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Nesterboy [21]
3 years ago
10

Describe a process in which energy changes forms at least twice

Physics
1 answer:
olga55 [171]3 years ago
7 0
That would be like dropping your cell phone on to the ground by accident. The object (cell phone)'s gravitational potential energy would be converted to kinetic energy or energy in motion more precisely. This is just a hypothetical example though.
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A 0.208 kg particle with an initial velocity of 1.26 m/s is accelerated by a constant force of 0.766 N over a distance of 0.195
eduard

Answer:

Explanation:

Initial kinetic energy of particle

= 1/2 m V²

= .5 x .208 x 1.26²

= .165 J

Work done by force = force x displacement

= .766 x .195

= .149 J

This energy will be added up .

Total final kinetic energy

= initial kinetic energy + work done on the particle

= .165 + .149 J

=  .314 J .

3 0
3 years ago
What causes the sodium channels to open successively?
11111nata11111 [884]
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4 0
3 years ago
If an atom had 35 protons in the nucleus how many electrons will it have orbiting the nucleus ?
Amanda [17]

Answer:

If it isn't an ion it should have 35 elektrons to cancel the positivity of the nucleus.

7 0
3 years ago
Read 2 more answers
Determine the average value of the translational kinetic energy of the molecules of an ideal gas at (a) 27.8°C and (b) 143°C. Wh
Alinara [238K]

Answer:

a) k_{avg}=6.22\times 10^{-21}

b) k_{avg}=8.61\times 10^{-21}

c)  k_{mol}=3.74\times 10^{3}J/mol

d)   k_{mol}=5.1\times 10^{3}J/mol

Explanation:

Average translation kinetic energy (k_{avg}) is given as

k_{avg}=\frac{3}{2}\times kT    ....................(1)

where,

k = Boltzmann's constant ; 1.38 × 10⁻²³ J/K

T = Temperature in kelvin

a) at T = 27.8° C

or

T = 27.8 + 273 = 300.8 K

substituting the value of temperature in the equation (1)

we have

k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 300.8  

k_{avg}=6.22\times 10^{-21}J

b) at T = 143° C

or

T = 143 + 273 = 416 K

substituting the value of temperature in the equation (1)

we have

k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 416  

k_{avg}=8.61\times 10^{-21}J

c ) The translational kinetic energy per mole of an ideal gas is given as:

       k_{mol}=A_{v}\times k_{avg}

here   A_{v} = Avagadro's number; ( 6.02×10²³ )

now at T = 27.8° C

        k_{mol}=6.02\times 10^{23}\times 6.22\times 10^{-21}

          k_{mol}=3.74\times 10^{3}J/mol

d) now at T = 143° C

        k_{mol}=6.02\times 10^{23}\times 8.61\times 10^{-21}

          k_{mol}=5.1\times 10^{3}J/mol

8 0
3 years ago
If the temperature of a volume of ideal gas increases for 100°C to 200°C, what happens to the average kinetic energy of the mole
mina [271]

Answer:\Delta E=2.0715\times 10^{-21} J

Explanation:

Given

Temperature of the gas is increased from 100 to 200

Also we know that average kinetic energy of the molecules is

E=\frac{3}{2}\cdot \frac{R}{N_A}T

Where

R=Gas constant

N_A=Avogadro's number

T=Temperature in kelvin

\frac{R}{N_A}=1.381\times 10^{-23}

So kinetic energy increases by

\Delta E=\frac{3}{2}\times 1.381\times 10^{-23}\left ( 200-100\right )

\Delta E=2.0715\times 10^{-21} J

8 0
3 years ago
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