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2 years ago

When a car is coasting downhill, how are its potential and kinetic energies changing?

1 answer:

3 0

When a car is coasting downhill, the kinetic and potential energies are increasing and decreasing respectively.

<h3>What are kinetic and potential energy?</h3>

Kinetic energy is the energy possessed by an object because of its motion, equal to one half the mass of the body times the square of its speed.

Potential energy, on the other hand, is the energy possessed by an object because of its position (in a gravitational or electric field), or its condition (as a stretched or compressed spring, as a chemical reactant, or by having rest mass).

According to this question, a car going downhill will begin to speed because there is lesser friction. This suggests that the kinetic energy increases while the potential energy decreases.

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You set your stationary bike on a high 80-N friction-like resistive force and cycle for 30 min at a speed of 8.0 m/s . Your body

stellarik [79]

A) The change in internal chemical energy is $1.15\cdot 10^7 J$

B) The time needed is 1 minute

Explanation:

First of all, we start by calculating the power output of you and the bike, given by:

$P=Fv$

where

F = 80 N is the force that must be applied in order to overcome friction and travel at constant speed

v = 8.0 m/s is the velocity

Substituting,

$P=(80)(8.0)=640 W$

The energy output is related to the power by the equation

$P=\frac{E}{t}$

where:

P = 640 W is the power output

E is the energy output

$t = 30 min \cdot 60 = 1800 s$ is the time elapsed

Solving for E,

$E=Pt=(640)(1800)=1.15\cdot 10^6 J$

Since the body is 10% efficient at converting chemical energy into mechanical work (which is the output energy), this means that the change in internal chemical energy is given by

$\Delta E = \frac{E}{0.10}=\frac{1.15\cdot 10^6}{0.10}=1.15\cdot 10^7 J$

From the previous part, we found that in a time of

t = 30 min

the amount of internal chemical energy converted is

$E=1.15\cdot 10^7 J$

Here we want to find the time t' needed to convert an amount of chemical energy of

$E'=3.8\cdot 10^5 J$

So we can setup the following proportion:

$\frac{t}{E}=\frac{t'}{E'}$

And solving for t',

$t'=\frac{E't}{E}=\frac{(3.8\cdot 10^5)(30)}{1.15\cdot 10^7}=1 min$

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3 0

3 years ago

A graph can be described as:

pantera1 [17]

D all of the above

....

6 0

3 years ago

Two representative elements are in the same period of the periodic table. Which statement correctly describes the atoms of the t

sveticcg [70]

1) They have valence electrons in the same energy level.

2) Helium atoms have 2 valence electrons, while atoms of the other elements in the group all have 8 valence electrons.

3) cesium (Cs)

4) There are two electrons in the first energy level and seven electrons in the second energy level.

In the periodic table, elements are arranged in order of increasing atomic number into groups and periods.

In the periodic table, we have an arrangement of atoms based on their respective groups and periods.

If I pick two representative elements in the same period, their number of valence electrons will differ by 1 but they will all be accommodated in the same energy level.

Helium is the first member of group 18. Unlike the other members of the group, its first shell is also its valence shell. It has two and not eight valence electrons like other members of the group.

The electronic configuration of Ca is [Ar] 4s², that of Li is [He] 2s1, that of Cs is [Xe] 6s1, that of Br is [Ar] 4s² 3d¹⁰ 4p⁵ and that of oxygen is [He] 2s² 2p⁴.

We can see that only Cs has valence electrons in higher energy orbitals than those of Ca.

Lastly, Fluorine is in group 17. There are nine electrons in this atoms. They arranged in such a way that there are two electrons in the first shell and seven electrons in the second shell.

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7 0

3 years ago

The graph is the path Beth took on a walk.

PIT_PIT [208]

Answer:

Beth went back toward her origin

Explanation:

Please, see the attached image that represent's Beth's path.

She started her walk 2 meters from the origin (0) and after 6 seconds she headed back to position zero (0) the origin. Notice the second half of the graph as a segment going down, thus reducing her distance from the origin, and getting closer to it.