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Darya [45]
3 years ago
15

The initial speed of a body is 4.24 m/s. What is its speed after 2.91 s if it accelerates uniformly at -1.76 m/s2

Physics
1 answer:
shutvik [7]3 years ago
6 0
  • Initial velocity=u=4.24m/s
  • Time=t=2.91s
  • Acceleration=a=-1.76m/s^2
  • Final velocity=v

Using 1st equation of kinematics

\\ \sf\longmapsto v=u+at

\\ \sf\longmapsto v=4.24+2.91(-1.76)

\\ \sf\longmapsto v=|4.24-5.12|

\\ \sf\longmapsto v=0.8m/s

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Coulombs law says that the force between any two charges depends on the amount of charges and distance between them. This force is directly proportional to the magnitude of the two charges and inversely proportional to the distance between them.

F=k\frac{|q_1| |q_2|}{r^2}

where q_1\hspace{1mm}and\hspace{1mm}q_2 are charges, r is the distance between them and k is the coulomb constant.

case 1:

q_1=-e\\ q_2=+3e\\ r=100pm\\ \Rightarrow F=k\frac{|-e||3e|}{(100pm)^2}=3ke^2\times10^8

case 2

q_1=-e\\ q_2=+2e\\ r=100pm\\ \Rightarrow F=k\frac{|-e||2e|}{(100pm)^2}=2ke^2\times10^8

case 3:

q_1=-e\\ q_2=+e\\ r=100pm\\ \Rightarrow F=k\frac{|-e||e|}{(200pm)^2}=0.25ke^2\times10^8

Comparing the 3 cases:

The maximum potential force according to coulombs law is between -1 charge and +3 charge separated by a distance of 100 pm.

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A, The water moving away

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The time interval for which the proton remains in the field is -

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We have a proton entering a uniform magnetic field which is in a direction perpendicular to the proton's velocity.

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<h3>What is the magnitude of force on the charged particle moving in a uniform magnetic field?</h3>

The magnitude of force on the charged particle moving in a uniform magnetic field is given by -

F = qvB sinθ



According to the question, we have -

Entering Velocity (v) = 20 i  m/s

Magnetic field intensity (B) = 0.3 T

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Now -

The entering and leaving velocity vectors have 90 degrees difference between them. Therefore, only a quarter of distance of the complete circular path of radius 'R' is traced by the proton. Therefore -

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