The magnetic fields of Earth and several other planets are produced by _____. magnetic substances conductive materials in motion
2 answers:
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Answer:
<em>the smaller particle moves with speed of 8.706 m/s in the opposite direction to the bigger particle.</em>
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Explanation:
Speed of the original particle = 13 m/s
We designate particles as A and B
The final weights of the component particles are
Particle A = 1.4 N
particle B = 1.9 N
The speed of the larger piece (particle B) = 29 m/s
We know that weight is the product of a body's mass and acceleration due to gravity g which is equal to 9.81 m/s^2, therefore, masses of the particles are
particle A = 1.4/9.81 = 0.143 kg
Particle B = 1.9/9.81 = 0.194 kg
The momentum of a body is the product of its mass and its velocity i.e
P = mv
This means that the mass of the particle before splitting is
0.143 kg + 0.194 kg = 0.337 kg
Momentum of the initial whole particle = mv
==> 0.337 x 13 = 4.381 kg-m/s
The bigger particle B remains horizontal, and has a momentum of
mv = 0.194 x 29 = 5.626 kg-m/s
<em>According to the conservation of momentum, the total initial momentum of a system must be equal tot the total final momentum of the system.</em>
Initial total momentum of the system = 4.381 kg-m/s (momentum of original particle before splitting)
Final total momentum of the system = Total momentum of the particles after splitting = 5.626 kg-m/s + ( 0.143 kg x )
where is the velocity of smaller particle A
final total momentum of the system = 5.626 + 0.143
Equating the two momenta of the system, we'll have
4.381 = 5.626 + 0.143
4.381 - 5.626 = 0.143
-1.245 = 0.143
= -1.245/0.143 =<em> -8.706 m/s</em>
<em>The negative sign indicates that the smaller particle moves in the opposite direction to the bigger particle</em>
Answer:
The electric field outside the sphere will be .
Explanation:
Given that,
Radius of solid sphere = R
Charge = q
According to figure,
Suppose r is the distance between the point P and center of sphere.
If be the volume charge density,
Then, the charge will be,
.....(I)
Consider a Gaussian surface of radius r.
We need to calculate the electric field outside the sphere
Using formula of electric field
Put the value from equation (I)
Hence, The electric field outside the sphere will be .
