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goblinko [34]
2 years ago
5

1. How is electric potential energy similar to gravitational potential energy? How is it different? Where will an electron bound

in an atom and have the largest electrical potential energy?
Physics
1 answer:
Korolek [52]2 years ago
7 0
Both ve similar equations 
<span>both are energies of one object w.r.t another </span>
<span>differences- electric pe is due to electrostatic force and gravitational pe is due to gravitational force </span>
<span>electric pe is > than gravitational pe since electrostatic force> gravitational force </span>
<span>electron bound in an atom ll ve largest potential enegy in its ground state. i think hope it helps</span>
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Of the following which is the largest body?
ankoles [38]

Answer:

Ganymede is the largest body

Explanation:

it is the satellite of jupiter

3 0
2 years ago
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On a snowy day, max (mass = 15 kg) pulls his little sister maya in a sled (combined mass = 20 kg) through the slippery snow. max
sesenic [268]

Work done by a given force is given by

W = F.d

here on sled two forces will do work

1. Applied force by Max

2. Frictional force due to ground

Now by force diagram of sled we can see the angle of force and displacement

work done by Max = W_1 = Fdcos\theta

W_1 = 12*5cos15

W_1 = 57.96 J

Now similarly work done by frictional force

W_2 = Fdcos\theta

W_2 = 4*5cos180

W_2= -20 J

Now total work done on sled

W_{net}= W_1 + W2

W_{net} = 57.96 - 20 = 37.96 J

7 0
3 years ago
What quantities determine the resistance of a piece of material? Choose all that apply.
bija089 [108]

Answer:

Option (a), (b) and (c)

Explanation:

The resistance of a conductor depends on the length of the conductor, area of crossection of the conductor and the nature of the conductor.

The formula for the resistance is given by

R = ρ x l / A

Where, ρ is the resistivity of the conductor, l be the length of the conductor and A be the area of crossection of the conductor.

So, It depends on the length, area and the type of material.

7 0
3 years ago
PLEASE HELP ME What does a moving charge experience when it is near a magnetic field? static friction gravity a stationary charg
attashe74 [19]

<u><em>A moving charge experiences a force as it moves close to a magnetic field present in a region.</em></u>

Explanation:

A charged particle when moves near a magnetic field is forced to move in a circular path. If there is a straight moving charged particle enters the region of the magnetic field, it's path is changed from straight path to the curved path.

The motion of the charged path inside or near the magnetic field will be a circular path and this circular path is due to the force experienced by the charged particle in the magnetic field.

The force experienced by the a charged particle as it moves near the magnetic field is termed as the Lorentz force. The expression for the Lorentz force experienced by the charged particle in the magnetic field is given by:

\boxed{F_m=Q(\overrightarrow{v}\times\overrightarrow{B})}

Here, F_m is the magnetic force, Q is the amount of charge, \overrightarrow{v} is the velocity vector and \overrightarrow{B} is the magnetic field intensity in the region.

Thus, <u><em>A moving charge experiences a force as it moves close to a magnetic field present in a region.</em></u>

Learn More:

1. Which one is true for the electromagnetic spectrum brainly.com/question/1619496

2. What is the current in resistor R2 brainly.com/question/3051098

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Answer Details:

Grade: High School

Subject: Physics

Chapter: Electromagnetism

Keywords:

moving, charge, straight, circular, path, magnetic, field, amount, force, velocity, intensity, Lorentz, expression, stationary.

6 0
3 years ago
A four-wheel-drive vehicle is transporting an injured hiker to the hospital from a point that is 30 km from the nearest point on
zepelin [54]

Answer:

D=200\ km

Explanation:

distance on terrain, d_t=30\ km

  • distance on the road, d_r=70\ km
  • speed on terrain, v_t=30\ km.hr^{-1}
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<u>time taken on the terrain,</u>

t_t=\frac{d_t}{v_t}

t_t=\frac{30}{30}

t_t=1\ hr

<u>time taken to cover the distance on the road:</u>

t_r=\frac{d_r}{v_r}

t_r=\frac{70}{130}

t_r=\frac{7}{13}\ hr

<u>Now the distance covered on terrain in the total time:</u>

D= v_r\times (t_r+t_t)

D= 130\times (\frac{7}{13}+1)

D=130\times \frac{20}{13}\

D=200\ km

<em>is the distance the vehicle must target on the road to minimize the time taken in going off the road.</em>

3 0
3 years ago
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