(10.4\,m)\cdot \left(2.8\,\frac{m}{s} \right)}

h_{3} = 3.8\,m

The depth of the resulting stream is 3.8 meters.

3 0
3 years ago
At a sudden contraction in a pipe the diameter changes from D 1 to D 2 . The pressure drop, Δ p , which develops across the cont
Thepotemich [5.8K]

Answer:

Velocity in the smaller pipe should not be included as an additional variable.    

Explanation:

$\Delta P = f(D_1, D_2, V, \rho, \mu)$

The dimensional formula of the variables are

$\Delta P = FL^{-2} , D_1 = L, D_2=L, V=LT_{-1}, \rho =FL^{-4}T^2, \mu = FL^{-2}T$

Now using Buchingham's Pi Theorem, 6 - 3 = 3 dimensional parameters are required.

Use, D_1, V, \mu as the repeating variables.

Therefore, $\pi = \Delta PD_1^aV^b \mu^c$

$(FL^{-2})(L)^a(LT^{-1})^b(FL^{-2}T)^c = F^0L^0T^0$

From this

1+c=0

-2+a+b-2c=0

-b+c=0

c=-1, b = -1, a = 1

Now, $\pi_1=\frac{\Delta PD_1}{V\mu}=\frac{(ML^{-1}T^{-2})L}{(LT^{-1})(ML^{-1}T^{-1})}=M^0L^0T^0$

For $\pi_2 = D_2D_1^aV^b\mu^c$

$F^0L^0T^0=L(L)^a(LT^{-1})^b(FL^{-2}T)^c$

c = 0

1 + a + b - 2c = 0

-b + c = 0

Therefore, a = -1, b = 0, c = 0

$\pi_2 = \frac{D_2}{D_1}$

For $\pi_3$

$\pi_3 = \rho, D_1^a V^b\mu^c$

$F^0L^0T^0 = (FL^{-4}T^2)(L)^a(LT^{-1})^b(FL^{-2}T)^c$

1 + c = 0

-4 + a + b - 2c = 0

2-b+c=0

c=-1, b=1, a = 1

Therefore, $\pi_3  = \frac{\rho D_1V}{\mu}$

Now checking,

$\pi_3 = \frac{(ML^{-3})(L)(LT^{-1})}{ML^{-1}T^{-1}} = M^0L^0T^0$

Therefore, $\frac{\Delta P D_1}{V \mu} = \phi (\frac{D_2}{D_1}, \frac{\rho D_1 V}{\mu})$

From continuity equation

$V\frac{\pi}{4}D_1^2 = V_s \frac{\pi}{4}D_2^2$

$V_s = V (\frac{D_1}{D_2})^2$

$V_s$ is not independent of $D_1,D_2, V$

Therefore it should not be included as an additional variable.

3 0
3 years ago
Complete the passage to summarize factors affecting the speed of a wave.
makvit [3.9K]

Answer:

The material or substance that a wave moves through is called a The medium affects the spee of the wave that passes through it.

Another factor that affects wave speed is the of the medium.

One factor that affects the speed of a wave is the of medium.

Explanation:

8 0
4 years ago
If the designers of submarines do not appropriately account for hydrostatic pressure,
andre [41]

Answer:

A) the pressure could crush the submarine

Explanation:

  • The submarines are mainly used for defense purposes.
  • In world war 1 and 2 the submarines played a vital role in the war.
  • The submarines work according to the principle of buoyancy.
  • According to this principle, when seawater is filled inside the tank in a required amount, the submarines sink.
  • When the seawater is expelled, the submarine rises to above the sea.
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  • If the hydrostatic pressure is not considered, the pressure could crush the submarine.
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3 years ago
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