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noname [10]
3 years ago
9

A student is subjected to a reaction force of 10 N northward from a 5 kg block while pushing the block over a smooth, level sufa

ce. Ignoring friction, what is the acceleration of the block?
Physics
1 answer:
harkovskaia [24]3 years ago
5 0

Answer:

The Analysis Model approach we focus on in this revision lays out a standard set of situations that appear in most

physics problems. These situations are based on an entity in one of four simplification models: particle, system,

rigid object, and wave. Once the simplification model is identified, the student thinks about what the entity is

doing or how it interacts with its environment. This leads the student to identify a particular Analysis Model for the

problem. For example, if an object is falling, the object is recognized as a particle experiencing an acceleration due

to gravity that is constant. The student has learned that the Analysis Model of a particle under constant acceleration

describes this situation. Furthermore, this model has a small number of equations associated with it for use in starting problems, the kinematic equations presented in Chapter 2. Therefore, an understanding of the situation has led

to an Analysis Model, which then identifies a very small number of equations to start the problem, rather than the

myriad equations that students see in the text. In this way, the use of Analysis Models leads the student to identify

the fundamental principle. As the student gains more experience, he or she will lean less on the Analysis Model

approach and begin to identify fundamental principles directly.

To better integrate the Analysis Model approach for this edition, Analysis Model descriptive boxes have been

added at the end of any section that introduces a new Analysis Model. This feature recaps the Analysis Model introduced in the section and provides examples of the types of problems that a student could solve using the Analysis

Model. These boxes function as a “refresher” before students see the Analysis Models in use in the worked examples

for a given section.

Worked examples in the text that utilize Analysis Models are now designated with an AM icon for ease of reference. The solutions of these examples integrate the Analysis Model approach to problem solving. The approach is

further reinforced in the end-of-chapter summary under the heading Analysis Models for Problem Solving, and through

the new Analysis Model Tutorials that are based on selected end-of-chapter problems and appear in Enhanced

WebAssign.

Analysis Model Tutorials. John Jewett developed 16

Explanation:

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A dolphin swims 1.85 km/h how far has the dolphin traveled after 0.60 h
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A high school physics student is sitting in a seat read-
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The equilibrium condition allows finding the result for the force that the chair exerts on the student is:

  • The reaction force that the chair exerts on the student's support is equal to the student's weight.

Newton's second law gives the relationship between force, mass and acceleration of bodies, in the special case that the acceleration is is zero equilibrium condition.

            ∑ F = 0

Where F is the external force.

The free body diagram is a diagram of the forces on bodies without the details of the shape of the body, in the attached we can see a diagram of the forces.

Let's analyze the force on the chair.

            N_{chair} - W_{chair} - W_{student} = 0 \\ \\N_{chair} = W_{chair} + W_{student}

Let's analyze the forces on the student.

          N_{student} - W_{student} = 0  \\N_{student} = W _{student}

           

In conclusion using the equilibrium condition we can find the result for the force that the chair exerts on the student is:

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Learn more here: brainly.com/question/18117041

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3 years ago
A sphere is originally at a temperature of 500°c. The sphere is melted and recast, without loss of mass, into a cube with the sa
aleksandr82 [10.1K]

Answer: The value of the celsius temperature of the cube is 472.2°c.

Explanation:        

The expression for the power radiated is as follows;

P=A\epsilon\sigma T^{4}

Here, A is the area, \sigma is the stefan's constant,\epsilon is the emissivity and T is the temperature.

It is given in the problem that A sphere is originally at a temperature of 500°c. The sphere is melted and recast, without loss of mass, into a cube with the same emissivity as the sphere.

Then the expression for the radiated power for the cube and the sphere can be expressed as;

A_{1}\epsilon \e\sigma T_{1}^{4}=A_{2}\epsilon \e\sigma T_{2}^{4}

Here, A_{1} is the area of the sphere, A_{2} is the area of the cube,T_{1}  is the temperature of the sphere and T_{2}  is the temperature of the cube.

The radiated powers and emissivity of the cube and the sphere are same.

A_{1}T_{1}^{4}=A_{2}T_{2}^{4}

The area of the sphere is A_{1}=4\pi \times r^{2}.

Here, r is the radius of the sphere.

The area of the cube is A_{2}=6\times a^{2}.

Here, a is the edge of the cube.

Put A_{1}=4\pi \times r^{2} and A_{2}=6\times a^{2}.

T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}  ....(1)

The masses and the densities of the sphere and the cube are same. Then the volumes are also same.

V_{1}=V_{2}

Here,V_{1},V_{1} are the volumes of the sphere and the cube.

\frac{4}{3}\pi r^{3}=a^{3}

\frac{r}{a}=(\frac{3}{4\pi })^{\frac{1}{3}}  

Put this value in the equation (1).

T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}T_{2}=T_{1}(\frac{2\pi }{3}\times ((\frac{3}{4\pi })^{\frac{1}{3}})^{2})^{\frac{1}{4}}

Put T_{1}=500°c.

T_{2}=(500)(\frac{2\pi }{3}\times (\frac{3}{4\pi })^{\frac{2}{3}})^{\frac{1}{4}}

T_{2}=472.2^{\circ}c

Therefore, the value of the celsius temperature of the cube is 472.7°c.    

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