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kvv77 [185]
4 years ago
5

Jafar's pretty practical, and the physics teacher seems to talk a lot about unreal things such as frictionless surfaces, point p

articles that represent whole objects (like his car), and ropes that do not stretch when they are used. Scott, one of Jafar's classmates, says that there's the world and then there's physics class and it's not real; Jafar should just memorize the unreal things, give the teacher what she wants, and forget them after the final examination, but Jafar is not comfortable with that. He wants to be an engineer like his older sister, and she told him that what they learned in physics does relate to her job. So giving the teacher what she wants and forgetting does not seem like a good option to him. Which, if any, of the following possible statements Jafar makes to Scott in response to his suggestion could be correct and may be able to persuade Scott to change his mind?
A. "Physics uses unreal situations as starting points because they are easy to visualize and do calculations with, then it adds reality little by little on to the simplified foundations; so points that represent objects aren't lies, they are just simplifications that give us a foothold onto the ideas that are important to consider."
B. "Physics gave the world the transistor and other devices that work in the computers in my car. Physicists have been able to land objects from Earth on Mars and have invented lasers and holograms. Something they are doing as scientists leads to these results, so we should both work to really understand how physics works."
C. "Engineers who built bathyscaphes to investigate the ocean depths use physics principles to keep occupants alive. Engineers who build bridges use physics so the cars do not plunge through even in heavy traffic. Engineers can create integrated circuits that do huge numbers of calculations each second. I think it must be important to understand how the basic physics leads to these useful results."
D. "My sister is an engineer, and she says learning physics is important. If she says so, we should learn O it."
Physics
1 answer:
iogann1982 [59]4 years ago
7 0

Answer:

Statements A,B,C are correct

Explanation:

For the first statement Jafar is trying to explain the reasons behind the use of unreal/ideal situations by the physics teacher while teaching in classrooms and that can make Scott change his mind about physics

The second statement shows the real applications of physics in every day Devices like the transistor and other devices seen in the world today

The third statement shows to Scott the application of physics in ensuring  the safety of lives and property  of people around the world

while statement D is not a correct statement to persuade Scott

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Which of the following is a region where magnetic force will affect other magnets?
Contact [7]
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3 0
3 years ago
Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1200 kg and was approaching at
GuDViN [60]

Answer:

a) v = 11.24 m / s ,    θ = 17.76º   b) Kf / K₀ = 0.4380

Explanation:

a) This is an exercise in collisions, therefore the conservation of the moment must be used

Let's define the system as formed by the two cars, therefore the forces during the crash are internal and the moment is conserved

Recall that moment is a vector quantity so it must be kept on each axis

X axis

initial moment. Before the crash

     p₀ₓ = m₁ v₁

where v₁ = -25.00 me / s

the negative sign is because it is moving west and m₁ = 900 kg

final moment. After the crash

      p_{x f}= (m₁ + m₂) vx

       p₀ₓ =  p_{x f}

       m₁ v₁ = (m₁ + m₂) vₓ

     vₓ = m1 / (m₁ + m₂) v₁

let's calculate

       vₓ = - 900 / (900 + 1200) 25

       vₓ = - 10.7 m / s

Axis y

initial moment

      p_{oy}= m₂ v₂

where v₂ = - 6.00 m / s

the sign indicates that it is moving to the South

final moment

     p_{fy}= (m₁ + m₂) v_{y}

     p_{oy} = p_{fy}

     m₂ v₂ = (m₁ + m₂) v_{y}

     v_{y} = m₂ / (m₁ + m₂) v₂

we calculate

    v_{y} = 1200 / (900+ 1200) 6

    v_{y}  = - 3,428 m / s

for the velocity module we use the Pythagorean theorem

      v = √ (vₓ² + v_{y}²)

      v = RA (10.7²2 + 3,428²2)

      v = 11.24 m / s

now let's use trigonometry to encode the angle measured in the west clockwise (negative of the x axis)

      tan θ = v_{y} / Vₓ

      θ = tan-1 v_{y} / vₓ)

      θ = tan -1 (3,428 / 10.7)

       θ = 17.76º

This angle is from the west to the south, that is, in the third quadrant.

b) To search for loss of the kinetic flow, calculate the kinetic enegy and then look for its relationship

      Kf = 1/2 (m1 + m2) v2

      K₀ = ½ m₁ v₁² + ½ m₂ v₂²

      Kf = ½ (900 + 1200) 11.24 2

      Kf = 1.3265 105 J

      K₀ = ½ 900 25²  + ½ 1200 6²

      K₀ = 2,8125 10⁵ + 2,16 10₅4

        K₀ = 3.0285 105J

the wasted energy is

        Kf / K₀ = 1.3265 105 / 3.0285 105

        Kf / K₀ = 0.4380

         

this is the fraction of kinetic energy that is conserved, transforming heat and transforming potential energy

5 0
3 years ago
Six artificial satellites complete one circular orbit around a space station in the same amount of time. Each satellite has mass
oee [108]

Answer:

The ranking of the net force acting on different satellite from largest to smallest is {F_E} > {F_F} > {F_A} = {F_B} = {F_D} > {F_C}

Explanation:

In order to get a good understanding of this solution we need to understand that the main concepts used to solve this problem are centripetal force and velocity of satellite.

Initially, use the expression of the velocity of satellite and find out its dependence on the radius of orbit. Use the dependency in the centripetal force expression.

Finally, we find out the velocity of the six satellites and use that expression to find out the force experienced by the satellite. Find out the force in terms of mass (m) and radius of orbit (L) and at last compare the values of force experienced by six satellites.

Fundamentals

The centripetal force is necessary for the satellite to remain in an orbit. The centripetal force is the force that is directed towards the center of the curvature of the curved path. When a body moves in a circular path then the centripetal force acts on the body.

The expression of the centripetal force experienced by the satellite is given as follows:

                    {F_{\rm{c}}} = \frac{{m{v^2}}}{L}

Here, m is the mass of satellite, v is the velocity, and L is the radius of orbit.

The velocity of the satellite with which the satellite is orbiting in circular path is given as follows:

                        v = \frac{{2\pi L}}{T}

Here, T is the time taken by the satellite.

The velocity of the satellite with which the satellite is orbiting in circular path is given as follows;

                    v = \frac{{2\pi L}}{T}

Since, all the satellites complete the circular orbit in the same amount of time. The factor of   \frac{{2\pi }}{T}   is not affected the velocity value for the six satellites. Therefore, we can write the expression of v given as follows:

Substitute  v = \frac{{2\pi L}}{T} in the force expression {F_{\rm{c}}} = \frac{{m{v^2}}}{L}   as follows:

                              \begin{array}{c}\\{F_c} = \frac{{m{{\left( {\frac{{2\pi L}}{T}} \right)}^2}}}{L}\\\\ = \frac{{4{\pi ^2}}}{{{T^2}}}mL\\\end{array}

Since, all the satellites complete the circular orbit in the same amount of time. The factor of \frac{{4{\pi ^2}}}{{{T^2}}}  not affect the force value for six satellites.Therefore, we can write the expression of {F_c}  given as follows:

        {F_c} = kmL

Here, k refers to constant value and equal to  \frac{{4{\pi ^2}}}{{{T^2}}}

    {F_A} = k{m_A}{L_A}

Substitute 200 kg for {m_A}   and 5000 m for LA in the expression                                  {F_A} = k{m_A}{L_A}

\begin{array}{c}\\{F_A} = k\left( {200{\rm{ kg}}} \right)\left( {5000{\rm{ m}}} \right)\\\\ = {10^6}k{\rm{ N}}\\\end{array}

The force acting on satellite B from their rocket is given as follows:{F_B} = k{m_B}{L_B}

Substitute 400 kg for {m_B} and 2500 m for in the expression {F_B} = k{m_B}{L_B}

\begin{array}{c}\\{F_B} = k\left( {400{\rm{ kg}}} \right)\left( {2500{\rm{ m}}} \right)\\\\ = {10^6}k{\rm{ N}}\\\end{array}

The force acting on satellite C from their rocket is given as follows:{F_C} = k{m_C}{L_C}

Substitute 100 kg for {m_C}and 2500 m for in the above expression  {F_C} = k{m_C}{L_C}

\begin{array}{c}\\{F_C} = k\left( {100{\rm{ kg}}} \right)\left( {2500{\rm{ m}}} \right)\\\\ = 0.25 \times {10^6}k{\rm{ N}}\\\end{array}

The force acting on satellite D from their rocket is given as follows:{F_D} = k{m_D}{L_D}

Substitute 100 kg for {m_D} and 10000 m for {L_D} in the expression{F_D} = k{m_D}{L_D}

\begin{array}{c}\\{F_D} = k\left( {100{\rm{ kg}}} \right)\left( {10000{\rm{ m}}} \right)\\\\ = {10^6}k{\rm{ N}}\\\end{array}

The force acting on satellite E from their rocket is given as follows:{F_E} = k{m_E}{L_E}

Substitute 800 kg for {m_E}  and 5000 m for  {L_E} in the expression {F_E} = k{m_E}{L_E}

\begin{array}{c}\\{F_E} = k\left( {800{\rm{ kg}}} \right)\left( {5000{\rm{ m}}} \right)\\\\ = 4.0 \times {10^6}k{\rm{ N}}\\\end{array}

The force acting on satellite F from their rocket is given as follows:{F_F} = k{m_F}{L_F}

Substitute 300 kg for {m_F} and 7500 m for {L_F} in the expression {F_F} = k{m_F}{L_F}

\begin{array}{c}\\{F_F} = k\left( {300{\rm{ kg}}} \right)\left( {7500{\rm{ m}}} \right)\\\\ = 2.25 \times {10^6}k{\rm{ N}}\\\end{array}

The value of forces obtained for the six-different satellite are as follows.

\begin{array}{l}\\{F_A} = {10^6}k{\rm{ N}}\\\\{F_B} = {10^6}k{\rm{ N}}\\\\{F_C} = 0.25 \times {10^6}k{\rm{ N}}\\\\{F_D} = {10^6}k{\rm{ N}}\\\\{F_E} = 4.0 \times {10^6}k{\rm{ N}}\\\\{F_F} = 2.25 \times {10^6}k{\rm{ N}}\\\end{array}

     The ranking of the net force acting on different satellite from largest to smallest is {F_E} > {F_F} > {F_A} = {F_B} = {F_D} > {F_C}

7 0
4 years ago
A solid disk of radius 5.50 cm and mass 1.25 kg , which is rolling at a speed of 1.50 m/s , begins rolling without slipping up a
shtirl [24]

Answer:

Explanation:

Deceleration of solid disk = g sin10/1 + k²/r² = g sin 10 / 1 + 1/2 = g sin 10 x 2/3

[ k is radius of gyration of disk which is equal to( 1/√2)x r ]

deceleration a = -1.1345 m/s²

v = u - at , t = u / a = 1.5 / 1.1345 = 1.322 s.

5 0
3 years ago
Which two options are forms of potential energy
lutik1710 [3]

Answer:

Energy is the ability to do work or cause change. There are two kinds of energy – kinetic energy when something is moving, and potential energy, which is energy that is stored and ready to be used. All objects have potential energy or stored energy if they are placed in a certain position.

5 0
4 years ago
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