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Tcecarenko [31]
4 years ago
13

Which statements can be supported by using a position-time graph? Check all that apply. A negative slope results when an individ

ual is moving toward the origin. A horizontal line on the graph means the individual is moving at the same velocity. A positive slope results when an individual is moving away from the origin. The slope of the line gives the velocity of the individual. The speed of an individual cannot be determined from this type of graph.
Physics
2 answers:
Naya [18.7K]4 years ago
8 0

A negative slope results when an individual is moving toward the origin.

A positive slope results when an individual is moving away from the origin.

The slope of the line gives the velocity of the individual.

Explanation:

In a position-time graph, the position is represented on the y-axis (distance from the origin), while the time is represented on the x-axis. Let's analyze each statement:

A negative slope results when an individual is moving toward the origin. --> TRUE. In fact, a negative slope means that the value on the y is decreasing as the time passes, so the distance of the individual is decreasing as the time increases, therefore the person is moving towards the origin.

A horizontal line on the graph means the individual is moving at the same velocity. --> FALSE. A horizontal line means that the position of the individual is constant, so the person is not moving.

A positive slope results when an individual is moving away from the origin. --> TRUE. In this case, the value on the y is increasing with time, so the distance of the person from the origin is increasing with time.

The slope of the line gives the velocity of the individual. --> TRUE. In a position-time graph, the velocity of the individual is equal to the ratio between the displacement and the time taken. But the displacement corresponds to the increase in the y-value, while the time corresponds to the increase in the x-value, so the velocity corresponds to the slope of the graph. In order to know velocity, we must also know the direction of the motion.

The speed of an individual cannot be determined from this type of graph. --> FALSE. The speed is equal to the slope of the position-time graph (in fact, the speed is the magnitude of the velocity)

Mademuasel [1]4 years ago
7 0

Answer:

A negative slope results when an individual is moving toward the origin.

A positive slope results when an individual is moving away from the origin.

The slope of the line gives the velocity of the individual.

Explanation:

Just took the quiz on ed

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Answer:

yes

Explanation:

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What happens when charged object is brought near uncharged object?<br> Attract or Repel ?
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When a charged object is brought near to but does not touch a neutral object, it causes the side of the neutral object that the charged object is near to become the other charge. It causes charge migration within the neutral object so the two charges (positive and negative) move to opposite sides of the object. Because the two objects do not touch, they do not repel each other, but rather have a slight attraction because of charge migration. If the two object were to touch then they would repel.
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3 years ago
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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
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Answer:

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Explanation:

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

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

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\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.

3 0
3 years ago
In the first stage of a two-stage Carnot engine, energy is absorbed as heat Q1 at temperature T1 = 500 K, work W1 is done, and e
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Answer:

Efficiency = 52%

Explanation:

Given:

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Heat released, Q₂ at temperature T₂ = 430 K

and the work done is W₁

Second stage

Heat released, Q₂ at temperature T₂ = 430 K

Heat released, Q₃ at temperature T₃ = 240 K

and the work done is W₂

Total work done, W = W₁ + W₂

Now,

The efficiency is given as:

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or

Work done = change in heat

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W₂ = Q₂ - Q₃

Thus,

\eta=\frac{(Q_1-Q_2)\ +\ (Q_2-Q_3)}{Q_1}}

or

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or

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\eta=1-\frac{(T_3)}{T_1}}

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Argon has 22 Neutrons, because Protons(18)+Neutrons(22)=Mass Number(40).

Hope I could help!
8 0
4 years ago
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