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Bas_tet [7]
2 years ago
15

A charged particle is accelerated in a uniform electric field. When its velocity is 2 m/s, its electric potential energy is 100

J and its kinetic energy is 10 J. What is the particle's potential energy when its velocity reaches 4 m/s?
Physics
1 answer:
zavuch27 [327]2 years ago
7 0

Answer:

particle's potential energy = 70J

Explanation:

From conservation of energy; K1 + Ue1 = K2 + Ue2

where K1 and K2 are the kinetic energies at two positions and Ue1 and Uue2 are the electrical potential energies at two positions.

k1 = 10J, Ue1 = 100J

K2 = 40J

substitute into K1 + Ue1 = K2 + Ue2

Ue2 = K1 + Ue1 - K2

= 10 +100 - 40

Ue2 = 70J

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In one cycle, a heat engine takes in 1000 J of heat from a high-temperature reservoir, releases 600 J of heat to a lower-tempera
Talja [164]

Answer:

η = 40 %  

Explanation:

Given that

Qa ,Heat addition= 1000  J

Qr,Heat rejection= 600 J

Work done ,W= 400 J

We know that ,efficiency of a engine given as

\eta=\dfrac{W(net)}{Q(heat\ addition)}

Now by putting the values in the above equation ,then we get

\eta=\dfrac{400}{1000}

η = 0.4

The efficiency in percentage is given as

η = 0.4  x 100 %

η = 40 %

Therefore the answer will be 40%.

4 0
3 years ago
A sportscar has a mass of 1500 kg and accelerates at 5 meters per second squared. What is the magnitude of the force acting on t
Hatshy [7]

Answer:

7500 Newtons

Explanation:

Mass of the sportscar= 1500 kg

Acceleration of the sportscar= 5m/s^2

Hence, let the Force acting on it be F

We\ know\ that,\\Force=Mass*Acceleration\\F=ma\\\\Here,\\F=1500*5\\=7500 kg m/s^2\ or\ 7500\ Newtons

4 0
2 years ago
Where are you on Earth if you experience each of the following? (Refer to the discussion in Observing the Sky: The Birth of Astr
Aloiza [94]

Explanation:

We know that the sky appears to us like a sphere called as celestial sphere which appears to rotate around an imaginary axis because of Earth's rotation. Since the axis cuts the celestial sphere at celestial poles all the object seems to circle around the celestial poles.

Condition 1: The stars rise and set perpendicular to the horizon

The observer is at the equator

Condition 2: The stars circle the sky parallel to the horizon

The observer is at the Pole of the Earth

Condition 3: The celestial equator passes through the zenith

The observer is at the equator

Condition 4: In the course of a year, all stars are visible

The observer is at the equator

Condition 5: The Sun rises on March 21 and does not set until September 21 (ideally)

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7 0
3 years ago
Which are characteristics of electromagnetic waves?check all that apply.
emmainna [20.7K]

Correct choices are marked in bold:

travel in straight lines and can bounce off surfaces  --> TRUE, normally electromagnetic waves travel in straight lines, however they can be reflected by objects, bouncing off their surfaces

travel through space at the speed of light  --> TRUE, all electromagnetic waves in space (vacuum) travel at the speed of light, c=3\cdot 10^8 m/s)

travel only through matter  --> FALSE; electromagnetic waves can also travel through vacuum

travel only through space  --> FALSE, electromagnetic waves can also travel through matter

can bend around objects  --> TRUE, this is what happens for instance when diffraction occurs: electromagnetic waves are bended around obstacles or small slits

move by particles bumping into each other  --> FALSE, electromagnetic waves are oscillations of electric and magnetic fields, so no particles are involved

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8 0
3 years ago
Read 2 more answers
A(n) 55.5 g ball is dropped from a height of 53.6 cm above a spring of negligible mass. The ball compresses the spring to a maxi
Serggg [28]

Answer:

The spring force constant is  k=243\ \frac{N}{m} .

Explanation:

We are told the mass of the ball is m=0.0555\ kg, the height above the spring where the ball is dropped is h=0.536\ m,  the length the ball compresses the spring is d=0.04897\ m and the acceleration of gravity is 9.8\ \frac{m}{s^{2}} .

We will consider the initial moment to be when the ball is dropped and the final moment to be when the ball stops, compressing the spring. We supose that there is no friction so the initial mechanical energy E_{mi} is equal to the final mechanical energy E_{mf} :

                                                    E_{mf}=E_{mi}

Initially there is only gravitational potential energy because the force of the spring isn't present and the speed is zero. In the final moment there is only elastic potential energy because the height is zero and the ball has stopped. So we have that:

                                                   \frac{1}{2}kd^{2}=mgh

If we manipulate the equation we have that:

                                                    k=\frac{2mgh}{d^{2} }

                                         k=\frac{2\ 0.0555\ kg\ 9.8\frac{m}{s^{2}}\ 0.536\ m}{(0.04897)^{2}m^{2}}

                                              k=\frac{0.58306\ \frac{kgm^{2}}{s^{2}}}{2.398x10^{-3}m^{2}}

                                                     k=243\ \frac{N}{m}

                                                   

                             

5 0
3 years ago
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