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ioda
2 years ago
3

An object initially at rest falls from a height H until it reaches the ground. Two of the following energy bar charts represent

the kinetic energy K and gravitational potential energy Ug of the object-Earth system at two positions. The first position is when the object is initially released, and the second position is when the object is halfway between its release point and the ground. Which two charts could represent the mechanical energy of the object-Earth system? Select two answers. by Ug K Ug к Mechanical Energy at Halfway Point Mechanical Energy at Release Point Mechanical Energy at Release Point Mechanical Energy at Halfway Point d. c. к Ug Ug K Mechanical Energy at Release Point Mechanical Energy at Halfway Point Mechanical Energy at Release Point Mechanical Energy at Halfway Point
Physics
1 answer:
Tom [10]2 years ago
6 0

Answer:

Initial:   bar  power U₀

Final:    bar  power U = U₀ / 2

             bar Kinetic energy  K = U₀ / 2

These two bars are half the height of the initial bar

Explanation:

To know which graph is correct, let's discuss the solution to the problem

Initial mechanical energy

      Em₀ = U₀ = m g H

The mechanical energy at the midpoint

     Em₂ = K + U₂

As there is no friction, mechanical energy is conserved

     Em₀ = Em₂

     U₀ = K + U₂

     K = U₀ - U₂

     K = m g (H - y₂)

Indicates that position 2 corresponds to y₂ = H / 2

     K = m g (H –H / 2)

     K = ½ m g H

     K = ½ Uo

Therefore the graph must be

Initial:   bar  power U₀

Final:    bar  power U = U₀ / 2

             bar Kinetic energy  K = U₀ / 2

These two bars are half the height of the initial bar

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kolezko [41]

Answer:

Part A - 3N/m

Part B - see attachment

Part C - 4.9 × 10-³J

Part D - E = 1/2kd² + 1/2mv² + mgh

Explanation:

This problem requires the knowledge of simple harmonic motion for cimplete solution. To find the spring constant in part A the expression relating the force applied to a spring and the resulting stretching of the spring (hooke's law) is required which is F = kx.

The free body diagram can be found in the attachment. Fp(force of pull), Ft(Force of tension) and W(weight).

The energy stored in the pring as a result of the stretching of d = 5.7cm is 1/2kd².

Part D

Three forces act on the spring-monkey system and they do work in different forms: kinetic energy 1/2mv² , elastic potential

energy due to the restoring force in the spring or the tension force 1/2kd², and the gravitational potential energy mgh of the position of the system. So the total energy of the system E = 1/2kd² + 1/2mv² + mgh.

8 0
2 years ago
Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to t
yarga [219]

Answer

The rate at which the magnetic field is changing is  [\frac{dB}{dt} ] =  0.000467 T/s

Explanation

From the question we are told that

   The electric field strength is E =  3.5mV/m =  3.5 *10^{-3} \ V/m

    The radius is  r =  1.5 \ m

The rate of change of the  magnetic  field  is mathematically represented as

        \frac{d \phi }{dt}  =  \int\limits^{} {E \cdot dl}

Where dl is change of a unit length

     \frac{d \phi}{dt}  =  A *  \frac{dB}{dt}

Where A is the area which is mathematically represented as

     A = \pi r^2

    So

    E \int\limits^{} {  dl} =  ( \pi r^2) (\frac{dB}{dt} )  

  E L  =  ( \pi r^2) (\frac{dB}{dt} )  

where L is the circumference of the circle which is mathematically represented as

     L = 2 \pi r

So

     E (2 \pi r ) =  (\pi r^2 ) [\frac{dB}{dt} ]

      E  =   \frac{r}{2}  [\frac{dB}{dt} ]

       [\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }

substituting values

      [\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }

      [\frac{dB}{dt} ] =  0.000467 T/s    

8 0
2 years ago
For nitrogen feel like with its temperature must be within 12.78 Fahrenheit of -333.22 Fahrenheit which equation can be used to
photoshop1234 [79]

Answer:

The following equation can be used.

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7 0
2 years ago
Read 2 more answers
You would like to know whether silicon will float in mercury and you know that can determine this based on their densities. Unfo
dolphi86 [110]

Answer:

Explanation:

To convert gram / centimeter³ to kg / m³

gram / centimeter³

= 10⁻³ kg / centimeter³

= 10⁻³  / (10⁻²)³ kg / m³

= 10⁻³ / 10⁻⁶ kg / m³

= 10⁻³⁺⁶ kg / m³

= 10³ kg / m³

So we shall have to multiply be 10³ with amount in gm / cm³ to convert it into kg/m³

2.33 gram / cm³

= 2.33 x 10³ kg / m³ .

3 0
2 years ago
At 213.1 K a substance has a vapor pressure of 45.77 mmHg. At 243.7 K it has a vapor pressure of 193.1 mm Hg. Calculate its heat
vlabodo [156]

Answer:

20.3125 kJ/mol

Explanation:

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P_{f} = final vapor pressure = 193.1 mm Hg

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T_{f} = final temperature = 243.7 K

H = Heat of vaporization

Using the equation

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ln\left ( \frac{193.1}{45.77} \right ) = \left ( \frac{-H}{8.314} \right )\left ( \frac{1}{243.7} - \frac{1}{213.1}\right)

H = 20312.5 J/mol

H = 20.3125 kJ/mol

8 0
2 years ago
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