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3 years ago

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Jack tries to place magnets on the door of his refrigerator. He observes that the magnets don’t stick. He guesses that the door

of the refrigerator is made of a nonmagnetic substance. What stage of scientific investigation does his guess represent?
A. ask questions
B. construct a hypothesis
C. explain the results
D. make observations

2 answers:

BigorU [14]3 years ago

6 0

His guess is the hypothesis.

zaharov [31]3 years ago

3 0

The answer is D.) Make observations

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A closed, rigid tank fitted with a paddle wheel contains 2 kg of air, initially at 300 K. During an interval of 5 minutes, the p

anzhelika [568]

Answer:

The final temperature of the air is $T_2= 605 K$

Explanation:

We can start by doing an energy balance for the closed system

$\Delta KE+\Delta PE+ \Delta U = Q - W$

where

$\Delta KE$ = the change in kinetic energy.

$\Delta PE$ = the change in potential energy.

$\Delta U$ = the total internal energy change in a system.

Q = the heat transferred to the system.

W = the work done by the system.

We know that there are no changes in kinetic or potential energy, so $\Delta KE = 0$ and $\Delta PE=0$

and our energy balance equation is $\Delta U = Q - W$

We also know that the paddle-wheel transfers energy to the air at a rate of 1 kW and the system receives energy by heat transfer at a rate of 0.5 kW, for 5 minutes.

We use this information to calculate the total internal energy change $\Delta U=W+Q$ using the energy balance equation.

We convert the interval of time to seconds $t = 5 \:min = 300\:s$

$\Delta \dot{U}=\dot{W}+ \dot{Q}\\=\Delta U=(W+ Q)\cdot t$

$\Delta U=(1 \:kW+0.5\:kW)\cdot 300\:s\\\Delta U=450 \:kJ$

We can use the change in specific internal energy $\Delta U = m(u_2-u_1)$ to find the final temperature of the air.

We are given that $T_1=300 \:K$ and the air can be describe by ideal gas model, so we can use the ideal gas tables for air to determine the initial specific internal energy $u_1$

$u_1=214.07\:\frac{kJ}{kg}$

Next, we will calculate the final specific internal energy $u_2$

$\Delta U = m(u_2-u_1)\\\frac{\Delta U}{m} =u_2-u_1$

$\frac{\Delta U}{m} =u_2-u_1\\u_2=u_1+\frac{\Delta U}{m}$

$u_2=214.07 \:\frac{kJ}{kg} +\frac{450 \:kJ}{2 \:kg}\\u_2= 439.07 \:\frac{kJ}{kg}$

With the value $u_2=439.07 \:\frac{kJ}{kg}$ and the ideal gas tables for air we make a regression between the values $u = 434.78 \:\frac{kJ}{kg},T=600 \:K$ and $u = 442.42 \:\frac{kJ}{kg}, T=610 \:K$ and we find that the final temperature $T_2$ is 605 K.

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2 years ago

Pls answer……………………..

UkoKoshka [18]

2nd egg experienced more impulse

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3 years ago

Consider the electric field lines shown in the diagram below. From the diagram, it is apparent that object A is ____ and object

Mariana [72]

Positive charge you gave the lines pointing away, negative charge is pointing toward. Don’t have a photo so I can’t fill in the blanks BUT I can tell you the logic

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2 years ago

An object is weighed at different locations on the

Step2247 [10]

Answer:

C

Explanation:

The weight will always be different while mass is described as the stuff inside an object, and that stays the same.

Such as it weighs differently in space.

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2 years ago

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What % of an object’s mass is above the water line if the object’s density is 0.82g/ml?

Sladkaya [172]

To develop this problem we will apply the Archimedes model. As well as the definitions of Weight based on mass and acceleration. The first in turn will be considered under the relationship of Density and Volume. From the values given we have to:

$\rho_w =1 g/mL \rightarrow \text{Water Density}$

$\rho_o = 0.82g/mL \rightarrow \text{Object density}$

Since it is in equilibrium, the weight of the object will have a reaction from the water, which will cause the sum of forces between the two objects to be zero, therefore

$\sum F= 0$

$F_w-F_o = 0$

$F_w = F_o$

$m_w g = m_og$

$\rho_w V_w g = \rho_o V_o g$

The value of gravity is canceled because it is a constant

$\frac{V_w}{V_o} = \frac{\rho_o}{\rho_w}$

$\frac{V_w}{V_o} = \frac{0.82}{1}$

$\frac{V_w}{V_o} = 0.82$

The portion of the object that is submerged corresponds to 82%, while the portion that is visible, above the water level will be 18%

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3 years ago