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Jet001 [13]
3 years ago
15

An electron volt is: A) the force acting on an electron in a field of 1 N/C B) the force required to move an electron 1 meter C)

the energy gained by an electron in moving through a potential difference of 1 volt D) the energy needed to move an electron through 1 meter in any electric field E) the work done when 1 coulomb of charge is moved through a potential difference of 1 volt
Physics
1 answer:
9966 [12]3 years ago
4 0

Answer: C) the energy gained by an electron in moving through a potential difference of 1 volt

Explanation: In order to explain this response we have to consider that the potential energy for any charge is given by:

Up=q*ΔV then if we consider an electron accelerate by  a difference potential of 1V, we have:

Up=e*V ( energy UNITS, which is equal to 1.6*10^-19 J).

The option A) and B) are wrong because elentron Volt has units of energy not force.

The option D) is also wrong because the the work done to move un electron one meter also depends of the magnitud of the potential difference ( electric field) where the electron is moving .

Finally the option D) is false because the one C*V is one J, is huge quantity of energy compared to the electron Volt (eV).

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A person with a mass of 15kg is walking uphill at a velocity of2m/s. What is the walker's momentum?
Nadusha1986 [10]

Answer:

30 kgm/s

Explanation:

The momentum of a body is given by the product of mass and  its velocity.

∴ momentum = mv

                        = 15 × 2

                        = 30 kgm/s

4 0
3 years ago
An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s1045 rad/s ). If a particular disk is s
kumpel [21]

To solve this problem we must keep in mind the concepts related to angular kinematic equations. For which the angular velocity is defined as

\omega_f =\omega_i-\alpha t

Where

\omega_f = Final angular velocity

\omega_i = Initial angular velocity

\alpha =Angular acceleration

t= time

In this case we do not have a final angular velocity, then

\omega_i = \alpha t

Re-arrange for \alpha

\alpha= \frac{\omega_i}{t}

\alpha = \frac{910}{0.167}

\alpha = 5449.1 rad\s^2

Therefore the mangitude of the angular aceleration is 5449.1rad/s²

6 0
4 years ago
A piston–cylinder assembly fitted with a slowly rotating paddle wheel contains 0.13 kg of air, initially at 300 K. The air under
olganol [36]

Answer:

Explanation:

From the first law of thermodynamics

Q=ΔU+W

Q=heat supplies to the system

ΔU=change in internal energy of the system

W= work done by the gas  by the system

Q =12 KJ

ΔU=U_{2} -U_{1}=(mR/k-1)ΔT

ΔU=\frac{0.3*0.287}{1.4-1}*(460-400)

     =14.924 KJ

Q=ΔU+W

W=-2.924 KJ

W_{net}=W_{air}+W_{paddle}

W_{air}=work done by the air

W_{air}= P(V2-V1)

W_{air}=mRT2-mRT1

W_{air}=mR(T2-T1)=0.138*0.287*(460-300)=

                           =6.33 KJ

W_{net}=W_{air}+W_{paddle}

                      =-2.924-6.33

W_{paddle}=-9.26  KJ                    

7 0
4 years ago
Please Help!!! An engine takes 3,000 J from a hot reservoir and expels 2,400 J to a cold reservoir. What is the thermal efficien
Murljashka [212]

Answer:

80%

Explanation:

Thermal efficiency is a measure of the extent to which a heat engine is able to convert to other form. It is measure in percentage.

Thermal efficiency = \frac{E_{out} }{E_{in} } × 100%

Where E_{in} is the input energy and E_{out} is the output energy.

Given that: E_{in} = 2400 J and  E_{out} = 3000 J, therefore:

Thermal efficiency = \frac{2400}{3000} × 100%

                               = 80%

The thermal efficiency of the engine is 80%.

4 0
4 years ago
- When an object is dropped what happens to potential and kinetic energy as it falls?
aev [14]
The answer is D: the GPE goes down and the KE goes up
7 0
3 years ago
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