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Ahat [919]
3 years ago
13

What are the differences between a magnetic and gravitational interaction

Physics
1 answer:
GREYUIT [131]3 years ago
3 0
Classically, gravitational attraction is due to the masses of objects. It is inversely proportional to the square of the distances between the objects.

Magnetic attraction is due to the magnetism in objects which is due to uncompensated electron spins in certain atoms.  The force due to magnetism less easy to put into one equation than gravity since magnetic fields can have different shapes, but the simplest one (the dipolar field) is inversely proportional to the cube of the distance between the magnetic dipoles.
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Two students, standing on skateboards, are initially at rest, when they give each other a shove! After the shove, one student (7
Elden [556K]

Answer:

The other student (59kg) moves right at 7.44 m/s

Explanation:

Given;

mass of the first student, m₁ = 77kg

mass of the second student, m₂ = 59kg

initial velocity of the first student, u₁ = 0

initial velocity of the second student, u₂ = 0

final velocity of the first student, v₁ = 5.7 m/s left

final velocity of the second student, v₂ = ? right

Apply the principle of conservation of linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(77 x 0) + (59 x 0) = (-77 x 5.7) + (59 x  v₂)

0 = - 438.9 + 59v₂

59v₂ = 438.9

v₂ = 438.9 / 59

v₂ = 7.44 m/s to the right

Therefore, the other student (59kg) moves right at 7.44 m/s

6 0
3 years ago
Can any juniors who go to Texas connections academy help me out with physics?
zimovet [89]

I'm not from that school but I can help you.

3 0
2 years ago
A 2.0 kg sphere with a velocity of 6.0 m/s collides head-on and elastically with a stationary 10 kg sphere
dmitriy555 [2]

Question: A 2.0 kg sphere with a velocity of 6.0 m/s collides head-on and elastically with a stationary 10 kg sphere, What is thier velocities after collision.

Answer:

v = 6 m/s, v' = 0 m/s

Explanation:

From the question,

For Elastic collision,

mu+m'u' = mv+m'v'......................... Equation 1

Where m = mass of the first sphere, m' = mass of the second sphere, u = initial velocity of the first sphere, u' = initial velocity of the second sphere, v = final veolocity of the first sphere, v' = final velocity of the second sphere.

Also,

The relative velocity before collision = relative velocity after collision

u-u' = v-v'............................ Equation 2

Given:  m = 2 kg, m' = 10 kg, u = 6 m/s, u' = 0 m/s

Substitute into equation 1 and 2

2(6)+10(0) = 2v+10v'

2v+10v' = 12.............. Equation 3

6-0 = v-v'

v-v' = 6 ................... Equation 4

Solve equation 3 and 4 simultaneously.

v = 6+v'............. Equation 5

Substitute equation 5 into equation 3

2(6+v')+10v' = 12

12+2v'+10v' = 12

12v' = 12-12

v' = 0/12

v' = 0 m/s.

Also substitute the value of v' into equation 5

v = 6+0

v = 6 m/s

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The image shows street lights powered by solar panels. Which sequence shows the energy transformations taking place in these lig
yawa3891 [41]

the answer is C. for plato

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Free runners jump long distances and land on the ground or a wall. How do they apply Newton’s second law to lessen the force of
Veseljchak [2.6K]

As we know that as per Newton's II law we have

F = \frac{dP}{dt}

here we will have

dP = change in momentum

dt = time interval in which momentum is changed

now in order to have least injury during jumping we need to have least force on the jumper

so in order to have least force we can say that the momentum must have to change in maximum time so that amount of force must be least

So we need to increase the time in which momentum of the system is changed

5 0
3 years ago
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