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stich3 [128]
3 years ago
12

A 1kg object loses 20 j of gpe as it falls how far does it fall

Physics
1 answer:
jarptica [38.1K]3 years ago
4 0
Gravitational potential energy (GPE) is energy within an object with respect to its height. The equation of gpe is:

GPE = mgh

where:
m = mass of object
g = gravitational constant, 9.81 m/s^2
h = height

For a given gpe of 20 J, the height can be calculated as follows:
*Remember to be consistent with units. Since Joule is derived from SI units, make sure to use SI units for the other variables

GPE = 20 J = 1 (9.81) (h)
h = 2.04 meters
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Answer:

Explanation:

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So despite your saying this is not a YES / NO question, the answer is NO the  acceleration is too low or not long enough to meet the required clearances.

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Which of the following has potential but not kinetic energy?
gavmur [86]

The answer would be option D "a ball sitting on a shelf." Potential energy is the amount of energy a object has while it's at rest.. (or not moving) Kinetic energy is how much energy a object is while it's moving. So in this case it's option D because a ball sitting on a shelf isn't moving therefore it has potential energy. It's not option A because thats a example of kinetic energy since how the roller coaster is moving. It's not option B because it's kinetic energy because the bike is moving. It's also not option C because it's kinetic energy because the bird is moving.


Hope this helps!

7 0
3 years ago
The interior space of large box is kept at 30 C. The walls of the box are 3 m high and have a ‘sandwich’ construction consisting
White raven [17]

Answer:

\frac{\dot Q}{A} =20.129\ W.m^{-2}

T_1=27.58\ ^{\circ}C & T_2=2.41875\ ^{\circ}C

Explanation:

Given:

  • interior temperature of box, T_i=30^{\circ}C
  • height of the walls of box, h=3\ m
  • thickness of each layer of bi-layered plywood, x_p=1.25\ cm=0.0125\ m
  • thermal conductivity of plywood, k_p=0.104\ W.m^{-1}.K^{-1}
  • thickness of sandwiched Styrofoam, x_s=5\ cm=0.05\ m
  • thermal conductivity of Styrofoam, k_s=0.04\ W.m^{-1}.K^{-1}
  • exterior temperature, T_o=0^{\circ}C

<u>From the Fourier's law of conduction:</u>

\dot Q=\frac{dT}{(\frac{x}{kA}) }

\dot Q=\frac{dT}{R_{th} } ....................................(1)

<u>Now calculating the equivalent thermal resistance for conductivity using electrical analogy:</u>

R_{th}=R_p+R_s+R_p

R_{th}=\frac{x_p}{k_p.A}+\frac{x_s}{k_s.A}+\frac{x_p}{k_p.A}

R_{th}=\frac{1}{A} (\frac{x_p}{k_p}+\frac{x_s}{k_s}+\frac{x_p}{k_p})

R_{th}=\frac{1}{A} (\frac{0.0125}{0.104}+\frac{0.05}{0.04}+\frac{0.0125}{0.104})

R_{th}=\frac{1.4904}{A} .....................(2)

Putting the value from (2) into (1):

\dot Q=\frac{30-0}{\frac{1.4904}{A} }

\dot Q=\frac{30\ A}{1.4904}

\frac{\dot Q}{A} =20.129\ W.m^{-2} is the heat per unit area of the wall.

The heat flux remains constant because the area is constant.

<u>For plywood-Styrofoam interface from inside:</u>

\frac{\dot Q}{A} =k_p.\frac{T_i-T_1}{x_p}

20.129=0.104\times \frac{30-T_1}{0.0125}

T_1=27.58\ ^{\circ}C

&<u>For Styrofoam-plywood interface from inside:</u>

\frac{\dot Q}{A} =k_s.\frac{T_1-T_2}{x_s}

20.129=0.04\times \frac{27.58-T_2}{0.05}

T_2=2.41875\ ^{\circ}C

4 0
3 years ago
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