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valkas [14]
3 years ago
12

Jade wanted to test the effect of ice on the weathering of rocks. She filled two containers with gypsum and placed a water ballo

on in one of the containers. She put both containers in a freezer for few hours and cut open the containers. Which of the following is an observation that Jade will most likely make after she cut open the containers?
NO LINKS PLEASE I NEED HELP FAST PLEASE
Engineering
1 answer:
Aleks [24]3 years ago
7 0

Answer:

 

The gypsum block with the water balloon has contracted.

Explanation:

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Water exiting the condenser of a power plant at 45 Centers a cooling tower with a mas flow rate of 15,000 kg/s. A stream of cool
MariettaO [177]

Answer: hello your question is incomplete below is the missing part

question :Determine the temperature of the cooled water exiting the cooling tower

answer : T  = 43.477° C

Explanation:

Temp of water at exit = 45°C

mass flow rate of cooling tower = 15,000 kg/s

Temp of makeup water = 20°C

Assuming an atmospheric pressure of = 101.3 kPa

<u>Determine temperature of the cooled water exiting the cooling tower</u>

Water entering cooling tower at 45°C

Given that Latent heat of water at 45°C = 43.13 KJ/mol

Cp(wet air) = 1.005+ 1.884(y1)

where: y1 - Inlet mole ratio = (0.01257) / (1 - 0.01257) = 0.01273

Hence : Cp(wet air) = 29.145 +  (0.01273) (33.94) = 29.577 KJ/kmol°C

<u>First step : calculate the value of Q </u>

Q = m*Cp*(ΔT) + W(latent heat)

Q = 321.6968 (29.577) (40-30) +  43.13 (18.26089)

Q =  95935.8547 KJ/s

Given that mass rate of water = 15000 kg/s

<u>Hence the temperature of the cooled water can be calculated using the equation below</u>

Q = m*Cp*∆T

Cp(water) = 4.2 KJ/Kg°C

95935.8547 = (15000)*(4.2)*(45 - T)

( 45 - T ) = 95935.8547/ 63000.    ∴ T  = 43.477° C

5 0
3 years ago
It is known that the connecting rod AB exerts on the crank BCa 2.5-kN force directed down andto the left along the centerline of
monitta

Answer:

M_c = 61.6 Nm

Explanation:

Given:

F_a = 2.5 KN

Find:

Determine the moment of this force about C for the two casesshown.

Solution:

- Draw horizontal and vertical vectors at point A.

- Take moments about point C as follows:

                         M_c = F_a*( 42 / 150 ) *88

                         M_c = 2.5*( 42 / 150 ) *88

                         M_c = 61.6 Nm

- We see that the vertical component of force at point A passes through C.

Hence, its moment about C is zero.

6 0
3 years ago
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Brums [2.3K]

Answer:

Cool song

Explanation:

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4 0
3 years ago
A 46.0-g meter stick is balanced at its midpoint (50.0 cm, zero point is a left end of stick). Then a 210.0-g weight is hung wit
Anna71 [15]

Clockwise torque due to 100g is 0.1029 Nm and 200g is 1.4406 Nm. Clockwise torque due to stick mass is 0.2254 Nm and Counter-clockwise torque due to normal force is 1.7689 Nm.            

<h3>What is clockwise torque?</h3>

The right-hand rule for cross products determines the direction of torque, which is calculated as the cross product of force and distance. Your thumb will point in the direction of the torque if you place your palm in the direction of the applied force and extend your fingers from the pivot point in that direction.

A related right-hand rule relates the direction of the rotation to the direction of the torque. Your fingers will curl in the direction of rotation if you point your thumb in the direction of the torque.

Positive torques cause counter clockwise rotation, while negative torques cause clockwise rotation.

The sum of all torques must be zero at equilibrium since an object in equilibrium has no net torque.

When the force is applied in a direction perpendicular to the line connecting the pivot and the force, the torque is at its greatest.

You can calculate the torque's magnitude using

                                             \begin{displaymath}\tau =rF_{\bot }=rF\sin \theta .\end{displaymath}

To solve problems involving torques, follow these eight steps: read the issue, create a free-body diagram, locate the pivot point, write down the expressions for all torques, For equilibrium conditions, set the sum of torques to zero, list all known variables, pick the desired variable(s), write down equations involving those variable(s), solve the equations, plug in numbers, and test your solution.

Clockwise torque due to 100 g                                                                         ⇒ T1 = 0.105* 9.8* 0.1 = 0.1029 Nm

Clockwise torque due to 200 g                                                                                                      ⇒ T2 = 0.210* 9.8* 0.7 = 1.4406 Nm

Clockwise torque due to stick mass                                                                               ⇒ T3 = 0.046* 0.5* 9.8 =0.2254 Nm

Counter-clockwise torque due to normal force                                                                             ⇒ T4 = (0.046 + 0.21 + 0.105)*9.8* 0.5 = 1.7689 Nm

Learn more about torque

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7 0
1 year ago
Why do giant stars become planetary nebulas while supergiant stars become supernovas when their nuclear fusion slows and is over
sashaice [31]

The reason why giant stars become planetary nebulas is  Supergiant stars do not have enough mass to generate the gravity necessary to cause a planetary nebula.

<h3>Why do giant stars become planetary nebulae?</h3>

A planetary nebula is known to be formed or created by a dying star. A red giant is known to be unstable and thus emit pulses of gas that is said to form a sphere around the dying star and thus they are said to  be ionized by the ultraviolet radiation that the star is known to releases.

Learn more about  giant stars from

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3 0
2 years ago
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